Physics-12-Ch 11 & 12 & 13-Solns

Unit 5 Matter-Energy Interface ARE YOU READY? (Pages 558–559)

Knowledge and Understanding 1.

2.

Frequency (Hz) 10 5.0 × 10 14 3.8 × 10 15 1.2 × 10 18 1.0 × 10

Product f × λ (m/s) 8 3.0 × 10 8 3.0 × 10 8 3.0 × 10 8 3.0 × 10

Wavelength (m) –3 6.0 × 10 –7 8.0 × 10 –7 2.5 × 10 –10 3.0 × 10

Spectrum Location radio, TV infrared ultraviolet X ray

(a) The product is the same same for each example of electromagnetic radiation, 3.0 × 108 m/s. (b) The results conclude that all electromagnetic waves travel at exactly the same speed. speed. (c) See table for spectrum spectrum location. 3. (a) The source of all electromagnetic waves waves is an oscillating electric or magnetic field. (b) Electromagnetic waves obey all of the laws of optics and travel through a vacuum. vacuum. 8 (c) 3.0 × 10 m/s 4. 1 – positive; 2 – negative; negative; 3 – no no charge charge –19 5. An electron accelerated through an electrical potential potential of 258 V would gain 258 eV or (257 eV)(1.6 × 10 J/eV). me = 9.11 × 10 –31 kg v = ? 1 EK = mv 2 2 v= =

2 E K m 2(258 eV eV)(1.6 × 10

−19

J/eV)

9.11× 10 −31 kg

v = 9.5 × 10 m/s The speed of the electron just before it hits the second plate is 9.5 × 106 m/s. 6. Law of conservation of energy: For an isolated system, energy can be converted into different forms, but it cannot be created or destroyed. 7. (a) The two quantities conserved conserved in elastic collisions are the total kinetic energy of the system system and momentum. (b) For an elastic elastic collision, we know that p = p 6

m1v1 + m2 v2 



= m1v1′ + m2v2′ 



Since m1 = m2 and v2 = 0, we can simplify to v1 = v1′ + v2′ 

Copyright © 2003 Nelson





Chapter 11 Einstein’s Special Theory Theory of Relativity 639

We can apply the conservation of momentum to the x the x-components: -components: ′ ′ v1 x = v1x + v2 x 10.0 10.0 cm/s cm/s = v1′ x (cos (cos 30°) + 5.0 5.0 cm/s m/s ( cos cos 60° ) v1′ x

= 9.05 cm/s

The speed of ball 1 after the collision is 9.05 cm/s. A scale vector diagram would yield the same result. 8 8. (a) The speed of light is a specific magnitude of speed (c (c = 3.00 × 10 m/s). A light-year is the distance light would travel in one year. 8 8 (b) v = (0.5)3.00  10 m/s = 1.5  10 m/s = ∆d = ∆t = =

(

4.3 ly = 4.3 y 3.00 × 10 10 m/s 8

 365 d   24 h    3600 s  = 4.068 × 10 16 10 m        h   y   d 

)

? v= t =

=

d t d v 4.068 ×1016 m 1.50 × 108 m/s

t = 2.712 × 108 s To convert to days:

 1 h   1 d  = 3.139 ×10 3 d, 3 d, or 3.14 ×10 d     3600 s   24 h 

2.712 × 10 s  8

103 days to travel to Alpha Centauri.  60 s  = 498 s (c) ∆t = = 8.3 min = 8.3 min    min  ∆d = v∆t It would take you 3.14



= (3. (3.0 00 × 108 m/s)( /s)(49 498 8 s) s)

∆d = 1. 1.49 × 1011 m, or 1.49 × 108 km The Sun is 1.49 9. (a) (b)

8

10 km from Earth.



→24 He + 82622 Rn 214 0 214 82 Pb →−1 e + 83 Bi 226 88 Ra

Math Skills 10. (a) y (a) y = = mx + mx + b 14 (b) f -intercept -intercept = 6.1 × 10 Hz E -intercept -intercept = –2.5 eV rise (c) slope = run =

5.5 eV 13.5 × 1014 Hz

slope = 4.1× 10−15 eV/Hz The slope of the line on the graph is 4.1 -15 (d) y = y = 4.1  10 x – x – 2.5

640 Unit 5 Matter–Energy Matter–Energy Interface Interface



–15

10

eV/Hz


11. (a) x (a) x = = 0.5 y t 1 = 1.0 h t 2

t 1

=

1−

=

y

2

3.6 × 102 m/s 1−

t 2

x 2

(0.5 y )

2

y 2

= 4. 8 × 1 0 2 s

The value of t 2 is 4.8  102 s. t

 1  4200 25 = 220    2 

(b)

log25

=

log220

+

t 4200

 1    2 

log 

log 25   log  log220   t = 4200   log  1     2      t = 13 177 a, a, or or 1.3× 104 a The value for t is is 1.3  104 a.

Technical Skills and Safety 12. Some precautions when when working with with a source of ultraviolet light: • Do not stare directly at the light or light source. • Wear protective eye wear, such as dark goggles or glasses with side shields that are specifically coated to block out UV light. 13. Some precautions when when using a high-voltage power supply: • Before connecting or disconnecting a high-voltage power supply, always make sure the power supply is in the OFF mode. • When plugging or unplugging cords, always grasp the base of the cord closest to the outlet, never yank on the cord. • Keep all cords and plugs off the floor to avoid stepping on them.

CHAPTER 11 EINSTEIN’S SPECIAL SPECIAL THEORY OF RELATIVITY Reflect on Your Learning (Page 560)

1. 2. 3. 4.

5.

All motion is relative. To say that the ball is moving at 16 m/s, you would have to also state what what it is relative to. For example, you could state that the ball is moving at 16 m/s relative to the boat, the water, or Earth. We must have have some reference point (position, size, personal experience, etc.) etc.) in order to relate most things and opinions we encounter in life. Thus, everything is relative to some frame of reference, whether physical or nonphysical. Student answers will will vary. What the student should understand understand from this question is that the perception of a time interval is always relative to the observer; each observer can have a different view of the rate of the flow of time. It is expected, under Newtonian physics, physics, that the velocity vectors would add together and and the speed speed of the laser light beam relative to Earth would be 0.9c 0.9 c + c = 1.9c 1.9c. For the speed of the laser light beam relative to the ship, it is expected that it would be c – 0.9c 0.9c = 0.1c 0.1c. This is actually not the case. On cursory analysis, the answer would be no. But, if it is known that the mass of the sun is slowly diminishing, the answer is yes, since mass is not conserved.


Chapter 11 Einstein’s Special Theory Theory of Relativity 641

6.

E =

energy mass 8 c = speed of light (3.00 × 10 m/s) The equation implies that as mass decreases or increases, energy inversely increases or decreases. Since energy cannot be created or destroyed, we conclude that mass is another form of energy. m =

Try This Activity: A Thought Experiment (Page 561) (a) Intuitive reasoning based on Newton’s laws of motion tells us that we should catch up with the light waves. (b) The light waves should appear stationary, that is, light will stand still relative to the observer. (c) Classically, the beam should be some form of a wave front of energy. (d) According to Maxwell’s theory, and all reliable observations, there is no such thing as stationary light; no one has ever held a stationary clump of light in the palm of his or her hand. Light always travels are 3.00 × 108 m/s in a vacuum— no slower, no faster. (And as we will see in this chapter, in all frames of reference.) (e) If the light is stationary relative to the observer, it would not flow to the observer nor be reflected from a mirror.

11.1 FRAMES OF REFERENCE AND RELATIVITY Section 11.1 Questions (Page 568)

Understanding Concepts 1. 2.

3.

4.

5. 6.

There are no physics experiments or measurements that will determine whether you are moving. This is because no physics experiment can determine whether one inertial frame i s moving with respect to another inertial frame. An inertial frame of reference is one at rest, or one moving with a constant velocity, while a noninertial frame of reference is one that is accelerating. The Newtonian laws of physics are valid in all inertial frames of references, but break down in noninertial frames. Earth is strictly a noninertial frame of reference, but its acceleration (rotating on its axis while moving along its Sun orbit) is so small that it may be considered very nearly inertial. A rapidly accelerating car, an accelerating elevator, or a merry-go-round are all noninertial frames of reference where the normal laws of physics are found not to hold. We often account for unexpected motions that occur in noninertial frames by introducing “fictitious” inertial forces into the picture. The centrifugal force in a rotating frame of reference is an example of such a fictitious force. Such forces are not necessary if we pick an inertial frame. The ball will fall in front of you. The ball will retain the horizontal component of velocity, v1, which the train had when the ball was thrown, and will travel a horizontal distance ∆d 1 = v1∆t while in the air. You and the train are slowing down while the ball is in the air. The ball will travel a horizontal distance ∆d 2 = vavg∆t , where vavg < v1. Therefore, the ball will land in front of you, since ∆d 1 > ∆d 2. The Michelson-Morley experiment was designed to detect the so-called “ether” or absolute frame of reference against which the speed of light could be measured. Its great significance lies in the fact that it failed to detect such an effect, allowing for the conclusion, by Einstein, that the speed of light was absolute, regardless of the frame of reference from which it is measured. (i) The laws of physics are valid in all inertial frames of reference. 8 (ii) Light travels through empty space with a speed of 3.00  10 m/s, relative to all frames of reference. Yes, in certain circumstances. Consider a stationary observer, Os, standing midway between two lampposts in a straight line, as shown. If the lamps come on at the same time, he will observe them as simultaneous events since he is equidistant from them. If observer Om is moving towards O s as shown, he may be at Om1 when the lights come on, and at Om2 when he observes them. But since O m2L1 = Om2L2 , he will see them as simultaneous events. This will only be true if O m is moving toward, or away from O s. If he were moving in any other direction, he would not see the two events as simultaneous.

642 Unit 5 Matter–Energy Interface


11.2 RELATIVITY OF TIME, LENGTH, AND MOMENTUM PRACTICE (Page 573)


Airline pilots’ watches are running slow relative to Earth. However, the change is so small that the watches do not record any change at all when the pilots return to Earth. 8 v = 2.0  10 m/s ∆t m = 1.6  10 –8 s ∆t s = ? ∆t s ∆t m = v2 1− 2 c

∆ts = ∆t m

1−

= (1.6 ×10

v2 c2

−9

 2.0 × 108 m/s  1−   8  3.0 ×10 m/s 

s)

∆t s = 1.2 × 10−8 s 3.

The particles’ average lifetime when at rest is 1.2 v = 0.60c ∆t m = 32.0 h v = 0.94c ∆t m = ?



10 –8 s.

΄

΄

First we must determine the time interval relative to Earth: ∆t s ∆t m = v2 1− 2 c ∆t s 32.0 h = 2 (0.60c ) 1− c2 ∆t s = 25.6 h To calculate the time interval relative to the Vulcans for their new speed: ∆t s ∆t m′ = v2 1− 2 c 25.6 h

=

2

1−

(0.94c ) c2

∆t m′ = 75.0 h The new time interval would be 75.0 h.


Chapter 11 Einstein’s Special Theory of Relativity 643

4.

8

v = 2.6  10 m/s

∆t s

∆t m =

1−

∆t m = ∆t s

v2 c2

1 1−

v

2

c2 1

=

(2.6 ×10

8

1−

2

m/s )

c2

∆t m =2 ∆t s Its lifetime has increased by a factor of 2.

PRACTICE (Page 576)

Understanding Concepts 5.

v = 0.90c Lm = 50.0 m Ls = ? Lm

Ls

=

1− Ls

c2

Lm

=

1−

v2

c2 50.0 m

=

2

1− Ls

v2

( 0.90c ) c2

= 114.7 m, or 115 m

The spaceship’s length at rest is 115 m. 6. v1 = 0.60c Lm = 8.0 ly v2 = 0.80c Lm = ? ΄

Lm

= Ls

Ls

=

1−

c2

Lm 1−


v12

v12 c2


Since Ls is the same in both cases: Lm 1−

Lm′

=

v12

1−

c2

8.0 ly 0.80

v2 2 c2

Lm′

=

0.60

Lm′ = 6.0 ly The friend measures the length of the trajectory to be 6.0 ly. 7. v = 0.65c 2 Lm =3.00  10 m Ls = ? Lm

= Ls

Lm

=

1−

2

c2

Ls 1−

v2 c2

3.00 × 10 2 m

=

2

1− Ls

v

(0.65c ) c2

= 3.95 × 10 2 m

8. L0 = 40.0 ly 8 (a) v = 1.00  10 m/s L = ? L = L0

1−

v2 c2 8

= 40.0 ly 1 −

2

(1.00 × 10 m/s ) c2

L = 37.7 ly You would determine the distance to be 37.7 ly. (b) t = ? For you as the observer, v = 1.00  108 m/s, which is t =

=

c 3

.

d v 37.7 ly

 c   3   

t = 113 a You would determine the journey to take 113 a.



9.

v1 = 0.40c v2 = ? Lm

= Ls

v2

1−

c2

Since Lm = Lm , and Ls = 2 Ls: ΄

2 Ls 1 − 2 1− 4−

v2 2 c2 v2 2 c2

4v2 2 c2

−

v2 2 c

2

= Ls =

1−

1−

v12 c2

0.40c 2 c2

= 0.84 =

0.84

4  0.84  c 2 c 2 − v 2 =    4  0.84  2 = 1 −  c 4   v = ±0.89c

v2

The speed of the faster spaceship is 0.89c relative to Earth.

PRACTICE (Page 578)

Understanding Concepts 10. v = 0.999c me = 9.11  10 –31 kg p = ? mv

p =

1−

v2 c2

(9.11×10−

31

=

kg ) (0.999c ) 2

1− p = 6.11× 10

(0.999c )

−21

c2 kg ⋅ m/s kg⋅m/s.

–21

The relativistic momentum of the electron is 6.11  10 11. m = 2.00 kg v = 0.40c p = ? mv p = v2 1− 2 c

=

( 2.00 kg ) 0.40c 2

1−

(0.40c ) c2

p = 2.62 × 108 kg ⋅ m/s 8

The relativistic momentum in the Earth frame is 2.62  10 kg m/s.


·


12. v = 0.60c m p = 1.67  10 –27 kg p = ? mv

p =

1−

v2 c2

(1.67 ×10−

27

=

kg ) (0.60c ) 2

1−

(0.60c ) c2

p = 3.76 × 10 −19 kg ⋅ m/s –19

The relativistic momentum in that system is 3.76  10

kg m/s. ·

Section 11.2 Questions (Page 579)


3.

The clocks run at the same time if they are in the same inertial frame. Moving clocks affect the time of an event, as measured by an observer in another inertial frame, where the object (with the clock) is moving relative to the observer. To an observer moving with the process, the observer is stationary relative to the clock measuring the time. Proper time for an observer moving relative to the process, the time interval ( ∆t m) will be longer. Given that

∆t m =

∆t s 1−

v2

and Lm

= Ls

1−

v2 c2

, if c approaches infinity, the expression

1−

v2 c2

approaches 1. In this

c2 case, there would be no relativistic effects for both time and length. 4. If the speed of light were 100 km/h, relativistic effects would become noticeable at much smaller speeds, and we would be confined to a much smaller space in which we would move more slowly and time would pass more slowly. Tr avelling at 60 km/h (0.6c) would introduce time and length changes that play havoc with daily life. For example, people in a car would age slower than those at rest, and the car could not exceed 100 km/h. That is, 110 km/h would be impossible. 5. As discussed in Section 11.2, exceeding the speed of light is impossible. Matter can exceed the speed of sound, but this is not true for light. –6 6. ∆t s = 2.2  10 s (a) v = 0.99c ∆t m = ? ∆t s ∆t m = v2 1− 2 c

=

−

2.2 × 10 6 s 2

1−

(0.99c ) c2

∆t m = 1.6 × 10−5 s The average lifetime in the laboratory inertial frame of muons is 1.6 (b) ∆d = ? ∆d = v∆t

–5

10 s.



= (0.99c ) ( 2.2 × 10−6 s ) ∆d = 6.5 × 102 m The muons could travel an average distance of 6.5 (c) Lm = ?




102 m in the laboratory inertial frame.


Lm

= Ls

1−

v2 c2

= 6.5 × 102 m 1 − Lm 7.

0.99c c

2

2

= 92 m

The muons travel at an average distance of 92 m in the laboratory frame. t m = 40 a v = 2.5  108 m/s ∆t s ∆t m = v2 1− 2 c

∆ts = ∆t m

1−

v2 c2 2

= 40 a 1 −

( 2.5 ×108 m/s ) c

2

∆t s = 22 a Therefore, her age upon return is 20 a + 22 a = 42 a. 8. v = 0.90c Lm = 50.0 m Ls = ? Since a stationary observer measures the relativistic length of the spaceship: Lm Ls

= Ls

1−

c2

Lm

=

1−

v2 2

c 50.0 m

=

2

1− Ls

v2

(0.90c ) c2

= 115 m

The tape measure shows a length of 115 m. 9. v = 0.80c Lm = 40.0 m 6 L0 = 2.0  10 m (a) Ls = ? Since a stationary observer measures the relativistic length of the spaceship: Lm Ls

= Ls =

1−

v2 2

c 40.0 m 2

1− Ls

c2

Lm 1−

=

v2

(0.80c ) c2

= 66.7 m

The tape measure gives a value of 66.7 m.



(b) L = ? Since the observer in the spaceship observes the relativistic diameter of the planet: v2

L = L0 1 −

c2 2

= 2.0 × 10

6

m 1−

( 0.80c ) c2

L = 1.2 ×10 m The resulting value of the diameter of the planet is 1.2 (c) t = 8.0 s t 0 = ? t 0 t = 2 v 1− 2 c 6

t0

= t 1 −

v



6

10 m.

2

c2 2

= 8.0 s 1 − t 0

(0.80c ) c

2

= 4.8 s

The astronaut considers the journey to take 4.8 s. 10. bs = 1.00 m v = 0.90c D = 2.70  103 kg/m3 (a) Only dimensions in the direction of motion are affected, therefore dimension b is affected. (b) V m = ? First we must calculate the relativistic length of dimension b, since it is the only one affected by the motion: bm

= bs

1−

v

2

c

2 2

= 1.00 m 1 − bm

(0.90c ) c2

= 0.436 m

To calculate volume: V = a 0 bc 0

= (1.00 m)(0.436 m )(1.00 m ) 3 V = 0.436 m 3

The relativistic volume of the cube is 0.436 m . (c) p = ? First we must calculate the rest mass of the cube: m = DV

= ( 2.70 × 103 kg/m3 )(1.00 m3 ) m = 2.70 × 103 kg



To calculate the relativistic momentum of the cube: mv p = 2 v 1− 2 c

=

( 2.70 ×103 kg ) (0.90c ) 2

1−

(0.90c ) c

2

p = 1.67 × 1012 kg ⋅ m/s The relativistic momentum of the cube is 1.67 11. m = 6.68  10 –17 kg v = 0.400c p = ? mv p = 2 v 1− 2 c

=

1012 kg m/s.



·

(6.65 ×10−17 kg ) (0.400c ) 2

1−

(0.400c ) c

2

p = 8.71× 10−9 kg ⋅ m/s

Applying Inquiry Skills 12. Table 1 v c

0.050 0.100 0.150 0.200 0.250 0.300 0.350 0.400 0.450 0.500 0.550 0.600 0.650 0.700 0.750 0.800 0.850 0.900 0.950

1−

v

2

c2 0.9987 0.9950 0.9887 0.9798 0.9682 0.9539 0.9367 0.9165 0.8930 0.8660 0.8352 0.8000 0.7599 0.7141 0.6614 0.6000 0.5268 0.4359 0.3122

∆t s

∆t m

Ls

Lm

100 s 100 s 100 s 100 s 100 s 100 s 100 s 100 s 100 s 100 s 100 s 100 s 100 s 100 s 100 s 100 s 100 s 100 s 100 s

100.13 s 100.50 s 101.14 s 102.06 s 103.28 s 104.83 s 106.75 s 109.11 s 111.98 s 115.47 s 119.74 s 125.00 s 131.59 s 140.03 s 151.19 s 166.67 s 189.83 s 229.42 s 320.26 s

100 m 100 m 100 m 100 m 100 m 100 m 100 m 100 m 100 m 100 m 100 m 100 m 100 m 100 m 100 m 100 m 100 m 100 m 100 m

99.87 m 99.50 m 98.87 m 97.98 m 96.82 m 95.39 m 93.67 m 91.65 m 89.30 m 86.60 m 83.52 m 80.00 m 75.99 m 71.41 m 66.14 m 60.00 m 52.68 m 43.59 m 31.22 m

(a) 0.89c (b) 0.40c (c) 0.10c to 0.15c; Objects that travel at this speed are typically


α particles.


(d) Graph of ∆t m versus v:

Graph of Lm versus v:

11.3 MASS AND ENERGY:

2

E = mc

PRACTICE (Page 583)


–27

m p = 1.67  10 E = ?

kg E

= mc2 2

= (1.67 × 10−27 kg )(3.00 ×108 m/s ) E = 1.50 ×10 −10 J To convert to electron volts: E =

1.50 × 10

−10

J

−19

1.60 × 10 J/eV E = 939 MeV –10 The rest energy of the proton is 1.50  10 J, or 939 MeV.



2.

2.00  102 MJ m = ?

E =

E

= mc2

m

=

E

= m

c

2

2.00 × 108 J 2

(3.00 ×108 m/s )

= 2.22 × 10−9 kg

The rest mass of the small object is 2.22  10 –9 kg. 3. v = 0.950c (a) E total = ? E total

mc

=

1−

2

v

2

c

2 2

=

(1.67 ×10−27 kg )(3.00 × 108 m/s ) 2

1− E total

( 0.950c ) c

2

= 4.813 × 10−10 J

To convert to MeV: 4.813 × 10−10 J

E total

=

E total

= 3.00 × 109 eV, or 3.00 × 103 MeV

1.60 × 10−19 J/eV

The total energy is 3.00  103 MeV. (b) E K = ? Since E K = E total – E rest, we must first calculate E rest: Erest

= mc2 2

= (1.67 × 10−27 kg )(3.00 × 108 m/s ) E rest

= 1.503 × 10−10 J

After converting to MeV, the rest energy is 939 MeV. To calculate kinetic energy:

= Etotal − E rest = 3.00 × 103 MeV − 939 MeV 3 E K = 2.07 × 10 MeV EK

The kinetic energy is 2.07  103 MeV.



4.

vi =

0c 0.990c E = ? vf =

E

= Etotal − E rest mc

=

1−

2

v

2

c

2

  2 = mc   

− mc 2

1 1−

v

2

c

2

  − 1    

  2 8 −31 = (9.11× 10 kg )(3.00 × 10 m/s )    −13 J E = 4.99 × 10 5.

  1  − 1 2 (0.990c )  1−  2 c 

The work required is 4.99  10 –13 J. 18 ∆ E = 9.80  10 J ∆m = ?

∆ E = ∆mc 2 ∆ E ∆m = 2 c

=

9.8 × 1018 J 2

(3.00 ×108 m/s)

∆m = 1.09 × 102 kg The amount of mass that would have to be converted to energy is 1.09  102 kg. 6. E p = 938.3 MeV E n = 939.6 MeV ∆m = ?

∆ E = ∆mc2 ∆ E ∆m = 2 c

= =

(939.6 MeV − 938.3 MeV) 2

(3.00 × 108 m/s) 1.3 MeV 2

(3.00 ×108 m/s ) (1.3 × 106 eV )(1.60 ×10−19 J/eV ) = 2 (3.00 ×108 m/s ) ∆m = 2.31× 10−30 kg The difference in their rest mass energy i s 2.31  10 –30 kg




Understanding Concepts

The equation E = mc2 is consistent with the law of conservation of energy in that it predicts that whenever an amount of mass m disappears, an amount of energy mc2 appears in some form or another. Mass is simply another form of e nergy to be considered when applying the law of conservation of energy. 2. m = 1.0 kg E = ? 1.

E

= mc2 2

= (1.0 kg ) (3.00 × 108 m/s) E

3.

= 9.0 × 1016 J

The amount of energy that can be produced is 9.0  1016 J. 24 mE = 5.98  10 kg 4 v = 2.96  10 m/s ∆m = ? First we must calculate the energy of Earth revolving around the Sun: 1 2 E

= =

E

mc

2 1

5.98 × 1024 kg )( 2.96 × 104 ) ( 2

2

= 2.62 × 1033 J

To calculate the amount of mass required:

∆m = =

∆ E c

2

2.62 × 1033 J 2

(3.00 ×108 m/s )

∆m = 2.91× 1016 kg The amount of mass required to accelerate Earth from rest to its orbital speed is 2.91  1016 kg. 4. E = 5.0 GW = 5.0  109 W ∆m = ? First we must convert the amount of energy produced per second in one year: 365 d  24 h  60 min  60 s  ∆ E = (5.0 × 109 J/s )       a  d  h  min 

∆ E = 1.577 × 1017 J To calculate the amount of fuel converted:

∆m = =

∆ E c

2

1.577 × 1017 J 2

(3.00 ×108 m/s )

∆m = 1.75 kg, or 1.8 kg 5.

The amount of fuel converted to energy in one year is 1.8 kg. v = 0.90c –27 m = 1.673  10 kg E K = ?



EK

= Etotal − E rest mc

=

1−

2

v

2

c

2

  2 = mc   

− mc2

1 1−

v

2

c

2

  − 1    

  2 = (1.673 × 10−27 kg )(3.00 ×108 m/s )     −30 kg E K = 1.95 × 10

   1 − 1 2  (0.90c ) 1−  2 c 

–10

The energy required to accelerate the particle is 1.95  10 6.

E =

1.60 MeV = (1.60 × 10 eV )(1.60 × 10

E T =

(a)

6

−19

J.

J/eV ) = 2.56  10

–13

J

3.20 MeV ?

m =

E

= mc2

m

= =

m

E c

2

2.56 × 10−13 J 2

(3.00 × 108 m/s)

= 2.84 × 10−30 kg –30

The rest mass of the particle is 2.84  10 (b) E K = ?

kg.

= Etotal − E rest = 3.20 MeV − 1.60 MeV E K = 1.60 MeV EK

The kinetic energy of the particle is 1.60 MeV.

Making Connections 7.

9.0  1016 J (from question 2) t = 1 h = 3600 s cost = ?

E =

Recall that a 1-W power source delivers 1 J per second.

= ∆E −16 W P = 9.0 × 10 P

Therefore,

∆ E ∆t ∆ E = P∆t = (9.0 × 1016 W ) (1 s ) P

=

∆ E = 9.0 × 1016 W ⋅ s



To convert to kW·h:

 1 h  1 kW  = 2.5 × 1010 kW ⋅ h   3  3600 s  1.0 × 10 W 

9.0 × 1016 W ⋅ s  To calculate the cost:

$0.15  = 2.5 ×1010 kW ⋅ h    kW ⋅ h  cost = $0.38 × 1010 cost

The cost of the rest energy would be $3.8  109. 8. Let the subscript g represent gas and the subscript a represent the Aspirin tablet. –4 m = 325 mg = 0.000325 kg, or 3.25  10 kg 8 E g = 1.05  10 J d g = 30.0 km d a = ? First calculate the energy provided by the Aspirin tablet:

∆ Ea = ∆ma c2 = (3.25 ×10−4

2

kg )(3.00 × 108 m/s )

∆ E a = 2.925 × 1013 J To determine how far the car could travel: da

= =

d a

∆ E a E g

(d g )

2.925 × 1013 J 1.05 ×108 J

(30.0 km )

= 8.36 × 106 km

The distance the car could travel on a single Aspirin tablet is 8.36  106 km. 9. E rest for deuteron = 1876 MeV Total rest mass of proton and neutron = 911 MeV + 941 MeV ∆ E = 1876 MeV – 1842 MeV ∆ E = 24 MeV The amount of energy liberated is 24 MeV.

11.4 THE LIFE AND TIMES OF ALBERT EINSTEIN PRACTICE (Page 586)

Making Connections (a) A description of Einstein’s early education, his lack of success in school, why he was not accepted at universities, why he ended up as a clerk in the patent office, etc. (b) This topic emphasizes his humanitarian interests and outlines his pacifism, with some emphasis on his intervention during the war years. (c) Although he was not involved in the project, he was instrumental in encouraging the project’s beginning and asking that its results (the atomic bomb) be demonstrated, not used on Japan. The role of others, such as Leo Szilard, who cajoled and convinced Einstein to write letters to President Roosevelt, would be part of the report. (d) In the post war period, with nuclear proliferation collimating in the testing of the hydrogen in 1951, many concerned scientists opposed these developments, often being accused as being communist sympathizers. This campaign and Einstein’s role should be the emphasis in this report. (e) An overview of what Einstein was attempting would begin this report, followed by why he was unsuccessful. An overview of developments since his death could complete the report.



CHAPTER 11 SUMMARY Make a Summary (Page 587)

(a) Although time will move at the same rate, as seen by the occupants of the spacecraft, the time on the space craft, relative to the people left behind, will be slower. As a result, the occupants of the spacecraft will “age” at a slower rate than their friends and relatives in Earth. (b) Barnard’s star is far enough away that the round trip would be over 12 ly. Assuming the spacecraft does not stop on the way, and it is travelling at close to the speed of light, the twin paradox will prevail. All your friends and relatives will be much older that you. (c) Although your spacecraft may end up travelling at close to the speed of light, it must accelerate to that speed, requiring more and more energy the faster it travels. To achieve the speed of light, an infinite amount of energy would be required, which is impossible. Nevertheless, the maximum speed will be limited by the large amount of energy required, and the inability to carry enough fuel to reach these speeds. (d) Since it is impossible to carry enough fuel or provide enough propulsion even from ion engines, somehow mass must be converted to energy, using the mass-energy conversion relationship. The fuel would have to be swept up from the free hydrogen in space and converted into energy in some type of thermal nuclear system. (Many physicists are skeptical that these types of space travel will possible at all, given the present state of knowledge.)

CHAPTER 11 SELF QUIZ (Pages 588–589)

True/False 1.

F The speed of light in water is less than the speed of light in a vacuum. This fact does not violate the speed-of-light postulate of the special theory of relati vity because it states “... light tr avels through empty space with a speed of c…” 2. T 3. T 4. (a) T (b) F Since both observers are moving relative to each other, and simultaneously moving relative to a clock, they are measuring two-position time, and will probably never agree on the rate at which the clock ticks. (c) T 5. F If you are in an inertial frame at a point a few thousand kilometres above the North Pole, the clock at the pole is stationary relative your inertial frame, while the equator is moving relative to your frame, As a result, there will be time dilation at the equator and the clock will run slower at the equator than at the North Pole, relative to your frame of reference. 6. F She could be addressing her son (see twin paradox, Section 11.2). 7. (a) T (b) T 8. T 9. T 10. F The rest mass energy is contained in the molecules of the spring. When the spring is compressed the mass does not change, only the potential energy in the spring changes (increases). 11. F The classical laws of conservation of energy and conservation of mass need to be modified mass-energy conversion, but this is usually only a factor at speeds greater than 0.1 c

Multiple Choice 12. (d) In Newtonian mechanics, no physics experiment can identify which inertial frame is truly at rest and which is moving. Since the train is windowless, there is no external frame of reference and thus there cannot be any relative velocity. 8 13. (c) Second postulate: “Light travels through empty space with a speed of c = 3.00 × 10 m/s in all inertial frames of reference” no more no less. Thus, (c) is the only correct option. 14. (d) Simultaneity is relative to the inertial frame of reference. Simultaneity is a relative concept (see Section 11.1). 15. (a) The Michelson–Morley experiment established that there is no observable ether wind. All of the other options are incorrect. 16. (b) Speed of light is constant in all inertial frames of reference (see question 13). Thus, (b) is the only correct answer.



17. (e) None of the options will inform you that you are moving, since none of the changes indicated will be observable to the astronaut. 18. (a) Since you are observing the duration of two events (two position time), there will be time dilation and the clock will tick slower, relative to the observer 19. (a) Proper time is the duration of a process measured by the observer who sees the entire process as occurring in one position. Thus, the correct option is (a). 20. (b) On the spaceship, time would run at the same rates as those seen by the occupants. Thus, the value does not change relative to those on the spaceship. In other words, the hearts beat at the same rate. Of course, to an observer in another inertial frame of reference, the rate could be slower. 21. (d)

∆t s

∆t m =

1−

2

c

2

∆t s

=

1−

(0.5c) c

2

2

∆t s

=

1 − 0.25 ∆t s

= ∆tm

v

0.877 = 1.2∆ts or 1.2T

The period is 1.2 T , therefore (d) is correct. 22. (b) Length contraction only occurs along the length of the moving object as seen by an external observer. Thus, (b) is the only correct answer. 23. (a)

∆ E = ∆mc ∆m =

= ∆m =

2

E c

2

3.7 × 106 J c

2

4.1 × 109 kg

Matter is converted to energy at the rate of 4.1

9

10 kg/s, therefore (a) is the correct answer.



CHAPTER 11 REVIEW (Pages 590–591)

Understanding Concepts 1. (a) The passenger measures the proper time since the passenger is stationary relative to the clock. (b) Again, the passenger is stationary relative to the train, and would measure the proper length. (c) You would measure the proper length between the railway ties, since you are stationary relative to the railway ties. 2. Only observers stationary relative to the event would record the proper time interval. (a) The spectator is stationary relative to the event and is in the same inertial frame. The spectator would record the proper time. (b) The fan sitting on the couch is stationary relative to the event, because the fan is watching the game through the TV camera’s eyes, which is also stationary relative to the event. The fan would record the proper time. (c) The short stop is moving relative to the event since he is running in to cover the play. He would not record the proper time. 3. Yes, all motion is relative to the observer. Therefore, stationary and moving can vary between an observer on the plane and an observer on Earth. 4. It means it does move slower. This is the actual behaviour, not just the perception of the observer. 5. As discussed in the text, the clocks in orbit will run slower by a factor predicted by the time dilation relationship. Although the difference will be small, the two clocks will no longer by synchronized—the clock on the airplane will be slower.



6.

To measure proper time, since the observer is at rest relative to the event, only one clock is needed. To measure time intervals, where the object or event is moving relative to the observer, the beginning and ending of the event are in two positions. Thus, two synchronized clocks are required. 7. One instance is the experiment where synchronized clocks were sent around the world on passenger jets and then compared with another at rest on Earth. Another instance is the ‘lifetime’ of high speed particles (muons) increased when accelerated to velocities approaching c. 8. v = 0.92c Lm = 48.2 m Ls = ? Lm Ls

= Ls

1−

c2

Lm

=

1−

v2

c2 48.2 m

=

2

1− Ls

v2

(0.92c ) c2

= 123 m

The ship is 123 m long at rest. 3 9. Ls = 3.20  10 km v = 0.70c Lm = ? Lm

= Ls

1−

v

2

c2 2

= 3.20 × 10 Lm

3

km 1 −

(0.70c ) c

2

= 2.28 × 103 km

The voyagers aboard the UFO measure the distance as 2.28 10. v = 0.90c Lm = 228 m Ls = ? Lm Ls = 2 v 1− 2 c

=

103 km.

228 m 2

1− Ls



( 0.90c ) c2

= 523 m

The length you now obtain is 523 m. 11. vX = 0.70c vY = 0.86c LX =? LY



LX LY

=

Lm Lm′

vX

Ls 1 −

2

c2

=

vY 2

Ls 1 −

c2

2

1−

(0.70c ) c

=

2

1−

= LX LY

2

(0.86c ) c2

0.714 0.510

= 1.4

The ratio of the values for the planet’s diameter is 1.4. 12. v = 0.866c p = ? mv p = 2 v 1− 2 c

=

(9.11×10−31 kg ) (0.866c ) 2

1−

(0.866c ) c

2

p = 1.58 ×10 −30 kg ⋅ m/s –30 The relativistic momentum of the electron is 1.58  10 kg·m/s. 6 –13 13. ∆ E = 4.9 MeV = 4.9  10 eV = 7.84  10 J ∆m = ? ∆ E ∆m = 2 c

(7.84 ×10−13 J ) = 2 (3.00 ×108 m/s ) ∆m = 8.7 × 10−30 kg –30

The rest-mass equivalent of the energy is 8.7  10 9 9 14. P = 3.00  10 W = 3.00  10 J·s  365 d  24 h   3600 s  = 3.15 × 107 s t = 1 a =      a  d   h  ∆m = ? E P = t 3.00 × 10 W 9

kg.

= 3.00 × 109 J/s

To calculate E total:

 3.00 ×109 J  =  (3.15 × 107 s ) s   16 E total = 9.45 × 10 J E total



To calculate the change in mass: ∆ E ∆m = 2 c

=

9.45 × 1016 J 2

(3.00 ×10 m/s ) 8

∆m = 1.05 kg The change in mass is 1.05 kg. 15. m = 9.11  10 –31 kg v = 0.20c E total = ? Total energy before the collision = E K electron + E K positron + E rest electron + E rest positron, but rest masses are equal and thus rest energies of the two particles are also equal. 1  Etotal = 2  mv 2  + 2 mc 2 2 

(

)

= mv 2 + mc 2 = m (v2 + c 2 ) = 9.11× 10 −31 kg ((0.20c )2 + c 2 ) = 9.84 × 10−14 J −14 9.84 × 10 J = 6.15× 105 eV, or 0.615 MeV −19 1.6 × 10 J/eV

E total Converting to MeV:

The energy of the electromagnetic radiation is 0.615 MeV. 16. E rest = 105.7 MeV E total = 106.7 MeV E K = ? EK = Etotal − E rest

= 106.7 MeV − 105.7 MeV E K = 1.0 MeV The kinetic energy of the muon is 1.0 MeV. 17. (a) m = 1.20  103 kg v = 28.0 m/s c = 47.0 m/s E K = ? EK = Etotal − E rest

=

mc 2 1−

v

2

− mc 2

c2

  2  = mc   

1 1−

v2 c2

  − 1    

   2 3 = (1.20 × 10 kg )(47.0 m/s )      E K = 6.49 × 105 J The relativistic kinetic energy of the car would be 6.49


   1 − 1 2  ( 28.0 m/s )  1− 2  ( 47.0 m/s )  5

10 J.




(b)

E K rel E K nonrel

=?

First we must calculate the nonrelativistic kinetic energy: 1 EK nonrel = mv 2 2 1 = (1.2 × 103 kg ) ( 28.0 m/s )2 2 E K nonrel

= 4.70 × 105 J

To calculate the ratio: E K rel E K nonrel E K rel E K nonrel

=

6.49 ×105 J 4.70 × 105 J

= 1.4

The ratio is 1.4:1. 68 18. E total = 1.00  10 J 1 67 E total = 5.00  10 J 2 m = 4.00  1030 kg number of stars = ? First we must calculate the change in mass: ∆ E ∆m = 2 c

=

5.00 × 10 67 J

(3.00 ×10

8

2

m/s )

∆m = 5.56 × 1050 kg To calculate the number of stars: number of stars

=

5.56 × 1050 kg 4.00 × 1030 kg/star

number of stars = 1.39 × 10 stars The number of stars half of the energy could create is 1.39 31 19. m = 1.97  10 kg 44 E = 1.02  10 J (a) ∆m = ? ∆ E ∆m = 2 c 20

=



20

10 stars.

1.02 × 10 44 J

(3.00 ×10

8

2

m/s )

∆m = 1.13 × 1027 kg 27

1.13  10 kg of mass are converted to energy in the explosion. ∆m =? (b) mstar

∆m mstar

∆m mstar

=

1.13 × 10 27 kg 1.97 × 1031 kg

= 5.75 ×10 −5

The ratio of the mass destroyed to the mass of the star is 5.75




–5

10 :1.


34

20. E = 1.00  10 J 33 (a) ∆ E = 1.00  10 J ∆m = ?

∆m = =

∆ E c2 1.00 × 1033 J

(3.00 ×10

8

2

m/s )

∆m = 1.11×1016 kg The decrease in the mass of the oceans would be 1.11 3 3 (b) Dw = 1.00  10 kg/m V = ? m V = D

=



16

10 kg.

1.11 × 1016 kg 1.00 × 103 kg/m 3

V = 1.11× 1013 m 3 The volume of water this corresponds to is 1.11 21. (a) energy consumption = 485 kW ·h m = 1.00 g efficiency = 40% t = ?



13

3

10 m .

 4.85 ×105 J ⋅ h   3600 s  = = 1.746 × 109 J/month    s    h  9 10 For 12 months total energy: 1.746 ×10 J × 12 = 2.095 ×10 J 485 kW ⋅ h

To calculate the change in energy:

∆ E = ∆mc 2 = (1.00 × 10−3 kg )(3.00 ×108 m/s )

2

∆ E = 9.00 ×1013 J To calculate time: t =

(

0.40 9.00 ×1013 J

)

1.746 ×109 J

t = 2.06 × 10 months 4 A mass of 1.0 g would last 2.06  10 months. (b) number of homes? 4

number of homes =

9.00 × 1013 J

= 4.30 × 103 homes

2.095 ×10 J/homes The number of homes that could be supplied with this energy is 4.30 10



103 homes.

Making Connections 22. Science fiction authors utilize some of the concepts of special relativity in their writing to suggest time as a fourth dimension, time standing still, space warp, and vapourization of mass into energy (ray guns). c 23. Since v = , and n =1.33 for water: n v of light in water =

299 792 500 m/s

= 230 609 315 m/s 1.33 Pavel Cerenkov had succeeded in speeding electrons up to 257 500 000 m/s by 1934, thus sending particles at speeds greater than light. β particles with kinetic energies of 0.26 MeV will travel at speeds in excess of 2.30 × 106 m/s in a dense medium such as water. Frank and Tamm also conducted this work in 1937 and the three shared the 1958 Nobel prize for their work in this area. Copyright © 2003 Nelson


While no particle can exceed the speed of light in a vacuum (3.00 × 10 m/s), it is possible for a particle to travel faster than the speed of light in certain media, such as water. This is not a violation of the Einstein’s prediction (Section 11.2) that no object can equal or exceed the speed of light since it refers to the situation in a vacuum or air, that 8 is, 3.00 × 10 m/s. When the charged β particle moves through the water it tends to “polarize” the water molecules in a direction. 8

Extension 24. p = ? First we must calculate the value for v: Etotal = 5E rest Etotal mc

2 2

1−

v

1−

v

2

c

2

1−

= 5mc 2 = 5mc 2

c2

v

2

c2 v

2

c2 v

2

c

2

=

1

=

1

5 25

= 1−

1 25

= 0.96

= 0.96c2 v = 0.98c v To calculate momentum: mv

p =

1−

v

2

c2

(9.11×10−

31

=

)

kg (0.98c )

1 5

p = 1.34 ×10

−21

kg ⋅ m/s

The momentum of the electron in that frame is 1.34 1 25. ts = t m 2 t m 1 =2= 2 t s v 1− 2 c 1−

v2 c

2

v

2

c

2

v

2

c2

=

1

=

3

–21

10



kg·m/s.

4 4

= 0.75

v = 0.75c v = 0.87c The value of v, expressed as a fraction of c is 0.87c.



26. pR = 3 p mv

pR =

1−

1−

1− 9c

2

v2 c

2

v

2

c

2

2

c2

p

3 p =

1−

v

=

1

=

1

v2 c2

3 9

− 9v = c 2 2

v=

8

c 9 v = 0.943c 8 The speed of the particle is 0.943c, or 2.83  10 m/s. 5 27. m = 1.40  10 kg 13 p = 3.15  10 kg·m/s v = ? mv p = v2 1− 2 c 3.15 × 10 kg ⋅ m/s = 13

(1.40 ×105 kg ) v 1−

v

=

v2

1−

c2

3.15 × 1013 kg ⋅ m/s 1.40 × 105 kg

c2

v v

1−

= 2.25 × 108 m/s

2

c2

v2 1−

v2 c

= ( 2.25 × 108 m/s )

2

2

v2 cv 2

(

v2

v 2 1.406 ×1017 m 2/s 2

2

= ( 2.25 × 108 m/s ) − ( 2.25 × 108 m/s ) = ( 2.25 × 108 m/s )

2

c2

2

v2 c2 2

− (2.25 × 108 m/s )

v2

) = 4.556 ×1033 m 2 /s 2

v = 1.80 × 10 m/s 8 The rocket is travelling at a speed of 1.80  10 m/s. 28. (a) In the special theory of relativity, Einstein concluded that there is no way for an observer to determine whether a given frame of reference is at rest or is moving at constant velocity in a straight line. Therefore, Einstein said, the laws of physics must be the same in different inertial r eference frames. To consider only uniformly moving reference fra mes is somewhat restricting. What about the general case of motion, where reference frames can be accelerating? It is in the general theory of relativity that Einstein tackled the problem of accelerating reference frames and developed a theory of gravity. You sense that you are accelerating when in a car moving around a sharp curve. You 8



could not prove by any experiment that in fact you weren’t simply experiencing the pull of a gravitational field. Conversely, we might think we are being pulled by gravity when in fact we are undergoing an “inertial” acceleration having nothing to do with gravity. For example, pilots making a steep bank often have this experience, and cannot tell in which direction Earth lies without their instruments. As a thought experiment, consider a person in a freely falling elevator near Earth’s surface. If an observer held out a ball and let go of it, gravity would pull it downward toward Earth. But, it would fall at the same rate 2 ( g = 9.8 m/s ) at which the person and elevator were falling. So the ball would float right next to the person’s hand. The effect is exactly the same as if the elevator was at rest and no forces were acting on it. On the other hand, suppose the elevator were far out in space where there is no gravitational field. If the person released the ball, it would float. If 2 instead, the elevator in space were accelerating upward at an acceleration of 9.8 m/s , the ball as seen by an observer 2 would fall to the floor with an acceleration of 9.8 m/s , just as if it were falling because of gravity on Earth. According to the principle of equivalence, the observer could not do an experiment to determine whether the ball fell because the elevator was accelerating upward at 9.8 m/s 2 in the absence of gravity, or because a gravitational field with g = 9.8 m/s2 was acting downward and the observer was at rest on Earth. The two descriptions are equivalent. The principle of equivalence is related to the concept of mass and to the idea that there are two types of mass. For any force, Newton’s second law says that Σ F = ma, where m is the inertial mass. The more inertial mass a body has, the less it is affected by a given force and the less acceleration it undergoes. You might say that inertial mass represents resistance to any type of force. The second type of mass is gravitational mass, present when one body attracts another by the gravitational force. Up to this point, inertial and gravitational masses are the same because no experiment, not even high-precision experiments, has been able to discern any measurable difference between inertial and gravitational mass. This, then, is another way to state the equivalence principle: gravitational mass is equivalent to inertial mass. (b) The principle of equivalence can be used to show that light ought to be deflected due to the gravitational force of a massive body. Let us consider a thought experiment to get the idea. Consider an elevator in free space where no gravity acts. If there is a hole in the side of the elevator and a beam of light enters from outside, the beam travels straight across the elevator and makes a spot on the opposite side if the elevator is at rest. If the elevator is accelerating upward, the light beam still travels straight across in a reference frame at rest. In the upward accelerating elevator, however, the beam is observed to curve downward. Why? Because during the time the light tr avels from one side of the elevator to the other, the elevator is moving upward at with an increasing speed. Now, according to the equivalence principle, an upwardly accelerating reference frame is equivalent to a downward gravitational field. Hence, we can picture the curved light path in as being the effect of a gravitational field. Thus we expect gravity to exert a force on a beam of light and to bend it out of a straight-line path. (c) That light is affected by gravity is an important prediction of Einstein’s general theory of relativity, and it can be tested. The amount a light beam would be deflected from a straight-line path must be small even when passing a massive –10 body. For example, light near Earth’s surface aft er travelling 1.0 km is pr edicted to drop only about 10 m, which is not detectable. The most massive body near us is the Sun, and it was calculated that light from a distant star would be deflected by 1.75” of arc as it passed near the Sun. However, such a measurement could be made only during a total eclipse of the Sun, so the Sun’s intensity would not overwhelm the starlight passing near its edge. An eclipse occurred in 1919 and scientists journeyed to the South Atlantic to observe it. Their photos of stars around the Sun revealed shifts in accordance with Einstein’s prediction, producing headlines in the popular press at the time and making Einstein famous beyond the scientific community.



CHAPTER 12 WAVES, PHOTONS, AND MATTER Reflect on Your Learning (Page 592)

1.

2.

3. 4. 5.

In both cases, if white light is emitted, most of the wavelengths of the visible spectrum originate in the source of light. However, the sources of energy are different. In the incandescent lamp, thermal energy agitates the atoms and molecules to the point that their changing electric and magnetic fields radiate electromagnetic energy. In the fluorescent lamp, changes in the energy levels within the atom produces radiation, not heat. It is this radiation that causes fluorescence in the phosphors on the inside of the tube, therefore producing white light. Light energy causes changes and eventually a breakdown in the colour pigments in the printing ink. A red colour printing appears red because most of the other wavelengths in the white light spectrum are absorbed, and the red wavelength is reflected. For a blue surface, most of the wavelengths of the white line spectrum are absorbed, and the blue wavelength is reflected. Since the emissions of red and green wavelengths from the surface are of a lower energy than that of blue, the source of energy reflected by these pigments or dyes require less energy to be damaged by the incident radiation, and thus “fade” sooner than that for blue and violet. Quantum is defined as an abrupt change, sudden increase, or dramatic advance. A quantum leap is an abrupt change or advance, particularly in method, information, or knowledge. An electron microscope is a microscope that uses electron rather than visible light to produce images. (a) The force holding the satellite in orbit is the centrally acting gravitational force between the satellite and Earth. (b) Friction on the satellite and irregularities in the orbit eventually cause its speed to decrease and it enters successively low orbits. Eventually, it enters Earth’s atmosphere and either burns up or crashes on Earth’s surface. (c) The Coulomb force of attraction between the positively charged nucleus and negatively charged electron holds the electron in orbit around the nucleus of the atom. The electron stays in a stable orbit and does not 'crash' into the nucleus.

Try This Activity: Discharging with Light (Page 593)

• The discharge times will vary, depending on the level of humidity in the air. • The negatively charged zinc plate will typically discharge within 2 min., as water molecules adjacent to the plate absorb electrons. • With the glass filter in place, the ultraviolet light incident on the zinc plate should have little affect and the discharge rate will be similar to that without the ultraviolet light. • Without the glass filter in place, the ultraviolet light causes the zinc plate to discharge quite quickly, typically in one-half to one-quarter of the time without the ultraviolet light. (a) For the negatively charged electroscope to discharge, the electrons on the zinc plate must have escaped into the atmosphere. (b) The ultraviolet light has provided energy to the electrons making it easier for them to escape.

12.1 FOUNDATIONS OF QUANTUM THEORY PRACTICE (Page 597–598)


Discrete quantities include money, matter (atoms), and scores because all are exact. Nondiscrete quantities include ti me, energy, and length because all are measured and therefore error is intro duced. 2. h = 6.63 × 10 –34 J·s 8 c = 3.00 × 10 m/s E = ?


Chapter 12 Waves, Photons, and Matter 667

–7 (a) λ = 941 nm = 9.41 × 10 m

E =

=

hc λ

(6.63 ×10−34 J ⋅ s )(3.00 ×108 m/s ) −7

9.41 ×10 m

E = 2.1 ×10 –19 The energy is 2.1  10 J, or 1.32 eV. (b) f = 4.4 × 1014 Hz E = hf

−19

J

= (6.63 × 10−34 J ⋅ s )(4.4 ×10−14 Hz ) E = 2.9 × 10 −19 J The energy is 2.9  10 J, or 1.8 eV –7 (c) λ = 435 nm = 4.35 × 10 m hc E = –19

λ

(6.63 ×10− J ⋅ s )(3.00 ×10 34

=

8

m/s )

4.35 × 10 −7 m

E = 4.57 × 10 –19 The energy is 4.57  10 J, or 2.86 eV. (d) f = 1.2 × 1018 Hz E = hf

−19

J

= (6.63 × 10−34 J ⋅ s )(1.2 ×1018 Hz ) E = 7.96 × 10 −16 J The energy is 7.96  10 –16 J, or 5.0 × 103 eV. 3. E = 3.20 × 10 –19 J λ = ? hc λ =

E

(6.63 × 10− J ⋅ s )(3.00 ×10 34

=

3.20 × 10

−19

8

m/s )

J

−7

The wavelength is 6.22 4. E = 2.25 eV

–7

λ = 6.22 × 10 m

10 m, or 622 nm. The colour is red (from Section 9.6).



To convert energy into joules: E = (2.25 eV)(1.60 × 10 –19 E = 3.60 × 10 J

–19

To calculate frequency: E

J/eV)

= hf

f =

=

E h 3.60 × 10 6.63 × 10

−19

−34

J

J⋅s

f = 5.43 ×10 Hz 14

The frequency is 5.43 × 10 Hz. –7 λ 1 = 3.80 × 10 m –7 λ 2 = 1.14 × 10 m 14

5.

λ 2 λ 1

= ?



Since E photon 1

=

hc λ 1

and E photon 2

E photon 1 E photon 2

= E photon 2

hc λ 2

,

 hc   λ  =  1   hc   λ   2  =

E photon 1

=

=

λ 2 λ 1 −7

3.80 × 10 m −

1.14 × 10 7 m 3.33 1

The ratio of a quantum of “soft” ultraviolet radiation to a quantum of “hard” ultraviolet radiation is 3.33:1.


Let the subscript R stand for radio waves, and C stand for cosmic rays. 2  10 m –16 λ C = 1.0  10 m hc hc Since, E R = and E C = , λ R = 1.0

λ R

λ c

 hc    E R  λ R  = E C  hc   λ   C  = = E R E C

λ C λ R

1.0 ×10 −16 m 1.0 ×10 2 m

= 1.0 × 1018

18

It would take 10 quanta of radio waves. Radio waves have a long wavelength (low frequency), thus they have low penetration and low energy. Cosmic radiation has a high freq uency, therefore, they have high energy and high penetrating power, which would be a biological hazard.

PRACTICE (Pages 604–605)


8.

Intensity is the quantity or number of protons. Frequency determines the energy ( E = hf ). For photoemission to occur, the photoelectron needs sufficient energy to be released from the surface. I f the photons, no matter how many, do not have sufficient energy, no photoemission occurs.



9.

The energy of the emitted photons is directly related to the energy of the incident photon ( E = hf ) by the constant h. Thus, all the graphs have the same slope (h). 10. According to classical wave theory, the energy would have to build up to the point where the photoelectron is ‘released’ from the surface, which isn’t the case. The photoelectron is released immediately if there is sufficient energy. 11. W = 2.4 eV f = ? E photon = W hf

= W

f =

=

W h

( 2.4 eV ) (1.6 × 10−19 J/eV ) 6.63 × 10

f = 5.8 × 10

−14

−34

J ⋅s

Hz

The minimum frequency of the photon is 5.8 × 1014 Hz. 12. W = 3.33 eV f = ? W f = h

=

(3.33 eV ) (1.6 ×10−19 J/eV ) 6.63 ×10

−34

J ⋅s

f = 5.02 × 10 Hz 14

The threshold frequency is 5.02 × 10 Hz. 13. W = 2.48 eV -7 λ = 450 nm = 4.50 × 10 m E K = ? 14

First we must calculate the energy of the incident photons: hc E = λ

(6.63 ×10− J ⋅ s )(3.00 ×10 34

=

8

m/s )

−7

4.50 × 10 m −19

E = 4.42 × 10 J Converted to electron volts, E photon = 2.76 eV. We can now calculate the kinetic energy: EK,max = Ephoton − W E K,max

= 2.76 eV − 2.48 eV = 0.28 eV

The maximum kinetic energy of the ejected electrons is 0.28 eV. 2 –7 14. λ = 3.50 × 10 nm = 3.50 × 10 m E K = 1.20 eV W = ? EK = Ephoton − W W

= E photon − E K =

hc λ

− 1.20 eV

(6.63 ×10− J ⋅ s )(3.00 ×10 m/s ) = − 1.20 eV (3.50 ×10− m )(1.6 ×10− J/eV ) 34 7

8

19

W = 2.35 eV The work function of the metal is 2.35 eV.



15. f = 8.0 × 1014 Hz W = 1.2 eV v = ? First we must calculate the kinetic energy: EK = Ephoton − W

= hf − W (6.63 ×10−34 J ⋅ s )(8.0 ×1014 Hz ) = − 1.2 eV −19 × 1.6 10 J/eV ( ) E K = 2.115 eV For the maximum energy of photoelectrons that reach the plate: 2.115 eV − 1.0 eV = 1.115 eV 2 E K

V =

m

( 2 )( 2.115 eV ) (1.6 ×10 −19 J/eV )

=

9.11 ×10

−31

kg

V = 6.3 × 105 m/s 5 The maximum speed with which an electron reaches the plate is 6.3 × 10 m/s.

Making Connections 16. Students’ answers will vary depending on their choice of device. Some choices may include smoke detectors, fibreoptic detectors, security systems, garage door openers, or automatic light switches.

PRACTICE (Page 607)

Understanding Concepts 17. The units of momentum are kg·m/s. Since, J = N ⋅ m = ( kg ⋅ m/s2 ) m = kg ⋅ m2 /s2 , therefore: h

=

λ

h

m

= kg ⋅ m/s

λ

Therefore, the units of

kg ⋅ m 2 /s 2 ⋅ s

h λ

are the units of momentum.

18. λ = 5.00 × 102 nm = 5.00 × 10 –7 m p = ? h p = λ

=

6.63 × 10

−34

J⋅s

−7

5.00 × 10 m

p = 1.33 ×10

−27

kg ⋅ m/s

The momentum of the photon is 1.33 × 10

–27


kg·m/s.


19. f = 4.5 × 10 Hz p = ? 15

p =

=

h λ

hf c

(6.63 ×10− J ⋅ s )(4.5 ×10 34

=

15

Hz )

3.00 × 108 m/s

p = 9.9 ×10

−27

kg ⋅ m/s

The momentum of the photon is 9.9 × 10 2 20. E = 1.50 × 10 eV p = ? E p = c

–27

=

(1.50 ×10

2

kg·m/s.

eV )(1.6 ×10

−19

J/eV )

3.00 ×10 m/s 8

p = 8.00 × 10−26 kg ⋅ m/s The momentum of the photon is 8.00 6 21. v = 1.0 × 10 m/s λ = ?



10 –26 kg·m/s.

First we must calculate momentum: p = mv

= (9.11× 10−31 kg )(1.0 × 10 6 m/s ) p = 9.11×10 −25 kg ⋅ m/s To calculate wavelength: λ =

=

h p 6.63 × 10

−34

9.11 ×10

λ = 7.3 × 10

−10

The wavelength of the photon is 7.3

J ⋅s

−25

m –10

10



m, or 0.73 nm.

Section 12.1 Questions (Pages 608–609)


The historical discoveries and interpretations that led to the confirmation of the photon theory of light are • blackbody radiation – ultraviolet catastrophe • quanta – Planck • photoelectric effect – Einstein • momentum of a photon – Compton 2. To escape, the photoelectron requires sufficient energy to overcome being bonded to the surface. Since E p ∝ f , the photoelectron is only emitted when the frequency is high enough, that i s when E p > W . 3. Students may include some of the following: • security system detectors • TV, audiosystem remote controls • outdoor security lights • automatic night lights • garage door detectors



• cameras – light control • digital still and movie cameras • scanners for computers 4. E K = the maximum energy of the photoelectron hf = the energy of the incident photon W = the work function, the energy binding the electron to the photoelectric surface 5.

Photoelectric Effect photon disappears low energy photon result – electron with kinetic energy

Compton Effect photon re-emerges with lower frequency high energy photon result – lower energy photon and electron with kinetic energy energy conserved

energy conserved 6.

Five interactions are reflection, the photoelectric effect, the Compton effect, an increase in the internal energy of an atom, and pair production. 7. (a) The surface temperature of the emitting source determines the relative magnitude of the colours of the spectrum present in the radiation. (The surface temperature is often called the “colour temperature” for this reason.) The chemicals in “daylight” colour film are chosen to match the relative spectral density of the light radiated from the Sun (6000 K), therefore only artificial light of similar composition will work. Since incandescent light and fluorescent light have lower colour temperatures, the radiation they produce is improperly balanced. For example, pictures taken with “daylight” film under incandescent light tend to have unnatural colours (skin tones are orange instead of white). (b) A common flash in cameras today, the xenon flash, provides radiation similar to the radiation produced by the Sun. The resulting colours on the film are correct. 8. λ = 122 nm E = ? hc E = λ

(6.63 ×10− J ⋅ s )(3.00 ×10 34

=

8

m/s )

−7

1.22 × 10 m −18

9.

E = 1.63 × 10 J –18 The energy of an ultraviolet photon is 1.63 × 10 J, or 10.2 eV. hc E = λ

(6.63 ×10− J ⋅ s )(3.00 ×10 m/s ) = − − λ ( nm ) × (10 m/nm ) × 1.6 × 10 J/eV 34

9

E (eV ) =

8

19

1.24 × 103 λ ( nm )

10. All three curves intersect on the x-axis and at this point the photocurrent is zero. The maximum retarding potential ( V 0) represents the maximum energy, and, as a result, the maximum velocities of the photoelectrons. Since all three intensity curves intersect at V 0, the maximum velocities of the photoelectrons are independent of the intensity of light directed at the photoelectric surface. 11. From Table 1 we obtained the following data: • sodium = 2.26 eV • copper = 4.46 eV • potassium = 1.60 eV • barium = 2.48 eV The work function represents the energy that binds the electron to the photoelectric surface. The minimum energy required for an electron to be released from the surface to become a photoelectron, is the value given to the work function for a particular metal. From the values above it is apparent that it takes 2.20 eV more energy to release a photoelectron from copper than it does for one from sodium. The larger the value of the work function, the larger the threshold frequency. We can therefore conclude that since barium has a higher work function than potassium, bar ium will have a higher threshold freq uency.



12. W = 7.2 × 10 f = ?

–19

J

Since the work function is the energy required to release an electron from an illuminated metal, we can use the following relationship: E = hf f =

=

E h 7.2 × 10 6.63 × 10

−19

−34

J

J ⋅s

f = 1.1×10 Hz 15

The minimum frequency is 1.1 × 10 Hz. 13. W = 4.52 eV E K = 1.68 eV λ = ? 15

First we must calculate energy: E = EK + W

= 1.68 eV + 4.52 eV E = 6.20 eV To calculate wavelength (using the relationship from question 9): E = λ =

=

1.24 × 103 λ (nm)

1.24 × 103 E 1.24 × 103 6.20 eV

λ = 2.00 ×10 nm 2

The wavelength of light required is 2.00 × 10 nm. 14. λ = 482 nm W = ? EK = Ephoton − W 2

= E photon − E K 1.24 × 103 = − 1.2 eV − 4.82 × 10 7 m W = 1.38 eV W

The work function of the metal is 1.38 eV. –7 15. (a) λ = 2.0 × 10 m f = ? c = f λ f =

=

c λ

3.00 × 108 m/s 2.0 × 10−7 m

f = 1.5 × 1015 Hz The frequency of the photon is 1.5 × 10 Hz. 15



(b) E = ? To calculate energy in electron volts: 1.24 × 103 E = λ ( nm )

=

1.24 × 103

200 nm E = 6.2 eV To calculate energy in joules: E = (6.2 eV ) (1.6 × 10

−19

J/eV )

−

E = 9.9 × 10 19 J The energy of the same photon is 6.2 eV, or 9.9 × 10 –19 J. (c) p = ? E p = c hf

=

c

(6.63 × 10− J ⋅ s )(1.5 ×10 34

=

15

Hz )

3.00 × 108 m/s

p = 3.3 × 10

−27

kg ⋅ m/s

The momentum of the same photon is 3.3 × 10 16. (a) λ = 2.50 × 10-9 m p = ? h

–27

λ =

p =

=

kg·m/s.

p h λ

6.63 × 10

−34

J⋅s

−9

2.50 × 10 m

p = 2.65 × 10 −25 kg ⋅ m/s The momentum of the photon is 2.65 × 10 -31 (b) me = 9.11 × 10 kg p = 2.65 × 10-25 kg·m/s v = ? p = mv

–25

v=

=

kg·m/s.

p m 2.65 ×10

−25

9.11× 10

kg ⋅ m/s −31

kg

v = 2.91× 105 m/s 5 The speed of the electron is 2.91  10 m/s. (c) E K = ? E photon = ? To calculate the kinetic energy of the electron: 1 EK electron = mv 2 2 2 1 = (9.11×10−31 kg )(2.91× 10 5 m/s ) 2 E K electron


= 3.85 × 10 −20 J


To calculate the energy of the photon: hc E photon = λ

(6.63 ×10− J ⋅ s )(3.00 ×10 34

= E photon

8

m/s )

2.50 × 10−9 m

= 7.96 × 10−17 J

The kinetic energy of the electron is 3.86 × 10 J, which is substantially smaller than the energy of the photon, which is 7.96 × 10 –17 J. 17. Let the subscript 1 represent the first radio station and 2 represent the second radio station. 2 5 f 1 = 5.70 × 10 kHz = 5.70 × 10 Hz f 2 = 102 MHz = 102 × 106 Hz E 1 = ? E 2 = ? –20

= hf : E 1 = (6.63 × 10−34 J ⋅ s )(5.70 × 10 5 Hz )

Using the relationship E photon

E 1

= 3.78 × 10−28 J

E 2

= (6.63 × 10 −34 J ⋅ s )(102 ×10 6 Hz )

E 2

= 6.76 ×10−26 J J, or 2.36 × 10 eV. The energy of the photons

–18

The energy of the photons emitted from radio station 1 is 3.78  10 –26 –7 emitted from radio station 2 is 6.76  10 J, or 4.23 × 10 eV.

–9

Applying Inquiry Skills 18. (a) Leaves containing chlorophyll appear green in white light because the red and violet photons are absorbed and not reflected, whereas the green photons are reflected and not absorbed. (b) As seen from the graph, green light has a much smaller effect on the rate of photosynthesis than does red and violet light. (c) As seen from Figure 2 on page 595, a source with a surface temperature of 2500 K is very weak in the blue region of the spectrum and does not produce any radiation in the violet region. The graph in this question shows the photosynthesis reaction is stimulated by light fro m the blue-violet region. Therefore, the r ate of photosynthesis produced by an incandescent lamp will be too low for most green plants and will be unable to maintain the health of the plant.

Making Connections 19. The higher the value of Planck’s constant, the steeper the slope and the lower the cutoff frequency. Thus, photoemission would occur for a broad range of materials, creating a state of change in many of them. On the other hand, the smaller the constant, the lower the slope and the higher the cutoff frequency. It is possible that the cutoff frequency is high enough that the photoelectric effect does not occur. Therefore, solar cells, cameras (picture and digital), and detectors would not work. –7 20. λ = 552 nm = 5.52 × 10 m –3 –3 P = 1.0 mW = 1.0 × 10 W = 1.0 × 10 J/s number of light quanta = ? To calculate the energy of each photon: hc E = λ

(6.63 ×10− J ⋅ s )(3.00 ×10 34

=

8

m/s )

−7

5.52 × 10 m −19

E = 3.60 × 10 J Since 15% of the solar energy is converted, 15%(3.60


–19

10



–20

J) = 5.4  10

J.


To calculate the number of photons: number of light quanta

= =

number of light quanta

photons E −

1.0 × 10 3 J 5.4 × 10

−20

J

= 1.85 × 10

16

Therefore, 1.85 × 10 light quanta would be needed by the calculator each second. -2 21. (a) λ = 1.0 cm = 1.0 × 10 m E = ? hc E = 16

λ

(6.63 ×10− J ⋅ s )(3.00 ×10 34

=

8

m/s )

−

1.0 × 10 2 m −

E = 2.0 ×10 23 J The energy of a single microwave quantum is 2.0 × 10 –23 J. –2 (b) λ = 1.0 cm = 1.0 × 10 m mw = 0.250 kg ∆T = 100°C – 20°C = 80°C c = 4.2 × 103 J/kg·°C number of quanta = ? First we must calculate the change in energy: ∆ E = mc∆T

= ( 0.250 kg ) (4.2 ×103 J/kg ⋅°C ) (80°C ) ∆ E = 8.4 × 104 J To calculate the number of quanta: 8.4 × 104 J

number of quanta

=

number of quanta

= 4.2 × 10 27

2.0 ×10 −23 J

Therefore, 4.2 × 1027 quanta are required to raise the temperature of the water. 22. All the detectors on the Hubble telescope are photodetectors since the information must be digitized for transmission to Earth. Some areas for the students to research are construction and type of detector, placement of detectors, and the sensitivity/range of detectors (infrared, ultraviolet, X ray). 23. Ultraviolet photons have shorter wavelengths and a higher energy than infrared photons. This energy is sufficient enough to kill the skin cells, which is the result we see when the skin is sunburned. On the other hand, the infrared photons have sufficient energy to warm the cells but not damage them. Students should be careful not to confuse individual photon energy with total energy. A few joules of ultraviolet radiation can do more damage than a few hundred joules of infrared radiation.

12.2 WAVE–PARTICLE DUALITY PRACTICE (Page 614)

Understanding Concepts 1. (a) m = 2.0 kg v = 15 m/s λ = ?



λ =

=

h mv 6.63 × 10

( 2.0

−34

J ⋅s

kg )(15 m/s )

λ = 2.2 × 10

−35

m The de Broglie wavelength is 2.2  10 –35 m. (b) v = 1.3  105 m/s λ = ? h λ =

=

mv 6.63 × 10

(1.7 × 10

λ = 3.0 × 10

−27

−34

J ⋅s

kg )(1.3 × 10 5 m/s )

−12

m The de Broglie wavelength is 3.0  10 –12 m. 4 (c) v = 5.0  10 m/s λ = ? h λ =

=

mv 6.63 × 10

(9.11×10−

31

−34

J ⋅s

kg )(5.0 × 10 4 m/s )

−8

λ = 1.5 × 10 m

The de Broglie wavelength is 1.5 2. E photon = 3.0 eV E electron = 5.0 eV λ photon = ? λ electron = ?



–8

10 m.

To calculate the wavelength of the photon: hc

=

λ photon

E photon

(6.63 ×10− J ⋅ s )(3.00 ×10 m/s ) = (3.0 eV ) (1.6 × 10− J/eV ) 34

8

19

= 4.1× 10 −7 m

λ photon

To calculate the wavelength of the electron, we must first calculate the speed of the electron: v=

=

2 E electron m

( 2 )(5.0 eV ) (1.6 ×10 −19 J/eV ) 9.11 ×10

−31

kg

v = 1.325 × 106 m/s To calculate the wavelength of the electron: h λ electron

= =

λ electron

mv 6.63 ×10

(9.11×10−

31

J ⋅s

kg )(1.325 × 10 6 m/s )

= 5.5 × 10−10 m

The de Broglie wavelength of the photon is 4.1


−34

–7

10 m, and of the electron is 5.5



–10

10



m.


3.

m = 0.50 kg 2 v = 5.00  10 m/s λ = ? λ =

=

h mv 6.63 ×10

The de Broglie wavelength is 2.7 –9 4. λ = 0.15 nm = 0.15  10 m E K = ? 1 2

2 E K

mv 2 , v =

m 10 –36 m.



2 E K

p = m p

−36

. Plugging this value for v into the relationship p = mv:

m

2

J ⋅s

(0.50 kg ) (5.00 ×10 2 m/s )

λ = 2.7 × 10

Since E K =

−34

m

= 2 EK m

E K =

p

2

2m 2

 h    =  λ  2m

2

 6.63 × 10 −34 J ⋅ s   0.15 × 10 −9 m   = −31 2 (9.11 ×10 kg )

E K = 1.07 ×10 −17 J The energy required is 1.07  10 –17 J, or 67 eV. 2 5. E K = 1.00  10 V = 100 eV λ = ? To convert E K to joules: E K = (100 eV ) (1.6 × 10 −19 J/eV ) E K = 1.6 × 10 −17 J To calculate wavelength (using the relationship found in question 4): E K = p = λ =

=

p 2 2m 2 EK m h 2 EK m 6.63 × 10

J ⋅s

( 2 ) (1.6 ×10 −17 J )(9.11 ×10 −31 kg )

λ = 1.23 × 10

−10

m The associated de Broglie wavelength is 1.23


−34

–10

10



m.


–10 6. (a) λ = = 1.0  10 m p = p = ?

p =

=

h λ −34

J ⋅s

−10

m

6.63 × 10 1.0 ×10

p = 6.6 × 10

−24

kg ⋅ m/s

The momentum of the electron is 6.6 (b) v = ? h λ =

v

= =

10 –24 kg·m/s.



mv h mλ 6.63 ×10

(9.11×10

−31

−34

J ⋅s

kg ) (1.0 ×10

−10

m)

v = 7.3 × 106 m/s The speed of the electron is 7.3  106 m/s. (c) E K K = ? 1 EK = mv 2 2 2 1 = (9.11×10 −31 kg ) (7.3 ×10 6 m/s ) 2 E K = 2.4 × 10

−17

J

The kinetic energy of the electron is 2.4

10 –17 J, or 1.5  102 eV.




Understanding Concepts 1. (a). Diffraction of electrons, because the behaviour is the same as for electromagnetic waves (X rays). rays). (b) The momentum of a photon. nλ 2. The equation sin θ n = for a single slit predicts the most probable location on the screen. For example, for w sin θ 1

3.

=

1.5λ

, the photon is most likely to land on the screen in a bright area, whereas sin θ 1

w land on the screen at that point (a dark area). Students may find it easier to summarize their their points in a table:

Property energy mass momentum

Photon 2 eV –19 (3.2  10 J) m = 0 p

= =

p

E c 3.2 × 10

3.00 × 10 m/s

w

will most likely not

=

2mE

= ( 2 ) ( 9.11 × 10−31 kg ) ( 3.2 × 10−31 kg)

J

8

λ

Electron 2 eV –19 (3.2  10 J) m = 9.11  10 –31 kg p

−19

=

p

= 7.6 × 10−25 kg ⋅ m/s

= 1.1 × 10−27 kg ⋅ m/s



Property speed

Photon v=c

Electron v

v = 3.00 × 10 m/s 8

= v

wavelength

λ =

=

hc

(6.63 × 10

−34

J⋅s

) ( 3.00 × 10 m/s)

3.2 × 10

8

−19

m

( 2 ) (3.2 × 10−19 J ) −31

9.11× 10

=

J

h p h 2mE

−7

λ = 6.2 × 10 m

kg

= 8.4 × 105 m/s

λ =

E

2 E K

=

=

6.63 × 10

( 2 ) (9.11 × 10

λ = 8.7 × 10

−31

−10

−34

kg

J⋅s

) ( 3.2 × 10− J) 19

m

4. (a) m = 1.67  10 –27 kg v = 1.5  104 m/s = ? λ = λ =

=

h mv 6.63 × 10

(1.67 × 10

λ = 2.6 × 10

−27

−34

J ⋅s

kg ) (1.5 × 10 4 m/s )

−11

m The de Broglie wavelength of the neutron is 2.6 –31 (b) m = 9.11  10 kg 6 v = 1.2  10 m/s λ = = ? h λ =

=

6.63 × 10

(9.11×10

−31

−34

J ⋅s

kg ) (1.2 ×10 6 m/s )

−10

m The de Broglie wavelength of the electron is 6.1 9 (c) E K K = 1.0  10 eV λ = = ? h

= =



–10

10

m.

p h 2mE K 6.63 × 10

−34

J ⋅s

( 2 ) (1.67 × 10 −27 kg ) (1.0 ×10 9 eV ) (1.6 ×10 −19 J/eV )

λ = 9.1× 10

−16

m The de Broglie wavelength of the proton is 9.1


m.

mv

λ = 6.1× 10

λ =

–11

10



–16

10



m.


5. E K 5.00  104 eV K = λ = = ? Since EK =

1 2

2 E K

mv 2 , then v =

m

.

Therefore, λ = =

h mv h

=

2 E K

m

m 6.63 × 10−34 J ⋅ s

= 9.11 × 10

−31

λ = 5.49 × 10

−12

kg

( 2 ) (5.0 ×10 4 eV )( )(1.60 ×10−19 J/eV )

m The de Broglie wavelength of the electrons is 5.49 –9 6. λ = = 0.20 nm = 0.20  10 m p = p = ? m = ?

9.11 × 10 −31 kg 

–12

10

m.

To calculate momentum: p =

=

h λ

6.63 × 10

−34

J ⋅s

−9

0.20 × 10 m

p = 3.3 × 10−24 kg ⋅ m/s To calculate mass:

p = mv p = mc (for a photon) m=

=

p c 3.3 × 10 −24 kg ⋅ m/s 3.00 × 108 m/s −

7.

m = 1.1× 10 32 kg The momentum of the photon is 3.3  10 –24 kg·m/s. The mass of the photon is 1.1 v = 1.2  105 m/s –14 λ = = 8.4  10 m m = ? h λ =

m=

=

10 –32 kg.



mv h vλ 6.63 × 10 −34 J ⋅ s

(1.2 ×10

5

m/s ) (8.4 ×10 −14 m )

−

−

m = 6.57 × 10 26 kg, or 6.6 × 10 26 kg –26 The mass of the microscopic object is 6.6  10 kg.



8. E K = 5.0 eV w = ? h

λ =

2mE K 6.63 ×10

=

nλ

=

J ⋅s

( 2 ) (9.11 ×10 −31 kg ) (5.0 eV ) (1.60 ×10 −19 J/eV )

λ = 5.5 × 10

For a single slit, sin θ n

−34

w

−10

m

. For significant diffraction,

λ

w

≥ 1 , or w ≤ λ. Therefore, slit width must be less than

–10

5.5  10 m. 7 9. v = 0.25c = 7.50  10 m/s (a) λ = ? h

λ =

mv 6.63 × 10

=

(1.67 × 10

λ = 5.29 × 10

−27

−34

J ⋅s

kg )(7.50 ×10 7 m/s )

−15

m The associated de Broglie wavelength is 5.29  10 –15 m. (b) E K = ? 1 EK = mv 2 2 2 1 = (1.67 × 10−27 kg )(7.5 ×10 7 m/s ) 2 −12

E K = 4.70 × 10 The kinetic energy is 4.70 (c) E = ?

10



E = 4.70 ×10

=

J

–12

J.

−12

J

4.70 × 10 −12 J 1.60 × 10 −19 J/eV

E = 2.94 ×107 eV Therefore, the proton was accelerated through a potential difference of 2.94 the same charge). –11 10. λ = 1.0  10 m h λ =

2

λ

=

E K =

=

107 eV (since electron and proton have

2mE K h2 2mE K h2 2λ 2 m 6.63 × 10

−34

J ⋅s

( 2 ) (1.0 ×10 −11 m )(9.11×10 −31 kg )

E K = 2.41× 10




−15

J


To calculate the potential difference: E = 2.41×10

=

−15

2.41× 10 1.60 × 10

J

−15

−19

J

J/eV

E = 1.5 × 10 eV 4 The potential difference is 1.5  10 eV. 4

Making Connections 11. The scanning tunnelling microscope (STM) allows scientists the ability of “atomic vision” where they can actually see atoms. An STM works by passing an electric current between an experimentally manipulated scanning probe the size of an atom and the surface of the material under study. The STM is able to “feel” the topography of the object by detecting small variations in the current. The STM can map the surface of atoms by probing a semi-conductive surface with a very small needle. When the electron shells overlap and a current is applied, the electrons participate in quantum leaps. If the surface of the atom has a depression, the electron shells are farther apart and the electrons do not leap. If the surface is raised, the electron shells are closer together and more electrons “jump” to the probe. By measuring the amount of electron jumps, the STM can pass the info to a computer which makes a false colour map of the surface The word tunnelling, as it relates to the STM, refers to a small current being applied to a separate materials, and artificially joining them together so that an electron is able to move from one atomic landscape to another. The image on the IBM website http://www.almaden.ibm.com/vis/stm/images/stm.gif shows the standing-wave patterns in the local density of states of a copper surface. 12. In a transmission electron microscope, exposure time must be limited. The high-energy electrons bombarding the specimen can cause physical changes that will affect the resulting image. In scanning electron microscopes it is desirable that the specimen be rendered electrically conducting, otherwise a sharp picture will not be obtained. Conductivity is usually achieved by evaporating a film of metal 50 to 100 angstroms thick onto the specimen in a vacuum (such a thickness does not materially affect the resolution of the surface details). 13. Biographical information for Willard Boyle: • physicist, Retired former Executive Director of Research, Communications Sciences Division, Bell Labs in New Jersey. • born: August 19, 1924 in Amherst Nova Scotia • resides in Wallace, Nova Scotia • degrees: BSc, McGill, 1947; MSc, McGill, 1948; PhD (Physics), McGill, 1950 • co-inventor of the Charge Coupled Device, first continually operating ruby laser • Awards: The Ballantyne Medal of the Franklin Institute, 1973; Morris Lieberman Award of the IEEE, 1974; Progress Medal of The Photographic Society of America; Breakthrough Award by the Device Research Conference of the IEEE; Co-winner, C&C prize of the NEC Foundation, Tokyo, 1999; Edwin H. Land Medal, Optical Society of America, 2001 Biographical information for James Hillier: • physicist, Retired Executive Vice President and Sr. Scienti st RCA Labs, Princeton, NJ • born: August 22, 1915 in Brantford, Ontar io • degrees: BA, U of Toronto, 1937, MA, U of T, 1938, PhD, U of T, 1941 • co-designer of the first commercially available electron microscope in North America • Awards: Inductee, National Inventors Hall of Fame, 1980; Albert Lasker Award, 1960; Commonwealth Award, 1980

12.3 RUTHERFORD’S MODEL OF THE ATOM Section 12.3 Questions (Page 625)


The atomic model must have a very small, but very massive, positively charged nucleus, and the remainder of the atom must be empty space. The electrons are kept in orbital paths by the Coulomb force of electric attraction between the positive nucleus and themselves.



3.

F g F e

= ? Gm1m 2 F g

d 2 kq1q 2

=

F e

d 2 =

=

F g

=

F e

Gm1m 2 kq1q 2

(6.67

×

11

−

10

N ⋅ m 2 /kg 2 )(9.11×10

(9.00 4.4 × 10

×

−

31

kg )(1.67 ×10

109 N ⋅ m 2 /C 2 )(1.6 × 10

−

19

−

27

kg )

2

C)

40

−

–40

Therefore, F g ≈ 10 F e, and we can ignore F g due to its negligible effect. 4. E K = 4.5 MeV = 4.5  106 eV m = 6.6  10 –27 kg q2 = +79e r 0 = ? At the point of closest approach, the initial kinetic energy has been totally converted into electric potential energy: kq q EK = E p = 1 2 r 0 To convert E K into joules: E K = 4.5 MeV =

(4.5

×

106 eV )(1.6 × 10

E K = 7.2 × 10

13

−

−

19

J/eV )

J

To calculate the closest distance: kq q r 0 = 1 2 E K =

=

r 0 5.

=

(9.0

×

109 N ⋅ m 2 /C 2 ) ( 2 ) (1.6 ×10 7.2 ×10

3.64 × 10

26

−

7.2 × 10 5.05 × 10

14

−

N⋅m

−

13

13

−

−

19

C ) (79 ) (1.6 ×10

−

19

C)

J

2

J

m, or 5.1× 10

14

−

m

The closest distance the α particle could approach a fixed gold nucleus is 5.1 r 0 = ? kq q r 0 = 1 2 E

10 –14 m.



α

r0 Au r0 Al

=

=

r 0Al

α

kq qAl

(If E is constant) α

α

=

r 0Au

kq qAu

=

qAu qAl 79 13 6.1

Therefore, the closest distance an α particle can approach the gold nucleus is 6.1 times the distance it can approach the aluminum.



Applying Inquiry Skills 6.

Place one magnet (N-pole upwards) at the centre of the air table. Lock or hold it in place. Direct the other magnet (N-pole upwards) at the fixed magnet from the left. Since like poles repel, the incident magnet will be deflected by the same magnetic field of the fixed magnet. For a straight on “hit,” the moving magnet will rebound along the same path or back to the same side of the air table. “Misses” will be deflected at various angles, with the smaller the aiming error, the larger the deflection. The angle of deflection can be measured. The pattern of deflections is similar to that for the Coulomb scattering of alpha particles, demonstrated by Rutherford, Marsden, and Hertz.


Students should report on Rutherford’s academic achievements and his work with Soddy and Geiger. They may also note that Neils Bohr, James Chadwick, and Robert Oppenheimer studied under him. Rutherford had many strong qualities including his ability to foster wonderful relationships with his colleagues, and his unpretentious demeanour. He often didn’t put his name on research papers that he initiated, even if he did the bulk of the work, and would give students the opportunity to complete the paper. He never put his name on Geiger and Marsden's paper announcing large angle scattering of alpha rays, nor on Chadwick's paper announcing the neutron, nor on Cockcroft and Wilson's paper announcing the splitting of the atom using a particle accelerator. He is remembered by colleagues to be an inspiring person.

12.4 ATOMIC ABSORPTION AND EMISSION SPECTRA PRACTICE (Page 630)


The lowest excitation energy level for free mercury atoms is 4.9 eV. Therefore, a 3.9-eV electron will have an elastic collision with a free mercury atom, and continue on with 3.9 eV of energy. The electron can lose only the amount of energy corresponding to the difference between ground state ( E 1) and any excited state ( E e). E2 − E 1 = 4.86 eV

E3

E4

3.

E e

=

9.00 eV − 4.86 eV

E e

=

4.14 eV

E 1

=

6.67 eV

E e

=

9.00 eV − 6.67 eV

E e

=

2.33 eV

E 1

=

8.84 eV

E e

=

9.00 eV − 8.84 eV

E e

=

0.16 eV

−

−

The energies with which it can scatter are 4.14 eV, 2.33 eV, and 0.16 eV. From data point (ii): E2 − E 1 = 1.8 eV − 0.3 eV = 1.5 eV From data point (iii): E2

−

E 1

=

5.2 eV − 3.7 eV = 1.5 eV

E3

−

E 1

=

5.2 eV − 1.9 eV = 3.3 eV

Therefore, the value of E 2 is 1.5 eV, and the value of E 3 is 3.3 eV.





λ =

589 nm = 5.89 E p = ?

10 –7 m



E p

=

=

E p

=

hc λ

(6.63 ×10

−34

J ⋅ s )(3.00 ×10 m/s ) 8

−

5.89 ×10 7 m 3.38 × 10

−19

J

–19

5.

The energy difference is 3.38  10 684 nm = 6.84  10 –7 m E = ? hc E =

J, or 2.11 eV.

λ =

λ

=

(6.63 ×10

−34

J ⋅ s )(3.00 ×10 8 m/s )

6.84 × 10−7 m −19

E = 2.91 ×10 J –19  The atom loses 2.91 10 J, or 1.82 eV of energy. 6. E 1 = 8.25 eV λ = 343 nm E f = ? First we must calculate the electric potential: E p

=

=

1.24 × 103 λ (nm)

1.24 × 103

343 nm E p = 3.62 eV To calculate the energy of the final energy level: Ef = E1 + E p = 8.25

E f 7.

eV + 3.62 eV

= 11.87

eV, or 11.9 eV

The energy of the final energy level is 11.9 eV. A 7.0-eV electron can excite a mercury atom to its first or second excited state (4.9 eV, and 6.7 eV above ground state). Therefore, the energies of emitted photons are: E2 → E 1 = 4.9 eV E3

→

E 1

=

6.7 eV

E3

→

E 2

=

6.7 eV − 4.9 eV = 1.8 eV

8. (a) The energy the mercury vapour absorbs from white light causes some electrons to be elevated to higher energy levels. They then quickly descend by giving off this energy in the photons of electromagnetic radiation. (b) Some of the excited mercury atoms will return directly to the ground state, giving off photons of identical wavelengths as those absorbed. Other atoms will return via intermediate levels, giving off longer wavelengths. All emitted photons will come off in random directions, so the dark band in the original white light will not be filled in. 9. One way atoms can receive energy is in collisions with other particles. An example of this is electrons colliding with mercury atoms. Since the mass of the electron is much smaller than the mass of the mercury atom, the electron bounces off and the mercury atom recoils very slightly. Another way atoms can receive energy is in collisions with photons. An example of this is a photon colliding with a mercury atom. The mercury atom absorbs the photon, which then raises the internal energy of the mercury atom.



10.

172 nm = 1.72  10 –7 m –7 λ 3–1 = 194 nm = 1.94  10 m –7 λ4–1 = 258 nm = 2.58  10 m (a) E 2–1 = ? E 3–1 = ? E 4–1 = ? λ 2–1 =

For any photon: E

=

hc =

hc λ

For any transition to the ground state the photon has energy: hc E n = λ n

E 2−1

=

=

E 2−1

=

hc λ 2 −1

(6.63 ×10

J ⋅ s )(3.00 ×10 8 m/s )

−34

2.58 × 10 −7 m 7.71× 10

−19

J

Performing similar calculations for to determine all three energies: E 2−1 = 4.82 eV

(b)

E 3−1

=

6.41 eV

E 4−1

=

7.22 eV

λ4–3 =

? λ3–2 = ? λ4–2 = ? Other spectral lines occur for transitions from E 4→ E 3, E 4→ E 2, and E 3→ E 2. ∆ E 4 − 3 = 7.22 eV − 6.41 eV =

0.81 eV

∆ E 4 − 3 = 1.30 × 10

−19

J

To calculate wavelength: λ 4 − 3 =

=

hc ∆ E 4 − 3

(6.63 × 10

−34

J ⋅ s )(3.00 ×10 8 m/s )

1.30 × 10

−19

J

−7

λ 4 − 3 = 1.52 × 10 m

Performing similar calculations to determine all three wavelengths: λ4–3 = 152 nm λ3–2 = 782 nm λ4–2 = 518 nm 11. λ = ? For its emission spectrum, each possible downward transition produces a line. From E 2→ E 1: ∆ E 2 −1 = 3.0 eV ∆ E 2 −1 =


4.8 × 10

−19

J


To calculate wavelength: λ 2 −1 =

=

hc ∆ E 2 −1

(6.63 ×10

−34

J ⋅ s )(3.00 ×10 8 m/s )

4.8 × 10

−19

J

−7

λ 2 −1 = 4.14 ×10 m

Similarly, E 2 → E 1 E 4 → E 1 E 3 → E 1 E 4 → E 3

λ =

414 nm λ = 207 nm λ = 249 nm λ = 1240 nm

For its absorption spectrum, only upward transitions from E 1 are possible. Therefore, λ = 414 nm E 1 → E 2 λ = 249 nm E 1 → E 3 λ = 207 nm E 1 → E 4



Source Pattern 2.

3.

Continuous Spectra heated solid

Line Spectra discharge in gas

Absorption Spectra discharge in gas

continuous spectrum

discrete bright lines

continuous with discrete dark lines

The experiment was so important because it showed that there were discrete energy levels within the atom, related to the energies and orbitals of electrons. This lead to a quantum interpretation of spectra, and verified the Bohr quantum/wave model of the atom. Spontaneous emission is energy absorbed from a photon, which raises the atom to a higher energy level. The photon emitted is equal to the original photon with no amplification. Stimulated emission is when an atom emits a photon identical to the passing photon.

4.

Source Spectra Coherence Polarized Power 5.

Ordinary Lamp heated solid

Laser excited gas

continuous emission

discrete emission (monochromatic)

noncoherent (divergent beam)

coherent (parallel beam)

no

yes

low

high

Since the laser light is coherent, it tends to stay in a narrow beam. The light from the street lamp obeys the inverse law, spreading the light energy over a large energy, which diminishes the intensity.



6. E p = 3.66  10 –19 J λ = ? E p

=

hc λ

λ =

hc E p

=

(6.63 ×10

−34

J ⋅ s )(3.00 ×10 8 m/s )

3.66 × 10 −19 J −7

λ = 5.43 × 10 m

The wavelength emitted by the laser is 5.43 –3 7. P = 5.00 mW = 5.0  10 J/s –7 λ = 632.8 nm = 6.328  10 m –2 ∆t = 2.50  10 s (a) E = ? E = P∆t =

–7

10 m, or 543 nm.



(5.0 ×10 J )(2.50 ×10 s ) −3

−2

E = 1.25 × 10 −4 J The energy radiated with each pulse is 1.25 (b) number of photons = ?

10 –4 J.



First we must calculate the energy of each photon: hc E photon = λ

=

E photon

=

(6.63 ×10

−34

J ⋅ s )(3.00 ×10 8 m/s ) −

6.328 ×10 7 m 3.14 × 10

−19

J

To calculate the number of photons: number of photons

=

1.25 × 10 −4 J 3.14 × 10

−19

J/photon

number of photons = 3.98 ×1014 photons 14

There are 3.98  10 photons per pulse. 8. (a) The electron will bounce off the atom, emerging with 4.0 eV. (b) Nothing will happen because the photon energy must be exactly equal to the atom energy to change the energy level. (c) The photon will be completely absorbed by the atom. (d) The atom will be ionized. 9. If light from a star passes through a spectrograph, the relative shape of its emission spectrum can be determined. By comparing its emission spectrum with those of a known temperature, the temperature can then be determined. The emission spectrum provides a sensitive “thermometer” for any emitting surface. Through careful analysis of the absorption spectra, specific lines in the spectrum indicate elements in the gas of the star.

Making Connections 10. The absorption spectra will have black lines at discrete points in the spectrum. These lines identify the gases present in the Sun’s atmosphere. 11. This is known as the Doppler shift. The shift in the spectrum can be measured and the speed and direction (away or towards Earth) of the star can be calculated. 12. Lasing has been observed in a large number of molecular gases. Of particular importance is the carbon dioxide (CO2) molecular gas laser, which has lasing energy transitions at several wavelengths in the infrared, principally around –9 –9 9.6 × 10 m and 10.6 × 10 m. CO2 lasers are capable of continuous, repetitively-pulsed, Q-switched, and mode-locked operation with high energy outputs and large working efficiencies (10%–30%). Furthermore, they emit light at a frequency exhibiting little atmospheric absorption.



An important area of industrial application for CO2 lasers has been in materials processing, including drilling holes in various substances, paper cutting and perforation, cloth cutting, scribing of semiconductor wafers, and welding. In addition, carbon dioxide lasers are being used in laser-induced fusion studies, experimental optical communications and tracking systems, and in environmental testing and monitoring. Areas to be discussed include power, types, physical construction with diagrams, lasering action, cooling, and applications. 13. (a) The Advanced Ballistic Laser (ABL) weapon system consists of a high-energy, chemical oxygen-iodine laser mounted on a modified 747-400F (freighter) aircraft to shoot down ballistic missiles in their boost phase. A crew of four, including pilot and copilot, is required to operate the airborne laser, which patrols in pairs at high altitudes (over 10 000 m), flies in orbits over friendly territory, and scans the horizon for the plumes of rising missiles. Capable of autonomous operation, the ABL acquires and tracks missiles in the boost phase of flight and illuminates the missile with a tracking laser beam. Simultaneously, computers measure the distance and calculate the missile’s course and direction. After acquiring and locking onto the target, a second laser with weapons-class strength fires a three- to fivesecond burst from a turret located in the 747’s nose, destroying the missiles over the launch area. The laser’s fuel consists of the same chemicals found in hair bleach and Drano—hydrogen peroxide and potassium hydroxide—which are then combined with chlorine, gas, and water. T he laser operates at an infrared wavelength of 1.315 µm, which is invisible to the eye. By recycling chemicals, building with plastics, and using a unique cooling process, the laser is comparably light and efficient. This has also increased its power by 400% over previous designs. For the operational ABL system, several modules are linked together i n series to achieve ABL’s required megawatt-class power level. (b) Atmospheric turbulence, which weakens and scatters the laser's beam, is produced by fluctuations in air temperature (the same phenomenon that causes stars to twinkle). To overcome this, a complex mirror is used to compensate for tilt and phase distortions in the atmosphere. The mirror has 341 actuators that change at a rate of about a 1000 per second. The key issues in the program will be the effective range of the laser and systems integration of a Boeing 747 aircraft. The fact that this type of missile defense only attacks missile as they rise from Earth and not when they descend, means that anti-missile missiles would also be required for those that get through the laser defenses. (c) If the system works, and it has yet to be tested successful, nations that have ballistic missile capability would probably build more missiles to ensure that some of thei r missiles will get thro ugh the laser defenses, thus provoking a worldwide proliferation of missiles.

12.5 THE BOHR MODEL OF THE ATOM PRACTICE (Page 642)


λ = 388 nm = 3.88

–7

10 m



n = ? E p

=

=

E p

hc λ

(6.63 ×10

−34

J ⋅ s )(3.00 ×10 8 m/s )

3.88 × 10−7 m

=

5.13 × 10

=

3.20 eV

−19

J

From examining the energy levels for hydrogen atoms, the two levels that are separated by 3.20 eV are n = 8, and n = 2. −13.6 eV = −0.21 eV E 8 = 82 −13.6 eV E 2 = = −3.40 eV 22 The values of n are n = 8, to n = 2. 2. (a) Since the ground state is –13.6 eV, the ionization energy is 13.6 eV. (b) When n = 3, the energy is –1.51 eV. Therefore, the ionization energy is 1.51 eV.



3.

λ = 656 nm

n = ? First we must calculate the energy: E =

1.24 × 103 λ (nm)

E = 1.89 eV To calculate the levels of transition: n3 → n2 = −1.51 − (3.40 ) = 1.89 eV The levels where the transition occurred were between n = 3 and n = 2. 4. E = ? (a) For n = 3 → n = 5: –0.54 eV – (–1.51 eV) = 0.97 eV The energy of the photon is 0.97 eV. (b) For n = 5 → n = 7: –0.28 eV – (–0.54 eV) = 0.26 eV The energy of the photon is 0.26 eV.



To calculate the de Broglie wavelength: h λ =

mv =

6.63 × 10

(9.11×10

λ = 3.32 × 10

−31

−19

−34

J ⋅s

kg )(2.19 ×10 6 m/s )

m

To calculate the circumference of the first orbit: C = 2π r =

( 2π ) (5.3 ×10

−11

m)

−

C = 3.3 × 10 10 m Therefore, the circumference of the first orbit in the hydrogen atom is equal to the de Broglie wavelength of the electron. 6. The energy levels of hydrogen are: E 1 = –13.6 eV E 2 = –3.4 eV E 3 = –1.5 eV E 4 = –0.85 eV E 5 = –0.54 eV Therefore, the energies of all the possible photons are: E 5 – E 4 = 0.31 eV E 5 – E 3 = 0.96 eV E 5 – E 2 = 2.86 eV E 5 – E 1 = 13.1 eV E 4 – E 3 = 0.65 eV E 4 – E 2 = 2.6 eV E 4 – E 1 = 12.8 eV E 3 – E 2 = 1.9 eV E 3 – E 1 = 12.1 eV E 3 – E 1 = 10.2 eV



7.

Electrons with energy 12.3 eV can excite an H atom to its first and second excited states; therefore, photon in the emission spectrum are: E3 → E 1 E p1

= −1.5

eV − ( −13.6 eV )

E p1

= 12.1

eV

E2

→

E 1

E p2

= −3.4

eV − ( −13.6 eV )

E p2

= 10.2

eV

E3

→

E 2

E p3

= −1.5

E p3

= 1.9

eV − ( −3.4 eV )

eV

For the wavelength of E p1: λ 1 =

hc E p1

=

(6.63 ×10

−34

J ⋅ s )(3.00 ×10 m/s ) 8

(12.1 eV ) (1.6 × 10

−19

J/eV )

−7

λ 1 = 1.03 ×10 m

8.

Similarly, λ 1 = 103 nm λ 2 = 122 nm λ 3 = 654 nm The wavelengths that could be emitted by hydrogen are 103 nm, 122 nm, and 654 nm. For the Balmer series, E 2 is the lower level. −13.6 E 4 = eV = −0.85 eV 42 −13.6 E 2 = eV = −3.40 eV 22 To calculate wavelength: E p

=

λ =

hc λ

hc E p

=

(6.63 ×10

−34

J ⋅ s )(3.00 ×10 8 m/s )

( 2.55 eV ) (1.6 ×10

−19

J/eV )

−7

λ = 4.88 × 10 m –7

9.

The wavelength is 4.88  10 m, or 488 nm. −13.6 eV E 7 = = −0.278 eV 72 E 1 = –13.6 eV −13.6 eV E 6 = = −0.378 eV 62 λ = ?



(a) For the most energetic photon: E p = E7 − E 1 = −0.278 = 13.32

E p

=

λ =

eV − ( −13.6 eV )

eV

2.13 × 10 −18 J

hc E p

=

(6.63 ×10

−34

J ⋅ s )(3.00 ×10 8 m/s )

2.13 ×10−18 J −8

λ = 9.33 ×10 m

The wavelength of the most energetic photon is 9.33 (b) For the least energetic photon: E p = E7 − E 6 = −0.278 =

E p

eV − ( −0.378 eV )

0.100 eV

= 1.6 × 10

λ =

–8

10 m, or 93 nm.



−20

J

hc E p

=

(6.63 ×10

−34

J ⋅ s )(3.00 ×10 8 m/s )

1.6 × 10−20 J −5

−5

λ = 1.24 × 10 m, or 1.2 × 10 m

The wavelength of the least energetic photon is 1.2 10. E p = ? λ = ?

–5

4

10 m, or 1.2  10 nm.



At room temperature, nearly all hydrogen atoms are in the ground state, with n = 1. E 1 = –13.6 eV For the least energetic photon absorbed, the hydrogen atom excited to the first level, with n = 2. −13.6 eV = −3.4 eV E 2 = 22 Therefore, E p

= −3.4

eV − ( −13.6 eV )

E p

= 10.2

eV

To calculate wavelength: λ =

hc E p

=

(6.63 ×10

−34

J ⋅ s )(3.00 ×10 8 m/s )

(10.2 eV ) (1.6 ×10

−19

J/eV )

−7

λ = 1.22 × 10 m

The energy of the least energetic photon is 10.2 eV. The wavelength of the least energetic photon is 1.22 122 nm.




–7

10 m, or


11. λ = ? The longest wavelength is the least energetic; therefore, it is the smallest transition. (a) n = 1 n = 2 → n = 1 = –3.40 eV – (–13.6 eV) = 10.2 eV λ =

1.24 × 103 E p (eV)

=

1.24 × 103

10.2 eV λ = 122 nm The longest wavelength in the hydrogen Lyman series is 122 nm. (b) n = 3 n = 4 → n = 3 = –0.85 eV – (–1.51 eV) = 0.66 eV λ =

1.24 × 103 E p (eV)

=

1.24 × 103 0.66 eV 3

λ = 1.89 × 10 nm

The longest wavelength in the hydrogen Paschen series is 1.89 –3 12. r n = 1.0 mm = 1.0  10 m n = ? E n = ?

3

10 nm.



For the hydrogen atom: rn

=

5.3 × 10 −11 n 2 m

n= = =

r n 5.3 × 10 −11 m 1.0 ×10 −3 m 5.3 × 10 −11 m 18.9 × 106

n = 4.3 ×103 To calculate the energy: E n

=

=

−13.6

eV

2

n −13.6 eV 2

(4.3 ×10 ) 3

−

E n = −7.4 × 10 7 eV The value of n is 4.3  103. The energy of an electron in that orbit is –7.4

10 –7 eV.





Theoretically, the spectral lines become closer and closer together as n increases. However, as the ionization level is approached, the electrons will begin to escape. 2. (a) n1 → n3 = –1.51 eV – (–13.6 eV) = 12.1 eV The energy of the absorbed photon is 12.1 eV.



(b) n3 → n2 = –1.51 eV – (–3.40 eV) = 1.9 eV n3 → n1 = 12.1 eV n2 → n1 = –3.40 eV – (–13.6 eV) = 10.2 eV The atom could emit photon energies of 12.1 eV, 10.2 eV, and 1.9 eV. −13.6 eV = −3.4 eV 3. For n = 2, E 2 = 22 Therefore, the energy required for ionization is 3.4 eV. This is not likely to happen, since an atom is not in the n = 2 state for any appreciable time and absorption of energy, in that level, is highly improbable. 4. λ = ? f = ? 13.6 eV E 6 = = −0.38 eV 62 E 2

13.6 eV

=

= −3.40

22

eV

First we must calculate the energy of the photon: E p = −0.38 eV − ( −3.40 eV ) E p

=

3.02 eV

=

4.83 × 10 −19 J

To calculate the wavelength: hc λ =

E p =

(6.63 ×10

−34

J ⋅ s )(3.00 ×10 8 m/s )

4.83 × 10 −19 J −7

λ = 4.12 × 10 m

To calculate frequency: f = =

c λ

3.00 ×108 m/s −

4.12 × 10 7 m

f = 7.28 × 1014 Hz, or 7.3 × 1014 Hz 5.

The wavelength of the fourth Balmer line is 4.12 –9 r = 0.847 nm = 0.847  10 m n = ? rn

=



–7

10 m, or 412 nm. The frequency of the same line is 7.3



14

10 Hz.

5.3 × 10 −11 n 2 m

0.847 ×10 −9 m = 5.3 ×10 −11 n 2 m −9

0.847 ×10 m

n=

5.3 × 10−11 m

n=4 The hydrogen atom can have this radius when n = 4. 6 6. v = 2.19  10 m/s (a) λ = ? h λ =

mv =

6.63 × 10

(9.11×10

λ = 3.32 × 10

−31

−34

J ⋅s

kg )(2.19 ×10 6 m/s )

−10

m –10 The de Broglie wavelength is 3.32 10 m. 



(b) r = 4.8  10 –10 m n = 3 rn 4.8 ×10 −10 m

=

5.3 × 10−11 n 2 m

= 5.3 ×10

n=

−11

n 2m

4.8 × 10

−10

m

5.3 × 10

−11

m

n=3 Therefore, when the radius of the Bohr orbit is 4.8 7. F e = ? –10 At the ground state, r 1 = 2.1  10 m. kq q ke 2 F e = 12 2 = 2 v r =

(9.0 × 10

9

N ⋅ m 2/C 2 )(1.6 ×10

( 2.1× 10 F e

=

10 –10 m, the quantum number is 3.



−10

− 19

2

C)

2

m)

5.2 ×10 −9 N –9

The Coulomb force of attraction is 5.2  10 N.

12.6 PROBABILITY VERSUS DETERMINISM Section 12.6 Questions (Page 653)


The Bohr model assumes precise values for orbital radius r , but the uncertainty principle says this means the radial velocity will be uncertain. The Bohr model also assumes precise values for orbital velocity v, but again the uncertainty principle says this means orbital position will be uncertain. Although the precise orbit s predicted by the Bohr theory are no longer accepted, the model paved the way for better theories. 2. A particle’s position may be known with an uncertainty ∆ x at the same time as its velocity is known with an uncertainty h . ∆v, only where ∆ x ∆v ≥ 2π m 3. m = 12.0 g = 1.20  10 –2 kg 2 v = 1.80  10 m/s (a) λ = ? h λ =

mv =

6.63 × 10 −34 J ⋅ s

(1.20 × 10

λ = 3.07 × 10

−2

kg )(1.80 ×10 2 m/s )

−34

m –34 The de Broglie wavelength is 3.07 10 m. –2 (b) ∆ x = 0.60 cm = 0.60  10 m h ∆ x ∆p ≥ 2π 

(0.60 × 10

−2

m ) ∆ p ≥

6.63 ×10

−34

J ⋅s

2π

∆ p ≥ 1.76 × 10

−32

kg ⋅ m/s

The minimum uncertainty in momentum is ±1.76  10 –32 kg·m/s.



4.

If Planck’s constant were much larger, the right hand side of the equation,

h 2π

the quantity of ∆ x∆ p would be larger, since it must be equal to or greater than

, would be much larger as well. As a result, h

. Therefore, uncertainty increases if the 2π value of Planck’s constant increases. When shooting an arrow at a fixed target, the uncertainty of its position ( ∆ x) would increase as well, making it harder to hit. When shooting at a moving target (a deer) its velocity ( ∆v) would also be more uncertain, making it even harder to hit.

CHAPTER 12 LAB ACTIVITIES Lab Exercise 12.1.1: Analyzing the Photoelectric Effect (Pages 654–656)

Analysis (b) The graph shows the relationship between photocurrent ( µA) and retarding electrical potential (V).

(c) When the photocurrents for low- and high-intensity violet light are compared, we see that the higher the intensity of the incident light, the larger the value of the maximum photocurrent. (d) The violet light has the largest cutoff potential. The values of the cutoff potential for both intensities of violet light are the same. This indicates that the intensity of the light source has no effect on the cutoff voltage. (e)

Colour of Light yellow green blue violet

Wavelength of Light (nm)

Cutoff Potential (V)

578 546 480 410

0.25 0.40 0.70 1.15

Maximum (eV) 0.25 0.40 0.70 1.15

E K of

Ejected Electrons (J) –20 4.0  10 –20 6.4  10 –19 1.1  10 –19 1.8  10

Frequency of Light (Hz) 14

5.2  10 14 5.5  10 14 6.3  10 14 7.3  10

Sample calculation to find E K in J: EK = eV 0 =

(1.60 ×10

E K = 4.00 ×10


−19

−20

)

C (0.25 J/C )

J


Sample calculation (using yellow wavelength) to find E K in eV: E K =

4.00 × 10−20 J −

1.6 × 10 19 J/eV E K = 0.25 eV Sample calculation to find frequency: c = f λ f =

c λ 8

=

3.00 × 10 m/s 578 ×10 −9 m

f = 5.2 × 1014 Hz Violet light has the highest-energy photons since the electrons ejected by violet light photons have the highest maximum kinetic energy. 14 –19 (f) The graph shows the relationship between frequency (10 Hz) and energy (10 J). For energy, students should –20 –20 –19 –19 convert 4.0  10 and 6.4  10 to 0.40  10 and 0.64  10 respectively.

Values for f are listed in the table above. As indicated by the straight-line graph, the maximum energy of the ejected photoelectrons is directly proportional to the frequency of t he incident light for each colour of li ght. In other words, it appears that EK ∝ f . (g)



To calculate slope: slope =

rise run

(1.6 × 10 = (7.0 × 10

14

=

3.1 × 10

) − (−1.5 ×10 Hz ) − ( 2.3 ×10

−19

−19

J

) Hz )

− 19

14

J

J

14

4.7 ×10 Hz slope = 6.60 × 10 The slope is 6.60

J ⋅s

–34

10



−34

J·s.

To calculate experimental error: % error =

measured value − accepted value accepted value 6.60 × 10

−34

J ⋅ s − 6.63 ×10

−34

× 100%

J ⋅s

×100% 6.63 × 10 J ⋅ s % error = −0.005% The measurement is acceptable within experimental error. –19 (h) Extrapolating the graph to the intercept on the y-axis, we obtain –3.0 × 10 J. The significance of the negative intercept means this value is considered the work function. An incident photon with a minimum energy of 3.0 × 10 –19 J (1.90 eV) is required for a photoelectron to be released from the surface of the cesium. =


−34


14

(i) The threshold frequency f 0 for cesium as measured by the x-intercept on the graph is 4.6 × 10 Hz. To calculate wavelength: c = f λ λ =

c f

=

3.00 × 108 m/s 4.6 × 104 Hz −7

λ = 6.5 × 10 m

–7

The wavelength of cesium is 6.5  10 m, or 650 nm. (j) Based on the information derived above, the general equation for the graph is: y = mx + b EK = (6.60 × 10 (k) From EK = hf

− W ,

we obtain hf

=

−34

J ⋅ s) f − 3.0 ×10

EK + W or E photon

−19

=

J

EK + W .

(l) After plotting the data for barium and calcium on the graph in step (f), the two intercepts can be read as follows: barium: W = 3.8  10 –19 J; f 0 = 5.95  1014 Hz calcium: W = 5.1  10 –19 J; f 0 = 8.00  1014 Hz 14 14 (m) Since the visible spectrum ranges from 4.0  10 Hz to 8.0  10 Hz, and the threshold frequency for calcium is 14 8.00  10 Hz, calcium will not produce photoelectric emission for the visible range of the spectrum. (n) The slopes of the graphs for the three substances are nearly the same, and each has a value very close to Planck’s constant –34 (6.63  10 J·s). This indicates that Planck’s constant is the same for all photoelectric surfaces.

Investigation 12.5.1: The Energy Levels of Hydrogen (Pages 656–657)

Observations and Calculations (Data based on the PSSC Apparatus) The table below shows answers the required calculations in step (a) and (b).

Colour of Spectral Line

Distance to Centre Line x

Average Value of x (m)

Wavelength λ =

d ∆x L

(nm)

Energy E =

hc λ

Transition Involved (eV)

(eV)

pale violet

left (m) 0.071

right (m) 0.0765

0.0738

419

2.96

E6

→

E 2 3.02

violet

0.0755

0.0805

0.0780

443

2.81

E5

→

E 2 2.86

green

0.085

0.091

0.0880

500

2.49

E4

→

E 2 2.55

red

0.122

0.1265

0.1243

706

1.76

E3

→

E 2

1.9

Sample calculations: d = 1.72 × 10 –4 cm = 1.72 × 10 –6 m L = 30.3 cm = 0.303 m



(a) To calculate the wavelength for pale violet:

 ∆ x    L  −6 (1.72 × 10 m)(0.0738 m)

λ = d 

=

0.303 m

−7 λ = 4.19 × 10 m

The wavelength for pale violet is 4.19  10 –7 m, or 419 nm. (b) To calculate energy for pale violet in joules: E

=

hc

= E

λ

(6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) −7

4.19 × 10 m

= 4.74 × 10−19

J

To calculate energy in electron volts: E

=

4.74 × 10

−19

J

−19

1.60 × 10 J/eV E = 2.96 eV The energy of the photons comprising pale violet is 2.96 eV. (c) Energy levels in hydrogen: −13.6 eV E n

=

E 2

=

E 6

=

n

2

−13.6 eV 22 −13.6 eV 62

= −3.4 eV = −0.38 eV

For E 6 → E 2, the energy of the emitted photon would be: E 6 → E 2 = –0.38 eV – (–3.4 eV) = 3.02 eV The experimental value was 2.96 eV. (d) Similar calculations gives results for the other three spectral lines observed. The values are reflected in Table 1. (e) See Table 1.

Evaluation (f) The percent error would be: percent error =

=

measured value − accepted value accepted value 2.86 eV − 3.02 eV

3.02 eV percent error = − 5.3%

× 100 %

×100%

Synthesis (g) To find the visible energy transitions for neon, a neon discharge tube could be placed in the apparatus, and similar processes and calculations to those used for hydrogen could be used.



CHAPTER 12 SUMMARY Make a Summary (Page 658)

CHAPTER 12 SELF QUIZ (Pages 659–660)

True/False 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

F The wavelength of a single quantum is directly proportional to its frequency. T F For a given photoelectric surface, the shorter the wavelength, the higher the cut off potential. T h F Photons have momentum whose value is given by p =

λ

F When light passes through a medium, its behaviour is best explained using its particle properties, whereas when light interacts with matter, its behaviour is best explained using its particle characteristics. T F Electrically excited gases produce an emission spectrum or line spectrum, while a continuous spectrum is produced from a heated solid. T T F In an atom we think of the electron as a particle moving in a circular orbit, whose wave properties predict the probability of it position and velocity.

Multiple Choice 12. (d) Increasing the frequency of the light incident on a metal surface: - has no affect on the threshold frequency for the emission of photoelectrons - increases the number of photoelectrons emitted - decreases the threshold frequency for the emission of photoelectrons (d) - increases the kinetic energy of the most energetic photoelectrons - increases the kinetic energy of the photoelectrons



13. (b) The light beam that produced graph B had the highest frequency because it required the largest negative potential difference to reduce its photoelectric current to zero 14. (c) The longest wavelength is associated with the lowest frequency. Thus, the beam having the lowest frequency must have produced curve C, because it fell to zero current with the lowest negative potential difference. 15. (b) Photons with highest momentum have the highest kinetic energy. These high energy electrons would have require the highest negative potential difference to be stopped. Thus, the light beam of graph B must have produced them. 16. (a) The formation of these rings is similar to that produced by waves passing through a small opening and would be similar to that produced by light passing through a grating, if the grating were rotated through 360 . Thus, the electrons diffraction pattern is similar to that for light, illustrating the wave nature of electrons scattered by a thin gold film. h 17. (e) Since p 2 2mE K , increasing the energy increases the momentum of the electrons. Since λ , increasing p p ˚

=

=

decreases λ . For diffraction grating,

λ

d

=

sinθ , decreasing λ decreases θ , and thus decreasing the distance of the rings

from the axis of the beam. 18. (d) As the 8.0 eV electron collides with the atom, energy can be absorbed as follows: 8.0 eV – 6.67 eV = 1.33 8.0 eV – 4.86 eV = 3.14 The electron could bounce off in an elastic collision, emerging with 8.0 eV. 19. (d) Since the 9.0 eV electron has energy exceeding that of all of the energy levels shown in the diagram, energies of 4.86 eV, 6.67 eV and 8.84 eV are all possible. 20. (e) For a photon to be absorbed, it must have exactly the same energy, as energy levels in the atom. It does not, so the atom cannot be excited by a 9.0-eV photon. 21. (d) Ionization energy is the energy required to move the electron from its lowest energy level completely away from the atom. (18 eV) 22. (d) For n , E 18 eV = ∞

For n

=

∞

1, E 1

=

=

0 eV

The ionization potential is the energy required to increase the potential energy of an electron from a lower energy level to n . ionization potential = E E 1 = ∞

∞

=

−

18 eV 0 eV −

ionization potential = 18 eV 23. (a) For n

=

3, E 3

For n

=

2, E 2

=

=

12 eV 8 eV Eemitted photon = E3 =

−

E 2

12 eV 8 eV −

E emitted photon = 4 eV 24. (b) The electron can transfer enough energy to the atom to raise the level from one state to another state, with a higher energy level. In this case, the only possible transition is from n = 1 (0 eV) to n = 2 (8eV). Thus, the electron transfers 8 eV to the atom and rebounds with 2 eV remaining.



Visible photons are the ones our eyes can detect. A dark room is full of photons with infrared wavelength and greater, but none have sufficient energy to activate the nerve cells in the retina of our eyes. Therefore, we cannot see in the dark because visible light photons do not exist. There may also be photons of ultraviol et wavelengths and shorter, but these are also invisible to our eyes. 2. Photographic film chemicals are chosen to respond to certain light wavelengths or photon energies. Some black and white films are insensitive to low energy red photons and thus are not exposed by the red light used in the dark room. Since colour film is sensitive to all visible photons a red light is not sufficient, and the film must be developed in complete darkness.



3.

The photon gives up its fixed amount of energy to an electron. However, the electron can lose any amount of this energy “jostling” its way to the surface. As long as it still has enough energy left to overcome the work function, it can escape. Hence the variations in the speeds of the photoelectrons. 4. E = 2.46 eV λ = ? 1.24 × 103 E = λ (nm) λ =

=

1.24 × 103 E 1.24 × 103

2.46 eV λ = 504 nm The longest wavelength of light is 504 nm, or 5.04 5. λ = 6.0  102 nm = 6.0  10 –7 m W = 2.3  10 –19 J E K = ? EK = hf − W

=

hc λ



− W

(6.63 ×10− J ⋅ s )(3.00 ×10 34

=

–7

10 m.

−7

6.00 × 10 m

=1.02 × 10

−19

8

m/s )

− 2.3 × 10−19 J

J

E K = 0.64 eV Therefore, the required potential difference is –0.64 eV. The value is negative because a negative potential is required to stop the electrons. 6. λ = 4.30  102 nm E K = 1.21 eV W = ? W = E p − E K 1.24 ×103

=

− 1.21 eV 4.30 × 10−2 nm = 2.88 eV − 1.21 eV W = 1.67 eV The work function of the surface is 1.67 eV. 7. λ = 410 nm = 4.10  10 –7 m p = ? h p = λ

=

6.63 × 10

−34

J⋅s

−7

4.10 × 10 m

p = 1.62 × 10 −27 kg ⋅ m/s The momentum is 1.62 –10 8. λ = 7.5  10 m E = ?



10 –27 kg·m/s.

EK =


1 2

mv

2

=

p

2

2m


Since p =

h λ

, 2

 h   λ    E = 2m

2

 6.63 ×10 −34 J ⋅ s   7.5 × 10−10 m   = −31 2 (9.11 ×10 kg ) = 4.29 × 10 −19 J E = 2.68 eV 9.

The energy required is 2.68 eV. 4 4 If the electron is accelerated through a potential difference of 1.50  10 V, it will gain 1.50  10 eV. 4 4 –19 –15 1.50  10 eV = (1.50  10 eV)(1.60  10 J/eV) = 2.40  10 J E K =

p 2 2m

p =

2 EK m

Plugging this value for p into the equation: h λ =

p h

=

2 EK m 6.63 × 10

=

−34

J ⋅s

( 2 ) (2.40 ×10 −15 J )(9.11 ×10 −31 kg )

λ = 1.00 ×10

−11

m The de Broglie wavelength is 1.00  10 –11 m. 10. v1 = 1.00  106 m/s 6 v2 = 0.50  10 m/s λ = ? First we must calculate the energy of the photon: 1 1 E p = ∆ EK = mv12 − mv2 2 2 2 1 = m (v12 − v2 2 ) 2 2 2 1 = (9.11× 10 −31 kg )  (1.00 ×10 6 m/s ) − (0.50 × 10 6 m/s )    2 E p

= 3.42 × 10−19 J

To calculate wavelength: λ =

hc E p

(6.63 ×10− J ⋅ s )(3.00 ×10 34

=

8

m/s )

3.42 × 10−19 J −7

λ = 5.82 × 10 m

The wavelength of the photon is 5.82


10 –7 m, or 582 nm.




11. (a) v = 2.52  106 m/s λ = ? First we must calculate momentum: p = mv

= (9.11× 10−31 kg )(2.52 × 10 6 m/s ) p = 2.30 × 10−24 kg ⋅ m/s To calculate wavelength: h λ =

=

p 6.63 × 10 2.30 × 10

λ = 2.88 × 10

−34

−24

J ⋅s

kg ⋅ m/s

−10

m –10 The wavelength of the photon is 2.88  10 m. (b) λ = ? h Since for an electron, λ = , the value will be 2.88 p



–10

10

m, the same as in part (a).

12. In the case of light, wave theory predicts the interference pattern if light is diffracted through a series of adjacent, equally spaced slits (see Section 10.2). When X rays (a form of electromagnetic wave) are diffracted by equally spaced atoms in a lattice structure in thin crystals, the interference pattern yields a circular pattern as illustrated in Figure 5(a), page 615. In other words wave theory predicts the interference pattern observed. De Broglie predicts that particles have a wavelength that will predict their behaviour. When electrons are directed at a equalled spaced atoms in the lattice structure of a crystal of nickel, the electrons are diffracted into a similar pattern to that produce by X rays in the same questions, as predicted by the de Broglie wavelength. In ot her words matter wave theory predicts the interference pattern observed. 13. Rutherford’s model suggests a continuous spectrum of light frequencies would be emitted as the spiralling electron collapsed into the nucleus. However, the observed spectrum consists of a distinct number of discrete frequencies only, with no evidence of atomic collapse. 14. To analyze the composition of a gas, an absorption spectrum is used. Since the light arriving from Mars or Venus must pass through Earth’s atmosphere, no conclusive results are possible since Earth’s atmosphere is 20% oxygen. 15. We don not observe any other series because the Lyman series of lines are in the ultraviolet section of the electromagnetic spectrum. 16. Since the electron is a circular orbit, it has centripetal acceleration. According to Maxwell, an accelerated charge moving at a constant speed should emit a continuous electromagnetic wave. 17. The energy of their photons is sufficient to ionize moving atoms, causing damage to constituents of matter, (e.g., human cells). 18. To understand a specific experiment, one must use either the wave or the photon theory, but not both. 19.

Water Waves low speed large slits interference patterns spread over a large surface diffraction is easy to utilize

Electrons high speed very small slits (spaces between atoms) interference pattern very small diffraction is difficult, very short wavelengths

Classical Mechanics positron, velocity, and momentum discrete location determined by classical laws of mechanics mass finite

Quantum Mechanics position, velocity, and momentum uncertain location predicted by wave properties mass interconvertable with energy

20.



=

4

3 . π r 1 3 2 In first excited state, r 2 = 2 r 1 = 4r 1, therefore: 4 V2 = π r 23 3 4 = π ( 4r 1 )3 3

21. In ground state, V1

V2

= 43V 1 = 64 V 1

The volume is greater by a factor of 64. −13.6 eV   22. Bohr’s relationship  E n =  yields the same energy levels for the hydrogen atom those discovered n2   experimentally for hydrogen gas excited by a high electric potential. Calculations of the wavelengths of the emitted wavelengths produce exactly the same values predicted mathematically by Bohr. 23. As the electron moves from one Bohr orbit to another, with a higher energy level: 1   (a) the speed will increase  vn ∝  n  

( ∝ n2 )

(b) the orbital radius will increase vn

 

(c) the energy will decrease in magnitude  E n =

−13.6 eV   , but increase in value relative to the ground state (–13.6 eV) n2 

(d) the de Broglie wavelength will decrease, since the energy has increased. 24. rn = 5.3 × 10 −11 n 2 m r 2 = (5.3 × 10 −11 m ) 2 2 r 2

= 2.1× 10 −10 m

The radius of the second excited state of the hydrogen atom is 2.1 2 25. intensity = 1.0 kW/m 2 –7 (a) λ = 5.50  10 nm = 5.50  10 m number of photons = ? intensity = 1.0 kW/m2

=



–10

10

m.

1000 W 104 cm 2

= 1.0 ×10−1 W/cm 2 2 −1 intensity = 1.0 × 10 J/cm ⋅ s First we must calculate the energy of the photon: hc E p = λ

(6.63 ×10− J ⋅ s )(3.00 ×10 34

= E p

= 3.60 × 10

8

m/s )

−7

5.50 × 10 m −19

J

To calculate the number of photons: number of photons

=

−

1.0 × 10 1 J/cm 2 ⋅ m 3.6 ×10

−19

J/photon

number of photons = 2.8 × 10 photons/cm 2 ⋅ m 17

The number of photons striking the area per second is equal to 2.8




17

10 .


(b) There are 2.8  1017 photons in a box with dimensions of 1 cm  1 cm  3.00  108 m. Therefore, the number of photons in a thimble with dimensions of 1 cm  1 cm  1 cm is: 2.8 × 1017 photons

= 9.3× 108 photons/cm 3

3.00 × 10 cm 26. Applying the principle of uncertainty to an electron orbiting a nucleus we say that there is uncertainty in measuring the electron’s position and velocity. Thus, it becomes impossible to say for any individual electron where it is now or where it will be at any future time. Rather than saying where it is, we must be satisfied with describing its location by stating the probability that it will be found near any point. One way of visualizing this si tuation is to think of the electron e xisting as a cloud of negative charge distributed around the atom, rather than as a particle moving in a circular orbit. The cloud is denser in areas of high probability and less dense in areas where the electron is less likely to be found. h 27. An object can never be truly at rest. The equation ∆ x∆p ≥ indicates that there is uncertainty in any measurement of 2π position, speed, and momentum. If the positio n is uncertain, then the object can never be at rest, at l east as we know it in classical mechanics. In reality, for macroscopic objects the uncertainty is so small that we do not notice it in normal, everyday life. 28. 8

3

14

29. (a) If x is f = 7.5 × 10 Hz on the graph, the y value will be 2.7 eV. In other words the kinetic energy of the photoelectrons is 2.7 eV. To reduce the current to zero a retarding potential of 2.7 eV would be required. 14 14 (b) The y-intercept is 2.7  10 Hz. Thus, light with a minimum frequency of 2.7  10 Hz is required before the work function is exceeded. (c) When the graph is extrapolated back to the y-axis, the value of 1.0 eV is obtained. This represents the work function. rise (d) slope = run 3.0 eV = 14 (1.0 − 2.7) × 10 Hz =

3.0 eV 7.3 × 1014 Hz

= 4.109 ×10

−15

eV

slope = 6.58 × 10 −34 J ⋅ s, or 6.6 × 10−34 J ⋅ s –34

The value for Planck’s constant is 6.6  10 J·s. 30. diffraction grating = 5000 lines/cm 1.00 cm w= = 2.00 × 10−4 cm = 2.00 × 10 −6 m 5000 lines r = 30.0 cm = 30.0  10 –2 m AB = 16.4 cm nλ sin θ n = w sin θ 1


=

λ

w


Since sin θ 1

=

x L

, x =

1 2

–2

–2

(16.4  10 m), and L = r = 30.0  10 m, therefore, λ =

=

xw L 1

16.4 × 10 − m )( 2.0 × 10 − m ) ( 2 2

6

30.0 × 10 −2 m

−7

λ = 5.47 × 10 m

The wavelength of the first-order green line of mercury is 5.47

–7

10 m, or 547 nm.



Making Connections 31. (a) When the sun's energy strikes a photoelectric surface, the energy 'knocks' electrons loose, allowing them to flow freely. Photovoltaic cells also all have one or more electric fields that act to force electrons freed by light absorption to flow in a certain direction. This flow of electrons is a current, and by placing metal contacts on the top and bottom of the cell, we can draw that current off to use externally. Topics could include: • electron structure of silicon • crystalline silicon and conductivity • doping, n-type, p-type, junction • photon action in the silicon (b) Unfortunately, the most a simple photovoltaic cell could absorb is around 25 percent, and more likely it is 15 percent or less. Since the light that hits the cell has photons of a wide range of energies, it turns out that some of them won't have enough energy to be useful and they pass through the cell as if it were transparent. Still other photons have too much energy. Only a certain amount of energy, measured in electron volts and defined by the cell material (about 1.1 eV for crystalline silicon), is required to knock an electron loose. If a photon has more energy than the required amount, then the extra energy is lost. These two effects alone account for the loss of around 70 percent of the radiation energy incident on our cell. There are other losses as well. Electrons have to flow from one side of the cell to the other through an external circuit. We can cover the bottom with a metal, allowing for good conduction, but if the top is covered, then photons can't get through the opaque conductor and we could lose much of the current. If the contacts are only at the sides of the cell, then the electrons have to travel an extremely long distance to reach the contacts. (Silicon is a semiconductor. It’s not nearly as good as a metal for transporting current and its internal resistance is fairly high—high resistance means high losses). To minimize these losses, a metallic contact grid that shortens the distance that electrons have to travel while covering only a small part of the cell surface covers the cell. Even then, the grid, which can't be too small, blocks some photons. The cell can't use photons that are reflected. For that reason, an anti-reflective thin film coating is applied to the top of the cell to reduce reflection losses to less than 5 percent. (c) Advantages • nonpolluting • continuously recycling energy source • uses free energy from the sun Disadvantages • expensive • low efficiency • solar panels take up large areas • low power output (d) On Earth • remote location power (e.g., weather sensors, emergency telephones, Third World) • low-power sensing devices • photovoltaic applications (security devices, lamp switches) In space • telescope power (Hubble) • powering television and telephone satellites • space station



(e) If the efficiency improves, solar cells could be used for supplementary power in a home, backup power for an electrical grid and in many remote locations requiring power, (Northern Canada, Third World) 6 32. Moving neutrons have de Broglie wavelengths. Fast moving ones (4  10 m/s) have de Broglie wavelengths of approximately 1.0  1013 m, while slow moving ones have wavelengths of approximately 2.0  10 –10 m. Since a typical nuclear dimension is 10 –13 m, we can see that the diffraction behaviour will change. There are other effects too. If we consider head-on collisions between moving particles and stationary targets we can use the equation to derive an expression for the fractional change in speed as follows: v1′ v1

m1 − m2

=

m1 + m2

For neutrons striking carbon-12 nuclei: v1′ v1

1 −12

=

1 + 12

=−

11 13

The minus sign indicates a reversal. Thus, the incident neutron moves on, after the collision, with

11

or 85% of the 13 original velocity. In order for the speed to be reduced by a factor of 2000, the number of collisions required would be: n

 11  = 1  13  2000   n  13  2000 =    11  n log

13 11

= log 2000

log2000 13 log 11 n = 46 This is what occurs in the nuclear reactor that has a graphite core. However, if the fast moving neutron strikes a deuterium nucleus (in heavy water) it takes only about seven such collisions to reduce the speed by the required factor of 2000. 33. Terms from science that have entered everyday language include: refractory, reflecting, echoing, mirroring, accelerating, impulse, virtual imagery (computer), inertia, turbulence, momentum, and transmutation. n=

Extension 34.

1.6  10 ° –22 p1 = 1.2  10 kg·m/s –4 θ 2 = 4.0  10 ° p2 = ? θ 1 =

Since sin θ1

–4

=

λ 1

w

, and λ 1

=

h p

, h sin θ 1

=

w=

p1 w h p1 sin θ 1

(Equation 1)

h


sin θ 2

=

p2

=

p2 w h w sin θ 2

(Equation 2)


Substituting Equation 1 into Equation 2: p2

=

= = p2 A magnitude of 4.8 35.

λ e λ p

h

 h   p sin θ  sin θ 2 1   1 p1 sin θ 1 sin θ 2

(1.2 ×10−

22

kg ⋅ m/s )(sin1.6 ×10 −40 )

sin 4.0 × 10

−23

kg ⋅ m/s

= 4.8 × 10−23 kg ⋅ m/s

10 –23 kg·m/s is necessary.



=?

Since E Ke = E Kp, 1 2

me ve 2 v p

2

ve

2

v p ve Since λ p

=

h m p vp

and λ e

=

h meve

=

1

=

me

2

mp vp 2

m p

=

me mp

, h λ e λ p

=

mev e h m p vp

= = = λ e λ p

m p vp meve m p

me

me

mp

1.67 × 10 −27 kg

9.11× 10 −31 kg

9.11× 10 −31 kg 1.67 ×10 −27 kg

= 2.33 × 102

The ratio of the de Broglie wavelengths is 2.33 6 36. v = 8.75  10 m/s –6 d = 5.0  10 m L = 1.0 m ∆ x = ?

102:1.



First we must calculate wavelength: λ =

=

h mv 6.63 × 10−34 J ⋅ s

(1.67 ×10−

λ = 4.537 × 10


27

kg )(8.75 × 10 6 m/s )

−14

m


To calculate the distance apart: λ  ∆ x = L    d   4.537 ×10 −14 m  = 1.0 m   −6  5.0 ×10 m  ∆ x = 9.07 ×10 −9 m, or 9.1× 10−9 m

The interference maxima will be 9.1  10 –9 m apart. 37. v = 0.05c = 1.5  107 m/s λ photon λ particle

=? h

λ particle

=

λ photon

=

λ photon

=

=

p

h mv

hc E p

But E p = E K, therefore, hc 1 2

λ photon λ particle λ photon λ particle

mv 2

   hc  1   mv 2     2 =  h   mv    =

c 1 2

v2

Since v = 0.05c, λ photon λ particle

=

c 1 2

=

c 1 2

=

1

λ particle

2

( 0.05)

c2

1 1 2

λ photon

2

( 0.05c ) c

2

=

v2

2

( 0.05)

c

= 2.7 × 10−6

The ratio of the photon wavelength to the de Broglie wavelength is 2.7 × 10 :1. 38. This is an order of magnitude question. Forgetting about the spaces between the spheres, you need to consider that you need a volume 1081 bigger than one proton so you need a radius of 10 27 times bigger. 10 –15 × 1027 = 1012, so we need a container that has a radius of approximately 10 12 m. (The empty voids would put in an error of two in the volume or about 30% in radius.) –6



CHAPTER 13 RADIOACTIVITY AND ELEMENTARY PARTICLES Reflect on Your Learning (Page 664)

1. 2.

3.

4.

5. 6.

Radioactivity refers to reactions that take place at the atomic level in which either α , β , or γ particles are emitted. These reactions can be detected using radio receivers. Living material tends to absorb small quantities of a radioactive form of carbon. When the material dies the radioactivity declines at a known rate. By comparing the present level of activity to the amount thought to exist when the material was alive the archaeologist can measure the age of the material. (a) The forces are the electromagnetic force, the gravitational force, the weak nuclear force, and the strong nuclear force. (b) It is assumed that when the golf club gets very close to the ball the charged particles—primarily the electrons—in one object tend to strongly repel the like-charged objects in the other object. This means that the force is the electromagnetic force. (c) The strong nuclear force is assumed to hold the nucleus together. (a) Physics sees a symmetry in particles that ensures that each type of particle has, associated with it, a corresponding antiparticle. The antiparticle is similar to the particle in some ways, particularly mass, however there are fundamental differences. For example, the anti-electron (also called a positron) has a positive charge. (b) When a particle collides with its corresponding antiparticle the reaction is called annihilation. In this reaction the matter transforms to energy. The two particles produce two photons, typically with very high energies, depending upon the initial kinetic energy and rest mass of the original particles. The four days is equal to two half-lives. One-half remains after one half-life and half of that amount remains after the second half-life. Overall, one-quarter of the original remains after two half-lives. Elementary particles are considered fundamental in that they cannot be subdivided into simpler particles. These are classified according to some well known properties as well as some lesser known properties that will be dealt with in this chapter. The well known properties include mass and charge. The lesser known properties are known as spin, strangeness, and baryon number.

13.1 RADIATION AND RADIOACTIVE DECAY PRACTICE (Page 673)

Understanding Concepts 1. (a) According to the periodic table, x = 82. In order to determine y, we must determine the values for the superscript a and the subscript b by balancing the equation: 212 82

212 83

Pb →

Bi + ba y

212 = 212 + a a=0 82 = 83 + b b = −1 0

The particle is designated as −1 y . Based on the values of N and Z, this is a beta particle. Therefore, 0 1

y = 212 82

−

e

212 83

Pb →

Bi +

0 1

−

e

(b) To determine x and y, we must balance the equation: 214 = x + 0 x = 214 83 = y + ( −1) y = 84 Therefore,

214 83

Bi


214 → 84

Po +

0 1

−

e.

Chapter 13 Radioactivity and Elementary Particles 715

(c) To determine x and y, we must balance the equation: x = 222 + 4 x = 226 y = 86 + 2 y = 88 Therefore, (d)

215 84

Po →

226 88

211 82

Ra

→

222 86

Rn + 24 He .

Pb + ba x

To determine x, we must determine the value of a and b by balancing the equation: 215 = 211 + a a

=4

84 = 82 + b b=2 From the periodic table we know that x is He, therefore (e)

3 1

215 84

Po →

211 82

Pb + 24 He .

→ ba x + γ

H

To determine x, we must determine the value of a and b by balancing the equation: 3= a+0 a

=3

1= b+0 b =1 From the periodic table, we know that x is H. Therefore, 31 H

→ 31 H + γ .



Type of Emission α

Mass (kg) 6.68



β

9.31



2. (a)

222 86

Rn

–27

+2

–31

β negative + β positive

10 10

kg kg

0

γ

→ 21884 Po +

Charge

–

0

Speed up to 6.67

7

6.67  10 m/s 8 to 3  10 m/s 8

7

10 m/s



3  10 m/s

Penetrating Ability thin paper (5 cm of air) thin metal (3–6 mm of aluminum) thick metal (30 cm of lead)

Ionization Ability low moderate

low

a b

x 222 = 218 + a a

=4

86 = 84 + b Therefore, x =

4 2

b=2 He , α decay.



(b)

141 57

La

→ 141 58 Ce +

a b

x 141 = 141 + a a

=0

57 = 58 + b Therefore, x = (c)

238 92

U→

0 −1

b = −1 e , β decay.

238 90

Th + ba x 238 = 234 + a a

=4

92 = 90 + b Therefore, x = (d)

141 56

Ba

4 2

b=2 He , α decay.

→ 141 La + 57

a b

x 141 = 141 + a a

=0

56 = 57 + b Therefore, x = (e)

35 17

Cl →

35 18

0 −1


Ar + ba x 35 = 35 + a x = 0 17 = 18 + b

Therefore, x = (f)

212 82

Pb →

212 83

0 −1


Bi + ba x 212 = 212 + a a

=0

82 = 83 + b Therefore, x = (g)

226 88

Ra

0 −1


→ 22286 Rn +

a b

x 226 = 222 + a a

=4

88 = 86 + b Therefore, x = (h)

215 84

Po →

211 82

4 2

b=2 He , α decay.

Pb + ba x 215 = 211 + a a

=4

84 = 82 + b Therefore, x =


4 2

b=2 He , α decay.


3. (a)

212 x

Pb →

212 83

a

Bi + b y

From the periodic table, x = 82. For y: 212 = 212 + a a

=0

82 = 83 + b b = −1 y = Therefore,

212 82

Pb →

212 83

0

−1 e

Bi +

0

−1 e

(b) 214 = x + 0 x = 214 83 = y + ( −1) y = 84 Therefore, 214 Bi → 83

214 84

0

Po +

−1 e .

(c) x = 222 + 6 x = 226 y = 86 + 2 y = 88 Therefore, (d)

215 84

Po →

226 88

211 82

Ra

→

222 86

Rn + 24 He .

Pb + ab x 215 = 211 + a a

=4

84 = 82 + b b=2 From the periodic table, x = 42 He . Therefore, 215 Po → 84

211 82

Pb + 24 He .

(e) 3 = a + 0 a = 3 1= b+0 b =1 From the periodic table, x = 31 H , therefore 31 H → 31 H + γ . (f) 141 = x + 0 x = 141 Therefore, 141 Ce → 58

141 59

Pr +

0

−1 e .

4. (a) The average binding energy per nucleon increases toward a maximum at about Z = 56. From there it declines again. This means that as Z increases beyond approximately 56, less and less energy is required to remove a nucleon from the nucleus. Thus, larger nuclei tend to be more unstable. (b) Because nuclei become increasingly unstable as Z increases, the elements with large atomic numbers have a tendency to decay through radioactive emission. 5. In alpha decay only one particle is produced. Obviously this carries off the released energy. In β decay, on the other hand, two particles are produced—the β particle and the neutrino, or antineutrino. In this instance , collectively, the two emitted particles carry off the released energy. The requirement i s that the sum is constant, not that any one particle should have a fixed proportion of the total. Because of this the emitted β particle can have an energy anywhere from practically 0% to practically 100% of the total energy released.



6.

n = 6 z = 14 mC-14 = 14.003242 number of neutrons mass of parent

14 – 6 = 8 14.003242 u

Daughter Products: 6 protons 8 neutrons total mass

mass difference

6 × 1.007276 u = 6.043656 u 8 × 1.008665 u = 8.069320 u 14.112976 u 14.112976 – 14.003242 = 0.109734 u

( ∆m )

E = ∆mc2

 931.4 MeV/c 2   = 102 MeV 1u  

(0.109734 u) c 2  Carbon-14 has 14 nucleons, so: E

=

102.2MeV

nucleon 14 The total binding energy is 102 MeV, or 1.02 6 7.30  10 eV. 7. (a) 146 C → ba x + −01 e + ν

= 7.30 MeV 108 eV. The average binding energy per nucleon is 7.30 MeV, or



14 = a + 0 a

= 14

6 = b + (−1) b=7 Using the values of N and Z, the unknown product is identified as nitrogen. Therefore, 14 C → 147 N + −01 e + ν 6 (b) The following masses are obtained from the periodic table. C-14 mass = 14.003242 u N-14 mass = 14.003074 u ∆m = (mass of C-14) − (mass of N-14)

= 14.003242 u − 14.003074 u ∆m = 0.000168 u (Note that with β – decay it is customary to use the masses of the non-ionized materials and to thus neglect the mass of the emitted β particle.) E

= mc2

 931.4 MeV/c 2  = (0.000168 u )c   1u   5 E = 0.156 MeV, or 1.56 × 10 eV 2

5

8.

The amount of energy released in the reaction is 1.56  10 eV. Radioactive decays release the binding energy as kinetic energy. Some of this kinetic energy will be transformed back to thermal energy in the substance.


One example of a device is a smoke detector, which contains a small quantity of a low-activity substance. The radioactive particles emitted ionize the air between two electrodes and permit a small current to flow in a circuit. Traces of smoke particles can interrupt this current. The circuit, in turn, can sense this interr uption and trigger an alarm. Another example is clock and watch dials, which contain traces of a low-level radioactive substance that emits photons of visible light. Three other examples of common household products are cosmetics, nonstick pans, and computer disks.



13.2 RATE OF RADIOACTIVE DECAY Try This Activity: Modelling Radioactive Decay (Page 677)

It is reasonable to assume that approximately one half of the coins will “decay” each time. The following data are reasonable:

Number of Trials Number of Coins

0

1

2

3

4

5

6

40

20

10

5

2

1

0

The graph shows the data. A smooth curve has been drawn through the points.

PRACTICE (Page 678)

Understanding Concepts 1. (a) Note the data point (10 000, 240) shown in the graph below. This indicates that the amount remaining after 10 000 years have passed is 240 µ g. (b) Note the data point (8000, 300) shown in the graph below. This indicates that the amount remaining after 8000 years have passed is 300 µ g.



2. (a) The table below lists the amount remaining after each number of half-lives.

0

1

2

3

4

5

6

0

1600

3200

4800

6400

8000

9600

0.005

0.0025

0.001 25

Number of Half-Lives Time (a) Amount (g)

0.000 625 0.000 313

0.000 156

7.81×10

-05

The graph shows the data. A smooth curve has been interpolated between the points.

(b) The data point (5000, 0.000 57) shows that the amount remaining after 5000 years is 5.7 × 10 g. (c) The data point (3700, 0.001) shows that when the amount remaining is 0.001 g, the elapsed time is 3700 years. –4


According to the mathematical model the amount will never be zero. As time progresses the graph approaches the horizontal axis. In mathematical terms you can say that the horizontal axis is a hoizontal asymptote. In practical terms the 1 fraction of the amount remaining after 6 half-lives is just of the original amount. After 10 half-lives (about 60 000 a) 64 the fraction remaining is approximately

1

of the original. Given that the whole of human existence spans a period of 1000 about 1 million years and also given the incredible number of atoms present in even a small sample, there will always be a some finite number of atoms present in a sample. The fraction of the original amount may, however, be an exceedingly small number.

PRACTICE (Page 680)

Understanding Concepts 4. N 0 = 5.0 mg t 1/2 = 29.1 a (a) t = 5 a

 1  t N = N 0    2 

t

1/ 2

5

1  29.1 = 5.0    2  N = 4.4 mg The amount of the sample remaining after 5 a is 4.4 mg.



(b) N = 0.1 mg t = ?

 1  t N = N 0    2 

t

1/ 2

t

 1  29.1 0.1 = 5.0    2  t

1  29.1 =   5.0  2   0.1  = t log  1  log      5.0  29.1  2  0.1

t =

29.1 log 0.02

 1    2 

log  t = 164 a The time required is 164 a. 5. (a) t 1/2 = 193 h t = 2.00 h 2 A0 = 5.00 × 10 µ g A = ?

 1  t N = N 0    2 

t

1/ 2

2.00

1  193 = 5.00 × 10 2    2  N = 496 µ g

The amount remaining after 2.00 h is 496 µ g. (b) t = 1 a 365 d 24 h × = 8760 h t 1/2 = 193 a = 193 a = a d N N 0

= ?

 1  t N = N 0    2 

t

1/ 2

t

N N 0

 1  t 1/ 2

=    2 

8760

 1  193

=    2  45.4 1   =    2  N

= 2−45.4 , or 2−45

N 0 The fraction of the original amount that remains after 1 a is approximately 2 –45, or 2.8  10 –14. (c) t 1/2 = 193 a t=? 1 N The fraction remaining is . This is . N 0 100



 1  t N = N 0    2 

t

1/ 2

t

1  193 =   100  2  1

t

 1  = log  1 193 log   2  100     1  = t log  1  log      100  193  2   1  193 log    100  t =  1  log    2  t = 1280 h 1

The time required for the amount to be

of the original quantity is 1280 h. 100 6. The lower half-life of I-131 indicates that the process of decay proceeds more quickly. In any given period of time atoms of I-131 have a higher probability of decay than do atoms of C-14. The sample of I-131 will thus register a higher activity. Note that chemistry students may notice a complicatin g factor: I-131 has a higher atomic mass. Since the two samples have the same mass then the C-14 sample actually has more radioactive atoms than does the I-131. If the samples actually had the same half-life then the C-14 would register higher. The difference in half-lives (when converted to hours the ratio 7 is 5 × 10 to 131) is so pronounced that this effect would be negligible here.

PRACTICE (Page 683)

Understanding Concepts 7. N = 0.25 µ g N 0 = 2.00 µ g t 1/2 = 5730 a t = ?

 1  t N = N 0    2 

t

1/ 2

t

 1  5730 0.25 = 2.00    2  t

1  5730 =   2.00  2  0.25

t

 0.25  = log  1 5730 log   2  2.00     0.25  = t log  1  log      2.00  5730  2   0.25  5730log    2.00  t =  1  log    2  4 t = 1.7 × 10 a

4 The time required for the amount remaining to drop to 0.25 µ g is 1.7 × 10 a. This is the approximate age of the sample.



8. A = 7.13 Bq A0 = 8.00 Bq t 1/2 = 5730 a t = ? t

 1  t 1/ 2

A = A0 

  2 

t

 1  5730 7.13 = 8.00    2  t

1  5730 =   8.00  2  7.13

t

 7.13  = log  1  5730 log   2  8.00     7.13  = t log  1  log      8.00  5730  2   7.13  5730log   8.00   t =  1  log    2  t = 9.5 × 10 2 a The age of the sample is approximately 9.5 9. N = 705 mg N 0 = 772 mg 9 t 1/2 = 4.468 × 10 t = ?



 1  t N = N 0    2 

2

10 a.

t

1/ 2

t

 1  4 .4 68×1 0 705 = 772    2 

9

t

1  4 .4 68×1 0 =   772  2  705

 705 

9

t

 1  4 .4 68×1 09

 = log  2   772    t  705  = 1  log  log    9  772  4.468 ×10 2 705 (4.468 ×109 ) log  772     t =  1  log    2  t = 5.9 × 108 a 8 The age of the sample is approximately 5.9 × 10 a. log 

10. When new, the knife would have contained only trace amounts of C-14. Dating the knife directly using C-14 would be impossible. A trained archaeologist might be able to determine whether the knife was consistent with the same culture as was the parchment. If this was the case, the parchment would be able to date the knife indirectly.



PRACTICE (Page 684)


238 92

U→

234 90

Th + 24 He 234 90

Th

→

234 91

Pa +

0

− 1e 234 91

Pa

→

234 92

0

U+

234 92

U→ 238 92

12. Using the chart, there is only one path from

−1 e 230 90

Th + 24 He

U to

214 83

Bi . From here there are two paths to the final product, thus there

are two possible unique ways by which a stable daughter can be produced.


Understanding Concepts 1. (a) A0 = 3.8 × 108 Bq 9 t 1/2 = 4.5 × 10 a 6 t = 1.00 × 10 a A = ?

 1  t A = A0    2 

t

1/ 2

1 .0 0×1 06

1  4 .5×1 0 = (3.8 × 108 )    2  A = 3.8 × 108 Bq 6 8 After 1.00 × 10 a the activity is 3.8 × 10 Bq. 9

(b) N = 0.100 mg N 0 = 1.00 mg 9 t 1/2 = 4.5 × 10 a t = ?

 1  t N = N 0    2 

t

1/2

t

 1  4 .5×1 0 0.100 = 1.00    2 

9

t

0.100 1.00

1  4 .5×1 0 =    2 

9

t

 0.100  = log  1  4 .5× 1 0 log   2  1.00     0.100  = t log  1  log      1.00  4.5 ×109  2  0.100 (4.5 × 109 ) log  1.00     t = 1   log    2  10 t = 1.5 × 10 a 9

The time required for the amount to decrease to 0.100 mg is 1.5 × 10 a.


10


2. A0 = 1.20 × 10 Bq 3 A = 1.00 × 10 Bq t = 1 h t 1/2 = ? 3

 1  t A = A0    2 

t

1/ 2

1

1.00 × 10

3

1.00 × 103 1.20 × 103

= 1.20 ×10

 1  t  2   

1/ 2

3

1

 1  t 1/ 2

=    2 

1

 1.00  = log  1  t log   2  1.20     1.00  = 1 log  1  log      1.20  t 1/ 2  2   1  log    2  t 1/ 2 =  1.00  log    1.20  t 1/ 2 = 3.8 h 1/ 2

3.

The half-life of sample X is 3.8 h. (Note that the time was given as “one hour.” This could be interpreted two ways with regard to the number of significant digits. The number could be assumed to be exact and, as such, not affect the answer. This was what was assumed. Alternatively the number could be taken as having one significant digit. If this were the case then the answer would be expressed as 4 h.) 9 t 1/2 = 4.468 × 10 a (from Appendix C) If 9.55% of the material is consistent with daughter products then you can infer that 100% – 9.55% = 90.45% of the N = 0.9045 . original atomic material remains. In other words N = 0.9045 N 0, which can be written as N 0

 1  t N = N 0    2 

t

1/ 2

t

N N 0

 1  t 1/ 2

=    2 

t

 1  4 .4 68×1 09

0.9045 = 

  2 

t

 1  4 .4 68×1 0 log ( 0.9045 ) = log    2 

9

log ( 0.9045 ) = t =

t 4.468 ×10

9

 1    2 

log 

4.468 ×109 log (0.9045 )

 1    2 

log  t = 6.47 × 108 a The age of the specimen is 6.47 × 108 a.



4. A = 4.00 Bq A0 = 18.0 Bq t 1/2 = 5730 a t = ? t

 1  t 1/ 2

A = A0 

  2 

t

 1  5730 4.00 = 18.0    2  t

4.00 18.0

1  5730 =    2  t

 4.00  = log  1 5730 log   2  18.0     4.00  = t log  1  log      18.0  5730  2   4.00  5730log   18.0   t =  1  log    2  t = 1.24 × 104 a 4 The age of the fragment is 1.24 × 10 a. 5. 6.

The transuranium elements have very short lives. They may have once existed, however they decayed very quickly afterwards and therefore these elements are not presently found in minerals. The clay would never have contained more than a trace of C-14. Chemical analysis might reveal the presence of another radioactive substance in the clay. If this were the case then radiometric dating might still be possible. Alternatives to radiometric dating would include reliance on other archaeological evidence. For example, it might be possible to compare the tablet to another, similar one, whose age is known.



It was likely that the dating process was based on the age of the rocks that the fossils were found in rather than using radiocarbon dating. This may have been done in absolute terms using radioactive substances present in the rocks. It may alternatively have been arrived at indirectly using relative dating based on the geologic principle of superposition, which states that older rocks tend to be found below newer rocks (once the effects of folding have been accounted for). Suppose, for example, the fossils were found in rock layer B, which was between layers A and C. Suppose further that the ages of the rocks in layers A and C were known. It would be possible to interpolate the age of the sample to be between that of the rocks of layers A and C.



9.

The radiology department of a local hospital would likely be able to supply approximate costs. These should include several factors such as the purchase cost, the annual maintenance, the costs of housing the device, and the human resources costs. If possible, ask the department to supply the typical cost of one such scan, with all factors included. 10. Look for general procedures used in the mining and refining process. A Web search will yield sufficient information in this respect. Ask in particular about the steps taken to protect the workers from the effects of the radiation. These include the use of shielding around equipment, protective clothing, and special badges that measure the amount of radiation a worker has received in any particular time.

13.3 WORKING WITH PARTICLES Explore an Issue: Funding Research on Elementary Particles (Page 692)

Understanding the Issue 1. Many of the sites, such as the one at Fermilab, have interactive tutorials that deal with other learning outcomes covered in Chapter 13. It is a good idea to browse these tutorials. 2. (a) Answers might include the following: • The Provincial Government: particularly the Provincial Department of Health since the radiation produced by synchrotrons is ideally suited to microscopy related to biological and medical research • The Federal Government: research centred around synchrotrons tends to raise the status of the country as a “world leader” in modern physics • Universities: given the costs involved, a consortium of many universities should be organized, each with research interests centered around the synchrotron • Electrical Companies: synchrotrons use massive amounts of electricity and, as such, constitute a significant customer • Municipal Governments: a large synchrotron would cross the boundaries of many municipalities, probably distributing wealth, through employment, throughout the municipalities • Private Companies: such as pharmaceutical companies, tend to make great use of the imaging capabilities of synchrotrons. As such they should be considered funding partners. (b) Sources of opposition might include: • Groups suspicious of the radiation or other potential health risks: An appropriate design, coupled with an effective public-relations campaign, should both minimize the real risks while showing the public the many benefits. • Groups who are opposed to the massive expenditure of money: An appropriate, globally-based funding plan should lessen the amount of public money that would have to be spent on the project. 3. The experience of the CLS has demonstrated two important factors that should be recommended. (a) Find and use as many funding partners as possible. (b) Include, in the funding model, provisions for the fact that synchrotrons now have important commercial applications. Include the financial benefits as part of the funding model. Form an Opinion 4. There is no “right” or “wrong” answer. The key term is “balance.” Clearly synchrotrons are extremely expensive to both build and operate. Experience has shown, however, that the device can be expected to provide important, perhaps critical, discoveries. In addition, the synchrotron has every possibility of being a device that can lead to the type of discovery that can revolutionize thinking in science. The key issue is to ensure that international co-operation is in place so that the massive costs do not need to be borne by one nation.

PRACTICE (Page 694)


As the speed of the particle approaches c, the mass increases greatly. The acceleration produced by a magnet (which alters its direction) or in an RF cavity (which alters its speed) depends upon both the charge of the particle and on its mass. This means that as the object’s relativistic mass increases, the magnetic field strength and the frequency of osculation of the electric field must be appropriately altered to ensure that accelerations act in a resonant fashion. 2. The TRIUMF is designed to produce beams of high-energy protons. Special equipment exists that can direct proton beams and cause them to collide with targets. These collisio ns release large quantities of mesons, particularly pions.



3.

When charged particles are accelerated they radiate photons. This is the cause of synchrotron radiation. The fraction of the kinetic energy radiated away depends upon the cube of the charge to mass ratio. The electron has the same (absolute value of the) charge as does the proton. Because its mass is smaller by a factor of about 2000, the fraction of the energy 3 radiated away is larger by about 2000 . 4. The operators of the CLS wish to ensure that the positive benefits to society are emphasized. As such they cannot be blamed for avoiding a term—radiation—that often car ries negative connotations in everyday language.


Understanding Concepts As the speed of the accelerating particle approaches c, its relativistic mass increases dramatically. This tends to cause the particle to fall out of synchronization with the periodic electric field, which is ca using it to increase speed. The end result is that particles in ordinary (classical) cyclotrons can only be accelerated to speeds at which the mass increase is negligible. There are two ways of dealing with this technological problem. One way is to modify the frequency across the RF cavity that causes the acceleration. Another method, one used by the TRIUMF cyclotron, is to modify the strength of the magnetic field that causes the particle to move in its circular motion. 2. The acceleration occurs in the RF cavity that exists between adjacent drift tubes. Because the particle is neutral the electric field will not accelerate it. 9 3. E initial = 8.0 GeV = 8.0  10 eV 12 E final = 1.0 TeV = 1.0 × 10 eV 6 ∆ E = 2.5MeV = 2.5 × 10 eV (for each revolution) (a) d = ? 1.

First we must look at the initial and final energies. Since E final >> E initial you can say that the amount of energy needed is E final. The number of revolutions is given by: 9

8.0 ×10 eV

n=

2.5 ×106 eV/revolution n = 3200 revolutions The distance travelled by the particle is equal to the circumference. The overall distance is the product of n and the circumference: d = nC d = n 2π r 3

= (3200)(2)(3.14)(1.0 × 10 m)

2.0 × 107 m 7 The proton travels 2.0 × 10 m while accelerating. (b) d = 2.01 × 107 m 8 v = c = 3.00 × 10 m/s t = ? d =

c=

d t

t =

d c

=

2.0 × 107 m

3.00 ×108 m/s t = 0.067 s The proton is accelerated in the synchrotron for 0.067 s.




The nature of the charge can be determined from the interaction with a magnetic field directed perpendicular to the motion. The diagram below shows one possible set-up. The × indicates that the magnetic field is coming out of the paper. As such, emitted particles moving upwards will be moving perpendicular to the field.

A right-hand rule shows that positive particles experience a force directed to the right. Similarly, negative particles experience a force directed to the left. Neutral particles experience no net force. Curved trails therefore indicate charged particles. The direction in which the particl es curve (with respect to the magnetic fi eld) can be used to determine the type of charge. 5. If you knew that the particles all had the same mass and velocity then you might just observe their behaviour in a magnetic field using a setup similar to the one in question 4. The particles with the greatest charge would be deflected the most. Unfortunately this is not feasible here since the varying masses and speeds could also account for different amounts of deflection. + A better method is to realize that the positively charged particles are most likely the results of either α or β emission. You can therefore distinguish between them based on their penetrating ability. First try some thin paper. If a detector on the other side of the paper shows no emissions then the material was only emitting α particles as these are readily stopped by paper. If the level of activity is reduced then you can say that it is emitting both, si nce the α particles are being stopped + while the β particles are being transmitted. If the level of activity is unchanged than you can say that it is only emitting β particles since these are not stopped by paper.


7.

If the mesh opening is smaller than the wavelength, the wave will tend not to pass through the mesh. Since there is negligible absorption, the energy will be reflected at the same wavelength. Waves with wavelengths between 3 m and –4 3  10 m are normally classified as microwaves. Obviously this is a rather “broad” classification. Typical mesh tends to have openings that range from about 0.5 cm to about 2 cm, which are comparable with the wavelengths of typical microwaves. For any given mesh size, say 1 cm, then microwaves with wavelengths longer than this will “see” the mesh as an opaque solid. –34 h = 6.63 × 10 J·s –27 m = 1.673 × 10 kg Owing to the enormous difference in mass that occurs at relativistic speeds, it is not feasible to directly estimate v using 1 2 EK = mv . Instead, an estimate of v = 0.9c will be used. 2 h

λ=

mv

(6.63 ×10 J s ) = (1.673 ×10 kg )(0.9 × 3.00 ×10 −34

−27

λ = 1.5 × 10

8

m/s )

−15

m –7 –7 Visible light has typical wavelengths between 4.0  10 m and 7.5  10 m. Clearly the typical wavelengths of the high-energy protons are much smaller than this value. Therefore, their resolving ability is much greater.



8.

Name cyclotron synchrocyclotron

linac synchrotron

Method for Accelerating Particles RF cavity between two “D” shaped magnetic fields RF cavity between two “D” shaped magnetic fields. Frequency of oscilation is varied to compensate for relativistic mass increase. RF cavity between drift tubes arranged in a line RF cavity between superconducting magnets

Particles Accelerated protons, electrons, α particles protons, electrons, α particles

Energies Achieved tens of MeV

protons, electrons, α particles protons, electrons, α particles, and others

50 GeV

1 GeV

> 1 TeV

9.

Experiment

Typical Particle Energies

Resolving Ability

gold foil scattering early experiments with cyclotrons expriments with synchrocyclotrons experiments with linacs tevatron experiments CREN LHC experiments

several MeV tens of MeV

>> 10 m –15 > 10 m

hundreds of MeV

10

GeV

< 10

TeV

< 10

tens of TeV

<< 10

–15

–15

m

–15

m

–15

m

–15

m

Types of Questions Addressed by Experiment What is the atom made of? What particles are in the nucleus? What particles are in the nucleus? What other particles exist? Are nuclear particles (prot on and neutron) fundamental? What particles exist? Are quarks fundamental? Are there other particles?

10. Student reports should include some of the following points: • in experiments with subatomic particles, information from the particle detectors flows to the computers • sometimes too much information is sent too rapidly • TRIUMF’s new microchips can temporarily store this information in the right sequence until a computer accepts it • technology is being transported to a firm that will manufacture the microchips for use in advanced radar, satellite communications, and similar situations where a huge volume of information must be processed very rapidly

13.4 PARTICLE INTERACTIONS PRACTICE (Page 701)




2.

PRACTICE (Page 703)


It was assumed—incorrectly as it turns out—that the meson was the carrier of the strong nuclear force. It was also assumed that the particle appeared from the quantum vacuum. Because the force was only active through the nucleus, the diameter of a typical small nucleus is a good estimate over the distance d that the particle could be expected to last. The uncertainty principle

( ∆ E )(∆t ) ≈

h

could be rearranged giving

2

hc

. The known values of h and c, as well 2π 2πd as the estimated value of d are then used to yield the value of the rest mass. 4. The particle discovered in 1936 was first thought to be Yukawa’s meson. However, closer examination of the particle’s properties revealed two reasons why the particle had to be eliminated as a contender. First, the rest mass was too low. The predicted mass of the Yukawa meson was 270 MeV and that of the newly-observed particle was only 207 MeV. Second, the particle’s interactions with matter were much too weak for it to be seriously considered to be the carrier of the strong interaction. mc

≈



The term “element” originated from an understanding of matter that saw certain substances as fundamental, or indivisible. Discoveries dating from the early part of the twentieth century led to the conclusion that these fundamental substances were, themselves, formed from particles. Originally, it was thought that there were three fundamental particles: the electron, the proton, and the neutron. Since then it has become apparent that two of these particles—the neutron and the proton—are composed of sub-particles. Clearly then, the so-called elements are not elemental. In this way the term is now outdated. 2. When dealing with particle interactions it is important to realize that much of the thinking is metaphoric. With this in mind, it is often said that the force-carrying particles are actually messenger particles. If a negatively charged particle exchanges a virtual photon with a positively charged particle then the message is “move closer.” If, on the other hand a negatively charged particle exchanges a virtual photon with another negatively charged particle then the message is “move away.” The following simple analogy is commonly used to help put attractive interactions in concrete form: the particles are exchanged by being grabbed from one another’s hands. 3. The particle is not observable using ordinary means. It is therefore virtual in the sense that its existence is indirectly inferred. 4. Under quantum theory, various interactions have associated probabilities that they will occur. These values tend to be quite low. When a compound event (first this, then this, then this, and so on) is examined, the probability that it will occur is the product of the individual probabilities of each event. Because the probability of any given photon exchange is small, the probability of a compound event that requires several of these exchanges is exceedingly small.



5. (a) Beta decay was the nuclear process. It could not be explained using the strong nuclear, electromagnetic, or gravitational forces. A fourth force, the weak nuclear force, was t herefore conceptualised. (b) The W boson is assumed to be the carrier of the weak nuclear force. In electroweak theory, which seeks to unite the electromagnetic and weak nuclear force, a second particle, the Z boson, is also a carrier. 6. (a)

(b)

7. (a) Gravity can be considered the odd one out in two important ways. First, the hypothetical charge carrier, the graviton, has never been observed. This contrasts sharply with the other three forces, in which the charge carriers have been observed. More importantly, at present there is no theory that can unite gravity with the other three forces. That is, there is no scale of activity in which gravity is indistinguishable from the other three. (b) The interactions between gravitons are quite complex, partly owing to the fact that they can interact with one another. Photons, for example, do not interact with one another and instead interact only with the types of particles that produce them. Gravitons, on the other hand, can possibly self-interact. The mathematics that must be devised to describe such situations in a probabilistic manner will have to be incredibly complex. 8. Electromagnetic (a) (b) (c) (d) (e)

Electron Positron Proton Neutron Neutrino


Weak Nuclear

Gravitational

yes

Strong Nuclear no

yes

yes

yes

no

yes

yes

yes

yes

yes

yes

no

yes

yes

yes

no

no

yes

no


9.

80.3 GeC/c2 = 80.3  109 eV/c2 2 9 2 mZ = 91.2 GeV/c = 91.2  10 eV/c d = ? mW =

For the W boson: hc

2

mW c ≈

2πd W hc

d W ≈

2π ( mW c 2 )

(6.63 × 10 J s )(3.00 × 10 m/s ) ≈ 2 (3.14 ) (80.3 × 10 eV/c )(1.6 × 10 J/eV ) −34

8

9

d W ≈

2

−19

2.5 × 10−18 m

For the Z boson: hc

2

mZ c ≈

d Z ≈

2πd Z hc

2π ( mZ c 2 )

(6.63 × 10 J s )(3.00 × 10 m/s ) ≈ 2 (3.14 ) (91.2 × 10 eV/c )(1.6 × 10 J/eV ) −34

8

9

d Z ≈

2

− 19

2.2 × 10−18 m –18

–18

The range of the weak force is 2.5  10 m to 2.2  10 m. 10. (a) A team of 100 scientists worked on the project. (b) The overall cost of the project is about $73 million, not counting the donation of about $300 million in heavy water from the Atomic Energy Commission of Canada. This funding came from the following institutions: Natural Sciences and Engineering Research Council of Canada, US Department of Energy, UK Science and Engineering Council, National Research Council of Canada, Northern Ontario Heritage Fund Corporation, Industry Canada, and Inco Limited. (c) One way in which neutrinos are detected is through the small amount of light produced when neutrinos interact with matter. A reaction unique to heavy water in which a neutrino changes a neutron to a proton and an electron can also be detected. Finally, now that salt had been added to the heavy water, a third neutrino reaction can occur in which all three types of neutrinos are found to interact. (d) To date, the research has provided evidence that the three types of neutrinos can transmute from one form to the other. Because the neutrinos can transmute they can have a small amount of mass associated with them. Note also that because neutrinos are produced in massive amounts by supernovae, the SNO is expected to be an effective monitor of such events.

13.5 THE PARTICLE ZOO PRACTICE (Page 707)

Understanding Concepts 1. (a) A gauge boson is described by gauge theory, which unites the electromagnetic force with the weak force. Gauge boson includes the photon, as well as the W and Z bosons. (b) Leptons do interact through the weak nuclear force, but do not interact through the strong nuclear force. Leptons include the electron, muon, and tau, as well as the respective neutrinos of the particles. (c) Baryons mainly interact through the strong nuclear force. Note that this classification includes a great many particles. 1 Baryons are subdivided into the mesons (small mass, spin 0) and the baryons (larger mass, spin ) 2



2.

3.

Certain particles, such as the K (kaon), the Λ (lambda) and the Σ (sigma) were found to display certain sets of behaviours. These included unexpectedly long decay rates and the fact that the particles were produced in pairs. The particles with these traits were said to have the quantity “strangeness”. Gell Mann’s work is an excellent example of the type of genius that draws from the logical left side of the brain as well as from the imaginative, intuitive, right side of the brain, at the same time. It is logical in that certain patterns and symmetries have been observed. It is also imaginative in that Gell Mann discerned the patterns from the basic data, that is, he was able to “see” the patterns in his mind’s eye.

PRACTICE (Page 709)

Understanding Concepts 4. (a) charge of u =

+

2 3 uuu

2

2

2

3 6

3

3

=+ + + =+

3 uuu = +2 The charge of uuu is +2. 2 (b) u = + 3 u =−

2 3 2 2 = + +  −   3  3  uu = 0 uu

The charge of uu is 0. 1 (c) charge of d = + 3 charge of u =

+

2 3

The charge of ddu is (d) charge of d =

−

1

charge of s =

+

1

4 3

1

1

2

3

3

3

qddu

=+ + +

qddu

=

4 3

.

3 3 1

1

3

3

ds

=− +

ds

=0

The charge of ds is 0. 5. (a) The antiproton is a baryon, with baryon number = –1 and so, must be composed of three antiquarks. The strangeness number is 0, which suggests that you should use a combination of u and d . The charge is –1. Based on Table 3 on page 708 the combination uud will work. To confirm this combi nation, we check the other quantum numbers:



 − 1  +  − 1  +  − 1  = − 1 (correct)  3  3  3        − 2  +  − 2  +  1  = − 3 = − 1 (correct)  3  3  3 3      

The baryon number is The charge is

(b) The antineutron is a baryon, with baryon number = –1, and so, it must be composed of three antiquarks. The strangeness number is 0, which suggests that you should use a combination of u and d . The charge is 0. Based on Table 3 on page 708 the combination udd will work. To confirm this combination, we check the other quantum numbers:  1  1  1 The baryon number is  −  +  −  +  −  = − 1 (correct)  3  3  3

 − 2  +  1  +  1  = 0 = 0 (correct)  3 3 3 3       0 The neutral pion (π ) is a meson and is composed of a quark-antiquark pair. The strangeness number is 0 and we are

The charge is (c)

limited to the combination of uudd . The possibilities are uu and dd . For both combinations, the charge is 0, the baryon number is 0 and the spin is 0 so, in the absence of furt her information, both combinations could for m a (π0). (d) K – kaon is an antiparticle (according to Table 2 on page 707), and a meson with spin 0, charge –1 and strangeness –1. Due to its strangeness number, one of its constituents must be s. The other particle must be d to ensure that the charge is 0. Therefore, the combination is sd .

PRACTICE (Page 711)


7.

The “chromo” in chromodynamics is indicative of the fact that a “colour” quantum number is used in the theory that describes the force interactions through gluons.

8.




Understanding Concepts 1. (a) To identify the particle, we must determine the quantum numbers. 1 1 For s: Q = − , B = , s = −1 3 3 For u :

2 1 Q = − ,B = − ,s = 0 3 3

For the combination of su : 2 +  −   = − 1 3  3  1  1  B = +  −  = 0 3  3  s = −1 + 0 = − 1 Q=−

1

The particle that has these quantum numbers is the anti-kaon ( Κ ). (b) To identify the particle, we must determine the quantum numbers. 1 For d: Q = − , B = 1, s = 0 3 –

For u :

2 Q = − , B = −1, s = 0 3

For the combination of du :

 

2 +  −  = −1  3  3 1  1  B = +  −  = 0 3  3  s = 0 + 0 = 0 Q = −

1

The particle that has these quantum numbers is the pi minus ( π – ). (c) To identify the particle, we must determine the quantum numbers. 2 1 For u: Q = ,B = ,s = 0 3 3 For d:

1 1 Q = − ,B = ,s = 0 3 3

1 1 Q = − , B = , s = −1 3 3 For the combination uds: 2  1  1 Q = +  −  +  −  = 0 3  3  3 For s:

B =

1

1

1

+ + =1

3 3 3 s = 0 + 0 + (−1) = −1

The particle that has these quantum numbers is the sigma zero ( ∑ ). (d) To identify the particle, we must determine the quantum numbers. 2 For u: Q = , B = 1, s = 0 3 0

For s:


1 Q = − , B = 1, s = −1 3


For the combination uus: 2 2  1  Q = + +  −  = + 1 3 3  3  B =

1

1

1

+ + = +1

3 3 3 s = 0 + 0 + (−1) = −1

The particle that has these quantum numbers is the sigma plus ( ∑ ). 1 It is assumed that the spins of u and d are each ± . The possible results for three quarks can therefore be given by 2 1 1 1 1 3 ± ± ± . Analysis of the possible sums give the results ± and ± . 2 2 2 2 2 A proton containing two red up quarks would contain two particles with identical sets of quantum numbers. According to the Pauli-exclusion principle this is impossible. Each type of quark comes in three colors: red, green and blue. Each type of antiquark also comes in three colours: antired, antigreen and antiblue. It is therefore assumed that the combinations (red and antired), (green and antigreen) and (blue and antiblue) are colour neutral. According to chromodynamics, the gluon is thought to be the carrier of the strong nuclear force. This was once thought to be the role of the meson. (a) Table 3 on page 708 summarizes the properties of each type of quark. (b) Presently, several important symmetries exist. First, for every quark discovered there exists experimental evidence of a corresponding antiquark. Second, the number three seems fundamental to quarks. Presently three families seem to exist. Thirdly, within each family there exists a pair of quarks. Family 1 contains u and d. Family 2 contains c and s. Family 3 contains T and B. Each member of each family now has a corresponding member in the other family, that is u, c, and T seem to correspond, as well as b, s and B seem to correspond. (a) K + is a meson with spin 0, charge +1, and strangeness +1. Since it is a meson, the K + particle is composed of a quark and an antiquark. A strangeness of +1 can only be obtained by the following quark-antiquark combinations: +

2.

3. 4.

5. 6.

7.

+

us and ds . The ds combination has a charge of 0, therefore the K particle’s quark composition is us . This combination produces a charge of +1. 1 0 0 (b) Σ is a baryon with spin , charge 0, and strangeness –1. Since it is a baryon, the Σ particle is composed of three 2 quarks. The combinations that produce a strangeness of –1 are: uus, dds, and uds . Since the uus produces a charge of +1, and dds produces a charge of –1, the of 0.

Σ0 particle’s quark composition is uds. This combination produces a charge


The diagram below shows one possible change that occurs:



Note that in this instance the gr een u quark and the blue d quark exchange a gluon. The Feynman diagram below represents the interaction.


Carbon 6 protons × 3 quarks/proton 6 neutrons × 3 quarks/neutron 6 electrons Total for carbon

= 18 = 18 = 6 = 72

Oxygen 8 protons × 3 quarks/proton 8 neutrons × 3 quarks/neutron 8 electrons Total for oxygen

= 24 = 24 = 8 = 96 × 2 atoms = 192

Total for molecule = 264 There are 264 different elementary particles in a CO 2 molecule.

13.6 CASE STUDY: ANALYZING ELEMENTARY PARTICLE TRAJECTORIES PRACTICE (Page 720)


= 12.3 cm − = 0.8 cm

l − π

s

π

(a)

q m

=? r =

l − 2 π

8 s

π

−

+

s

π

−

2 2

=

(12.3 cm )

8(0.8 cm)

+

0.8 cm 2

r = 24.0 cm, or 0.24 m



q m

=

1.9 × 108 C ⋅ m/kg r 1.9 × 10 C ⋅ m/kg 8

= q m

0.24 m

= 7.9 × 108 C/kg

The charge-to-mass ratio of the π − particle is 7.9 ×108 C/kg . (b) m = ? q m

= 7.9 × 108 C/kg

m=

=

q 7.9 × 108 C/kg 1.6 ×10

−19

C

7.9 × 10 C/kg

m = 2.0 × 10

8

−28

kg

The rest mass of the π − particle is 2.0 ×10−28 kg .

13.7 THE STANDARD MODEL AND GRAND UNIFIED THEORIES Section 13.7 Questions (Page 732)

Understanding Concepts 1. (a) Table 1 shows 6 quarks. Recall that each quark has an antiquark associated with it. This gives 12 quarks. According to quantum chromodynamics, each of these quarks can exist in one of 3 colour states. This gives a total of 24 fermions. Since each fermion has an associated antiparticle, this gives us a total of 48 fermions. (b) In part (a) it was noted that there are 48 fermions. Additionally, Table 1 lists all 6 leptons. This gives a total of 54 fermions. Furthermore, counting the 8 different types of gluons and 3 different types of bosons for the weak force there are 13 different bosons. Overall there are 54 + 13, or 67 different types of particles. 2.

Fermions Leptons Name electron electron neutrino

Charge –1 0

Quarks Mass (me) 1 ~0

Name up down

Charge 2/3 –1/3

Mass (me) 20 20

Bosons Name photon + 0 W , W , and Z bosons gluons (8 different types) graviton 3.

Spin 1 1 1 2

Force electromagnetic weak nuclear force strong nuclear force gravitational force

The standard model has been constructed to fit the available data in a similar way that a line of best fit is calculated to best fit data obtained from measurements. While it does have powerful predictive ability regarding what reactions occur, it is weak with regard to answering the question of why the events occur.



4. (a) The GUTs seek to unify the electromagnetic, strong nuclear, and weak nuclear forces. (b) These extend symmetries to the particles and forces that already exist. It is assumed that leptons and quarks are members of one class of particle. At distances below the unification scale of 10 –30 m, it is assumed that there is no distinction between the types of forces and of particles. At this point, for example, quarks can become leptons (and vice versa) through the exchange of X bosons. 14 –30 5. The X particle is massive (about 10 times heavier than a proton) and exists over a short distance (< 10 m). It therefore has an extremely small probability of coming into existence. Even if it did, it would only exist for such a short period of time that it would leave an almost imperceptibly small detection signature. 6. It is possible (although extremely unlikely) that an X boson can come into existence inside a proton. It could come close enough to be exchanged between nearby quarks. If this happened, the two quarks would produce an antiquark (which could annihilate the remaining quark) and a positron (which could annihilate an orbital electron). The end result of this is that all matter is unstable. 7. Since there is no direct evidence that supports the instability of matter, there is some doubt on the current GUTs. Additionally the GUTs omit one major interaction—gravity. As a result, some scientists seek to develop “theories of everything” with the goal of uniting all four forces. 8. The strings are assumed to be extremely small. Because they are thousands of times smaller than the unification distance, they appear as point-like particles. 9. The strings are assumed to have more than four dimensions. One version of superstring theory, for example, assumes the existence of 10 dimensions—the strings are vibrating in all 10 dimensions. Different particles are therefore represented by different vibrational modes. –30 10. Assume that the particles need to travel close to c. For the X boson, the distance is about 10 m and for the strings the –35 distance is about 10 m. For the X boson:

∆ E ∆t = ∆ E = =

h 2π h 2π∆t h

 x    c  − 6.63 ×10 34 J ⋅ s =  10−30 m  2π   8  3.00 ×10 m/s  ∆ E = 3.2 × 104 J 2π 

To convert to eV: 3.2 × 104 J 1.6 × 10

−19

J/eV

= 2.0 × 1023

eV = 2.0 × 1017 MeV

For the string:

∆ E ∆t = mc 2

= =

h 2π h 2π∆t h

 x    c  − − 6.63 × 10 34 J ⋅ s ÷ (1.6 ×10 19J/eV ) =  10−35 m  2π   8  3.00 ×10 m/s  ∆ E = 2.0 × 1028 eV 2π 

The particle energies that would have to be produced to observe the X boson is 2.0 28 2.0  10 eV.


17

10 eV. For the strings it is




11. Using the uncertainty uncertainty principle: h ∆ p∆x = 2π h ∆ p = 2π∆ x hc c∆p = 2π∆ x hc ∆ E = 2π∆ x

(6.63 ×10−

34

=

J s ) (3.00 ×10 m/s ) 8

2 (3.14 ) (1.0 ×10 −

30

m)

∆ E = 3.167 × 104 J To convert to eV: −

3.167 ×10 4 J

= 2.0 × 1023

eV 1.6 × 10 J/eV 23 17 Therefore, the unification distance corresponds to 2.0  10 eV, or 2.0  10 MeV. −19

Using de Broglie’s formula: h λ= mv h = mc hc = 2 mc hc = E hc E =

λ

=

hc 2π∆ x

(6.64 ×10− J s ) (3.00 ×10 = 2π (1.0 × 10− m ) 34

8

m/s )

30

E = 3.170 ×104 J

To convert to eV: 3.170 ×10 J 4

= 2.0 × 1023

eV 1.6 × 10 J/eV 23 17 The unification distance corresponds to a value of 2.0  10 eV, or 2.0  10 MeV. −19

CHAPTER 13 SUMMARY Make A Summary (Page 734)

The following key points should be included: • Three types of radioactivity were discovered in the late 1900’s. One type, α radiation, radiation, was explained making reference to one new force, the strong nuclear force. A second type, β radiation radiation was explained making reference to a second new force, the weak nuclear force. • New technologies resulted in the production of nuclear reacti ons with ever-increasing energies. These technologies were accompanied by new methods to both detect and analyze the reactions. The experiments gave evidence of many more particles. • The work of Dirac and others suggested that each type of particle had a corresponding antiparticle.



• So many particles were discovered that the task of classifying them became important. Scientists searched for satisfactory • • • • • •

models. One notable example was Murray Gell Mann’s eight-fold way. Further developments led to the realization that protons and neutrons were not elementary but were instead composed of other particles called quarks. Modern classification systems start with two basis types of particles: bosons (force carriers) and fermions (matter-formers). Bosons are of four basic types, one type for each force. Photons mediate the electromagnetic force, W and Z bosons the weak nuclear force, gluons the strong nuclear force, and gravitons the gravitational force. Fermions are further divided into leptons and quarks. There are six of each. Grand unified theories seek to unite the strong nuclear, weak nuclear and electromagnetic forces. Theories of everything seek to unite all four forces. So far, these tend to be mainly speculative.

CHAPTER 13 SELF QUIZ (Page 735)

True/False 1. T 2. F The average binding energy per nucleon increases for Z for Z = = 1 to about Z about Z = = 56. From that point forward, the average binding energy per nucleon decreases. 3. T 4. F α decay decay can be explained in terms of the strong nuclear force. β decay decay can be explained in terms of the weak nuclear force. 5. T 6. T 1 7. F Electrons always have spin of . 2 8. F According to quark theory, three quarks are needed to form a baryon and two quarks are needed to form a meson. 9. F When a quark absorbs or emits a gluon, its mass remains unchanged.

Multiple Choice 10. (b) 11. (c) 214 = x + 0 x = 214 82 = y + ( −1) y

= 83

 1  t 12. (d) N = N 0    2 

t

1/ 2

573

1  5730 = 10.0 mg    2  N = 9.30 mg 13. (a) The frequency with which a particle revolves in a cyclotron varies as follows: - directly with both the charge and the magnetic field strength. - inversely with the mass. Since the mass is increased by increased by a factor of 2 the frequency therefore decreases by decreases by a factor of 2. 14. (a) 15. (e) 16. (a)


Chapter 13 Radioactivity and and Elementary Particles Particles 743



Type α –

β + β γ

Composition 2 protons and 2 neutrons; essentially a helium nucleus electron positron photon

Charge +2 –1 +1 0

Penetrating Ability thin paper thin metal thin metal thick metal (lead)

2. (a) x = 212 + 0 x = 212 y = 83 + (−1) y = 82 (b) 238 = x + 4 x = 234 92 = y + 2 y = 90 (c) 84 = x + 2 x = 82 (d) Since γ has no quantum numbers to balance with the rest of the equation: x = 49 y = 116 (e) 30 = 30 + y y = 0 15 = 14 + x x = 1 Therefore, z Therefore, z = = e. (f) 13 = y + 0 y = 13 7 = x + 1 x = 6 From the periodic table, z table, z = = C (carbon). 3.

Strong Nuclear Force can be attractive can be repulsive acts on the proton acts on the neutron does not act on the electron active over distances of –15 the order of 10 m 4.

Electric Force can be attractive can be repulsive acts on the proton does not act on the neutron acts on the electron can be felt at any distance

The alpha particle consists of two protons and two electrons. It is therefore impossible for 12 H to under undergo go α decay decay because it does not have the constituent parts of an α particle particle to begin with.



5. (a)

239 94

Pu

→

235 92

4

U + 2 He

239 94

or

Pu →

235 92

U+α

(b) parent: Pu-239 = 239.052157 u daughters: U-235 = 235.043924 u α = 4.002602 u total = 239.046526 u ∆m = 0.005631 u Therefore,

 931.4 MeV   u  

E = 0.005631u 

E = 5.24 MeV When one atom of Pu-239 decays to U-235 by α emission, 5.24 MeV of energy is released. 6. (a) N 0 = 22.0 mg t = 24 h t 1/2 = 10.64 h (from Appendix C) N = ?

 1  t N = N 0    2 

t

1/ 2

24

 1  10.64

= 22.0    2  N = 4.61 mg

The amount of lead remaining after 24 h is 4.61 mg. (b) N 0 = 22.0 mg t 1/2 = 10.64 h N = 5.0 mg t = ? t t 1/ 2

N

1  = N 0    2 

t

 1  10.64 5.0 = 22.0    2  t

1  10.64 =   22.0  2  5.0

t

 5.0  = log  1 10.64 log   2  22.0     5.0  = t log  1  log      22.0  10.64  2   5.0  10.64log   22.0   t =  1  log    2  t = 23 h The amount of time required for the amount to decrease to 5.0 mg is 23 h. 7. A = 89 Bq A0 = 105.4 Bq t 1/2 = 5730 a t = ?



t t 1/ 2

 1    2 

A = A0 

t

 1  5730 89 = 105.4    2  t

1  5730 =   105.4  2  89

t

 89  = log  1  5730 log   2  105.4     89  = t log  1  log      105.4  5730  2   89  5730 log   105.4   t =  1  log    2  t = 1382 a The estimated age of the sample is 1382 a. 8. Some examples of medical application are described on pages 684–685. These include: • The use of radiation to kill cancerous cells. This might involve placing many tiny radioactive samples close to the tumour, or bombarding the area with radiation from several external sources. • The use of radioactive samples as medical tracers. For example these can be injected into the bloodstream and external monitors can be used to follow their progress through the system. • The combination of positron emitting tracers in addition to gamma cameras to produce PET scans. 9. (a) The electric field is timed precisely to oscillate in such a way that the particle will be accelerated across the gap as it emerges. With each successive acceleration across the electric field between the dees, the particle gains kinetic energy. Owing to mass-energy equivalence, its mass also increases as the particle’s kinetic energy increases. Because the frequency of rotation through the dees also depends upon the mass, the particle will become out of synchronization with the oscillating field between the dees. (b) The synchronization problem can be overcome in two ways. One way is to continuously alter the frequency of oscillation of the field to keep the particles synchronized with it. Another is to alter the strength of the magnetic field to compensate for the increase in mass. 10. Particles in a synchrotron are continuously being accelerated, even when they are not in the RF cavity. This is because superconducting magnets are used to bend their path into a circle. When charged particles are accelerated they tend to emit radiation. The fraction of the kinetic energy that is radiated away varies with the cube of the charge to mass ratio. Because the electron is 2000 times less massive than the proton, it tends to radiate away a much higher proportion of its kinetic energy. It is worth noting that while this synchrotron radiation can be a nuisance for applications that require very high energy particles, it can also be useful. The synchrotron radiation is a full spectrum of electromagnetic radiation than includes significant amounts of radiation that are particularly well suited for use as imaging tools. The Canadian Light Source (CLS) is an example of a synchrotron devoted to the production of synchrotron radiation to be used mainly for imaging purposes. 11. The linear accelerator only takes a particle through its path one time. On the other hand, the synchrotron is capable of accelerating the particle many times. All other things being equal, the synchrotron is therefore capable of obtaining the highest energies. It is worth noting, however, that the linac’s acceleration is all linear. Accelerating particles are therefore less susceptible to the phenomenon of reradiating away the kinetic energy. Less massive articles such as electrons are therefore easier to work with in linacs than in synchrotrons. 12. Each part has a different purpose. Some parts (the calorimeters) measure the kinetic energy through the particle’s penetrating ability, while other s track, or measure, position. Additionall y, because various particles respond differently, there are separate components for different types of particles. For example, photons are not detected by tracking detectors, and therefore must be detected by other means.



13.

2

14. mπ = 135.0 MeV/c Assume that this rest mass comes equally from the kinetic energies of the two protons. This means that each proton contributed one-half the total, a minimum of 67.5 MeV. 15. (a) To identify the particle, we must first determine the quantum numbers. 2 1 For the u: Q = ,B = ,s = 0 3 3 1 1 Q = − , B = , s = −1 3 3 For the combination uss: 2  1  1 Q = +  −  +  −  = 0 3  3  3 For the s:

B =

1

1

1

+ + =1

3 3 3 s = 0 + (−1) + (−1) = − 2

0

The xi zero (Xi ) has these quantum numbers. (b) To identify the particle, we must first determine the quantum numbers. 2 1 For the u: Q = ,B = ,s = 0 3 3 2 Q = − , B = −1, s = 0 3

For the u :

For the combination u u : 2 +  −   = 0 3  3  1  1  B = +  −  = 0 3  3  s = 0 + 0 = 0 Q=

0

2

0

Both pi zero (π ) and eta zero (η ) have these quantum numbers. 16. The bosons are the particles that mediate the various forces. One way to think of the interaction that results in a force is to think of the boson as carrying a message such as “move closer” or “move away”. Because the particles can be seen as conveying information, they are sometimes referred to as messenger particles. 1 17. (a) Fermions, which all have a spin of , are the building blocks of matter. Bosons, which have a spin of 1 or 2 2 (gravitons), are the force-carriers. (b) Leptons are found outside the nucleus and are considered fundamental. Hadrons, which are found inside the nucleus, are not considered fundamental and are instead considered to be made up of quarks. (c) The hadrons are classified into two types. Mesons are formed of quark-antiquark pairs and have spins of 0 or 1. 1 Baryons are formed from three particles, each of which can be a quark or an antiquark. The baryons have a spin of . 2



18.

Charge

Mass (me)

2

20

First Family up

3

down

−

Second Family charm

20

1 3

3 000

2 3

strange

−

Third Family top

300

1 3

350 000

2 3

bottom

−

11 000

1 3

19. (a) Spin measures the angular momentum of a particle. The spin quantum number can be considered as a multiplication h factor that measures the angular momentum in units of . 2π (b) Some particles have been found to exhibit consistently strange behavior in that they have unusually long decay times and tend to be produced in pairs. The strangeness number reflects this property. 20. l π − = 11 cm

sπ − (a)

= 0.6 cm q m

=? r =

l − 2 π

8 sπ −

+

sπ − 2 2

=

(11 cm )

8(0.6 cm)

+

0.6 cm 2

r = 26 cm, or 0.26 m q m

=

1.9 × 108 C ⋅ m/kg r 1.9 × 10 C ⋅ m/kg 8

= q m

0.26 m

= 7.3 × 108 C/kg

The charge-to-mass ratio of the π − particle is calculated to be 7.3 × 108 C/kg .



(b) m = ? q m

= 7.3× 108 C/kg

m=

=

q 7.3 × 108 C/kg 1.6 ×10

−19

C

7.3 × 10 C/kg 8

m = 2.2 × 10 −28 kg The calculated mass of the π − particle is 2.2 ×10 −28 kg . 21. The helium atom has two protons (each is uud), and two neutrons (each is udd), as well as two electrons.

22. (a) The standard model is sophisticated and complete in that it has the ability to make accurate predictions about the behaviour of particles. Furthermore, every particle postulate d to exist according to the model has been detected. (b) The standard model is crude because it lacks the quality of simple mathematical elegance. Furthermore, it is more concerned with describing the behaviour of particles than it is with explaining the behaviour. 23. It is assumed that at distances smaller than the unification distances, the types of forces are indistinguishable. The X –30 boson assumes that the matter particles are different for ms of the same thing. Within distan ces smaller than 10 m, objects can theoretically interact through the exchange of an X boson. In doing so, they can transmute from one type of particle to another. 24.

Applying Inquiry Skills 25. The table below shows the activity over a one-year (12 month) period.

Years Months Activity (Bq)

0

0.083

0.167

0.25

0.33

0.417

0.5

0

1

2

3

4

5

6

7

8

240

238

235

232

230

227

225

222

220


0.583 0.667 0.75

0.833

0.917

1

9

10

11

12

217

215

213

210


The table can be used to construct either of the graphs that follow. Notice that the elapsed time in relation to the half-life is very short. It is only about 0.2 half-lives so only a small fraction actually decays. The graph is therefore fairly linear in shape.

26. (a) The graph is shown, along with a curve of best fit.

(b) The initial activity is 1000 Bq. When the half-life occurs, the activity will have declined to 500 Bq. Inspection of the graph shows that this occurs at 35 h. The half-life is therefore 35 h.



(c) A = 1000 Bq t 1/2 = 35 h (from part (b)) t t 1/ 2

 1    2 

A = A0 

t

 1  35 A = 1000    2  In the above equation the units of t are hours and the units of A are Bq. 27. (a) The rate of detection was lower when the tube was covered in aluminum foul. (b) The most likely explanation is that the tube is being bombarded by radiation from netural sources. This “background radiation” is blocked by the foil, hence lower readings occur when the window of the tube is covered.

Making Connections 28. (a) Scientific discoveries in particle physics have followed from technological breakthroughs in both accelerators and detectors. For example, the development by Glaser of the cloud chamber resulted in many discoveries about particles because their motions could be analyzed much more accurately and precisely. (b) An example of this is the devices that have been developed based on superconducting magnets. The magnets were primarily designed to facilitate t he development of high-energy accelerators. The resulting knowledge and availability of the superconducting materials helped create machines such as MRI scanners. An interesting example of this is the CLS, which is based on the technology used in high energy pure-physics particle research. It is likely that the most interesting discoveries and developments will exist in other fields, particularly biotechnology. 29. Unlike the standard model, only indirect experimental evidence exists in favour of superstring theory. As a result, most of the work is still speculative and not testable in the way that scientific knowledge is expected to be. 30. (a) It is expected that the behavior of the virtual particles is in keeping with the uncertainty principle. That is h h or the equivalent ( ∆ E )( ∆t ) > . ( ∆ p )( ∆x ) >∼ ∼ 2π 2π (b) The conditions imply that the more massive the particle, the shorter the time (and hence the shorter the distance) over which it can act. Likewise, the less massive the particle, the longer the time (and hence the longer the distance) over which it can act. (c) As the mass of the particle decreases, the distance over which it can act increases. A particle without mass, such as a photon or a graviton, could therefore act over an unli mited distance.

Extension

2

31. m proton = 938.3 MeV/c 2 mneutron = 939.6 MeV/c (a) m of parent particles = ? m = 2m proton

+ E K

= 2(938.3 MeV/c2 ) + E K 2 m = 1876.6 MeV/c + E K The mass of the parent particles is 1876.6 MeV/c2. (b) m of daughter particles = ? mrest = mproton + mneutron + mπ

mrest

= 938.3 MeV/c 2 + 939.6 MeV/c 2 + 139.6 MeV/c 2 = 2017.5 MeV/c 2 2

The mass of the daughter particles is 2017.5 MeV/c . (c) E K = ? Assuming conservation of mass-energy you can equate the two expressions 1876.6 + E K = 2017.5 E K = 140.9 The total E K is 140.9 MeV. If this is divided equally among the two initial protons then the E K of each proton must be 70.45 MeV before the collision.



UNIT 5 PERFORMANCE TASK THE PHOTOVOLTAIC (SOLAR) CELL (Pages 738–739)

The Unit 5 Performance Task requires students to study the application of the photoelectric effect in photovoltaic cells (solar cells). The performance component of this task allows student to choose one of the two following options: • Task 1: Make a Photovoltaic Cell • Task 2: Using Solar Energy in a Remote Location In Task 1, students construct a working solar cell, while in Task 2, students conduct research on the technology required to power an isolated cottage or retreat. In either case, students are required to submit a written report for assessment/ evaluation. Although students select either Task 1 or Task 2, they are all required to conduct library and Internet research on the following topics (this will be called Part 1): • How solar cells use the photoelectric effect to convert solar energy into electricity • The efficiency of solar cell technology, and improvements to their efficiency • The differences between photovoltaic technology and conventional electricity-generating technologies with particular attention paid to environmental impacts • Advantages and disadvantages of solar power generation

Task 1: Make a Photovoltaic Cell In this option students construct a photovoltaic cell and test its output. They follow the instruction given in a suitable “kit” provided by the teacher. Commercially produced kits are available fro m most science supply companies. Students are required to develop a procedure for testing and measuring the electrical output of the cell they construct. A schematic diagram of the process is required in the submitted for mal report. Also, students are required to use their knowledge of wave optics to propose a way of controlling the intensity or colour of the light used to power the photovoltaic cell they’ve constructed, and devise a process for measuring the relationship between intensity or col our and output. They should graph the relationship if possible. A formal scientific report containing sections titled Materials, Procedure, Analysis, and Evaluation must be submitted for assessment/evaluation.

Task 2: Using Solar Energy in a Remote Location This task requires students to conduct extensive library and/or Internet research on the technology used to power an isolated home (retreat). A written report will be submitted for assessment/evaluation. The scenario described in the text provides the student with a context for conducting the research. It outlines the electrical needs that will be used as a basis for determining the following: • minimum number of hours of sunlight expected in the darkest months for the location • average power requirements in a 24-hour period • size and type of solar panels needed, the power of the inverter, and the type of storage batteries required • type of water turbine needed to provide supplemental power during the winter months The submitted report must describe: • equipment needed • costs of materials and maintenance • physics principles associated with the system Relevant Web sites include: http://www.newenergy.org/ http://www.mrsolar.com/education.htm http://www.newenergy.org/re_guide_toc.html http://www.newenergy.org/organizations.html http://www.solarbuzz.com/Canada.htm


Unit 5 Performance Task 753

Assessment The report containing the common research (Part 1) will be assessed and/or evaluated on the basis of the following criteria: • adequacy and accuracy of researched information • level to which the applicable physics principles seem to have been understood • referencing • quality of the diagrams • quality of the written communication

Task 1: Make a Photovoltaic Cell This task will be assessed and/or evaluated on the basis of the following criteria: • demonstrated ability to carefully follow instructions • viability of the manufactured solar cell • quality of the written report

Task 2: Using Solar Energy in a Remote Location This task will be assessed and/or evaluated on the basis of the following criteria: • adequacy and accuracy of researched information • level to which the applicable physics principles seem to have been understood • referencing • quality of the diagrams • quality of the written communication

754 Unit 5 Matter-Energy Interface


UNIT 5 SELF QUIZ (Pages 740–741)

True/False 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.

T F It is impossible for a particle with a nonzero mass to be accelerated to the speed of light. Otherwise the particle would have to infinite momentum, or the energy required to accelerate it to c would have to be infinite (see Section 11.2). T T F The intensity of the light does not affect the threshold frequency. T kq q F Coulomb’s law, F e = 12 2 , holds for forces between small charged particles even at distances smaller than the size of r atoms. This was shown first to be true by Rutherford and his colleagues. T F An α particle is also called a helium nucleus. T F For any given energy, less synchrotron radiation results in an accelerator when more massive particles are used. F The strong nuclear force can only be attractive. T F All hadrons are baryons or mesons T T

Multiple Choice 17. (a) According to the second postulate, the speed of light has an absolute value of c in all inertial frames of reference. 18. (c) Since you are stationary relative to the moving spaceship, your measurements of the time on the moving spaceship will be in two-position time and wil l be dilated or the clock will t ick at a slower rate relative to your fr ame of reference. 19. (a) Classical physics only provides and explanation for (a). All the others required quantum physics. 20. (a) In β decay, the neutrino was postulated to account for the conservation of energy and momentum. See Section 13.1. 21. (a) Gamma radiation is a photon that travels as an electromagnetic wave, not a particle with a non-zero mass. 22. (e) Since α particles have a positive charge and β particles have a negative charge, they will be deflected in opposite directions by an external magnetic field. 23. (a) Since a positron is produced, the Feynman diagram represents β + decay. 24. (b) The quantum quantity called strangeness was postulated because the observation revealed unexpectedly long decay times and opposite charges . 25. (e) The baryon number must add up to 0 or 1. 1 The spin must add to 0 or . 2 The strangeness must sum to 0 The only option where the rules do not hold up is (e).

Completion 26. 27. 28. 29. 30. 31. 32. 33. 34. 35.

inertial, noninertial the ether does not exist length rest mass quanta different lower probability 0, 1.0 heated solid, electrically excited gases


Unit 5 Self Quiz 755

36. 37. 38. 39.

not deflected decreases, increases, decreases 400 a spin

Matching 40. Scientist α-scattering experiment radioactivity diffraction of particle energy levels in the hydrogen atom energy levels in an excited gas matter waves particle classification momentum of a photon photoelectric effect planetary model of the atom quanta uncertainty


Discovery or Innovation Rutherford Becquerel Davisson Bohr Frank de Broglie Gell-Mann Compton Einstein Rutherford Planck Heisenberg


UNIT 5 REVIEW (Pages 742–745)


2.

3.

In your car, you are “moving” relative to the other car, whether you view it from the frame of reference of your car, or in the frame of reference of the other car. In other words there is motion relative to the second car, and motion of your car relative to the second car, even if you are at rest. We can conclude that all motion is relative and there is no absolute motion in either frame of reference. For the observer moving relative to the duration of a process (two position time), the time will be dilated or seem longer. One position time, that is, where the observer is moving with the process, is the proper time. Lm

5.

1−

v

2

c

2

and

∆t s

∆t m =

1−

→ c, then v 2 → c 2

v2 c

2

and the expression 1 −

v

2

→ 0. In the case of length contraction, the relativistic length will also c2 approach zero at c, which is impossible. Thus, speeds of c are impossible. In the case of time dilation, the expression ∆t s will approach infinity, which impossible. Thus, speeds of c are also impossible. v2 1− 2 c Applying a constant force to accelerate a nonzero mass to c would require an infinite energy. The faster the mass goes, the larger its relativistic momentum. At c the momentum (and mass) would be infinite which is impossibility. 8 v = 2.8  10 m/s –6 ∆t m = 4.86  10 s ∆t s = ? ∆t s ∆t m = v2 1− 2 c If v

4.

= Ls

∆ts = ∆t m

1−

v2 c2 2

= 4.86 × 10 −6 s

( 2.80 ×108 m/s ) 1− 2 (3.00 ×108 m/s )

∆t s = 1.74 ×10 −6 s The lifetime of the particle at rest is 1.74  10 –6 s. 8 11 6. ∆d = 5.9  10 km = 5.9  10 m v = 0.67c (a) ∆t s = ? ∆t m = ? To calculate the time you claim it to be: ∆d ∆t s = v

= =

5.9 × 10 m 11

0.67c 5.9 × 1011 m

(

0.67 3.00 × 108 m/s

)

= 2.935 ×103 s ∆t s = 48.9 min Copyright © 2003 Nelson

Unit 5 Review 757

If the local time at launch was 12:00 noon, the time you claim it is when you reach Jupiter is 12:00 P.M. + 48.9 min = 12:49 P.M. To calculate the time mission control claims it to be: ∆t s ∆t m = 2 v 1− 2 c 48.9 min

=

2

1−

( 0.67c )

c ∆t m = 65.9 min

2

If the local time at launch was 12:00 noon, the time mission control claims it is when you reach Jupiter is 12:00 P.M. + 65.9 min = 1:06 P.M. (b) Lm = 5.9  108 km Ls = ? Lm Ls

= Ls

1−

1−

v2 c2

5.9 × 108 km 2

1− Ls

c2

Lm

=

=

v2

(0.67c ) c2

= 7.947 ×108 km, or 7.9 × 108 km

The distance from Earth to Jupiter, in the inertial frame of your ship, is 7.9 (c) ∆t = 2.935  103 s m = 1.0 × 10−3 kg/s 2.935 × 10 3 s

(

∆ E =

)(

108 km.



) = 2.935 kg

?

∆ E = ∆mc 2 2

= ( 2.935 kg ) (3.00 ×108 m/s ) ∆ E = 2.64 × 1017 J

If the usable energy is converted at an efficiency of 10%, the usable power generated is (2.64 2.6  1016 J. 7. v = 0.850c Lm = 52.2 m Ls = ? Lm Ls

= Ls =

1−

v2 c2

v2 2

c 52.2 m 2

1− Ls

17

10 J)(0.10) =

Lm 1−

=



(0.85c ) c2

= 99.1 m

The length of the spaceship when at rest on Earth would be 99.1 m.



8.

v = 0.60c p = ? mv

p =

1−

=

v

2

c2

(1.67 ×10−27 kg ) (0.60c ) 2

1−

( 0.60c ) c

2

p = 3.757 × 10−19 kg ⋅ m/s, or 3.8 × 10−19 kg ⋅ m/s The relativistic momentum of the proton is 3.8  10 –19 kg·m/s. 9. v = 0.866c p = ? mv p = 2 v 1− 2 c

=

(9.11×10−31 kg ) (0.866c ) 2

1−

(0.866c ) c

2

p = 4.73 ×10 −22 kg ⋅ m/s –27 The relativistic momentum of the electron is 4.73  10 kg·m/s. 8 10. v = 2.8  10 m/s 3 ∆d = 3.0 km = 3.0  10 m (a) ∆t s = ? ∆d ∆t = v

=

3.0 ×10 m 3

2.8 × 10 m/s 8

∆t = 1.07 ×10−5 s ∆t is ∆t m since

it is two position time. Therefore, ∆t s ∆t m = v2 1− 2 c

∆ts = ∆t m

1−

v2 c2 2

= 1.07 × 10−5 s

( 2.8 ×108 m/s ) 1− 2 (3.00 ×108 m/s )

∆t s = 3.84 × 10−6 s, or 3.8 × 10−6 s It takes the electron 3.8


–6

10 s to reach the target.



Unit 5 Review 759

(b) p = ? mv

p =

1−

=

v2 c2

(9.11×10−31 kg )(2.8 ×108 m/s ) 2

(2.8 ×108 m/s ) 1− 2 (3.00 ×108 m/s ) p = 7.1× 10

−22

kg ⋅ m/s

The electron’s momentum when it hits the target is 7.1  10 –22 kg·m/s. –27 –31 11. The mass of a proton is 1.673  10 kg, while that of an electron is 9.11  10 kg. The proton is over 1800 times more massive than an electron. To accelerate a particle from rest to a speed approaching c requires a certain magnitude of force. Since ∑ F = ma , this force is much larger for a proton than an electron. Thus, it is easier to accelerate an electron than a proton. 1 1 2 2 Alternately, from the relationship Wtotal = ∆EK = mvf − mvi , the energy required to accelerate a particle will be 2 2 equal to the change in the kinetic energy of the particle. Since ∆ EK ∝ m and m p >> me, a great deal more energy will be 12.

required to accelerate a proton than is required for an electron. 8 ∆ E = 2.3  10 J ∆m = ?

∆ E = ∆mc 2 ∆ E ∆m = 2 c

=

2.8 ×10 m/s 8

2

(3.00 ×108 m/s )

∆m = 3.1× 10−9 kg 13. 60 W = 60 J/s ∆m = ? To calculate the amount of energy required per year: 60 J 3600 s 24 h 365 d ∆ E = × × × s 1h 1d 1a

∆ E = 1.89 ×109 J To calculate the amount of mass:

∆ E = ∆mc 2 ∆ E ∆m = 2 c

=

1.89 × 109 J 2

(3.00 ×108 m/s )

∆m = 2.1× 10−8 kg The amount of mass that must be converted is 2.1 18 14. f = 2.15  10 Hz E = ? E = hf

–8

10 kg.



= (6.63 ×10−34 J ⋅s )(2.15 ×1018 Hz ) E = 1.42 × 10−15 J The energy associated with a single quantum of X radiation is 1.42


–15

10



J.


15. E

= mc 2

or ∆m =

∆ E c

2

. In other words, mass and energy are interconvertible and mass can be expressed in energy units.

∆m (kilograms) =

From the standpoint of units,

∆ E (joules)

. But, the energy in joules is a very small number in particle 2 c –19 physics, so the energies in eV and MeV are used after converting joules to electr on volts (1.60 × 10 J/eV). 16. Let the subscript s represent the “soft” radiation, and h represent the “hard” radiation. –7 λs = 3.80  10 m –7 λh = 1.14  10 m E s =? E h Since E s

=

hc

λ s

hc

=

and E h

λ h

, hc

E s E h

=

λ s hc

λ h

= =

λ s λ h 1.14 × 10−7 m

3.80 × 10−7 m E s E = 0.300, or h Eh E s

= 3.33

The ratio of the two energies is 3.33:1. –19 17. E = 3.20  10 J λ = ? hc E =

λ

λ =

hc E

(6.63 ×10− J ⋅ s )(3.00 ×10 34

=

3.20 × 10

−19

8

m/s

)

J

−7

λ = 6.22 × 10 m The wavelength of the photon is 6.22  10 –7 m, or 622 nm. hc 18. Since E = hf = , if the wavelength is lower the energy of the incident photons is lower. To overcome the energy λ

binding the electron to the surface ( work function W ), requires a minimum amount of photon energy. If no photoelectrons are released, the energy of the incident photons is not high enough to release the photoelectron since the wavelength is too long. hc 19. Given that E = hf = :

λ

(a) To release photoelectrons the photon energy must be equal or greater than the binding energy (work function). This energy will be different for each metal and thus the frequency of the required incident light will be unique for each metal. (b) Although the red light is intense, the energy of the incident photons is only determined by its frequency. High intensity red light has a larger number of photons, but the energy of each photon is insufficient to liberate photoelectrons However, the blue light, although of low intensity, has photons each with a higher energy (higher frequency), and enough energy to liberate photoelectrons. (c) As noted above, the more intense the light the larger the volume of photons incident on the metal. If the frequency is equal or higher than the threshold energy, each photon releases a photoelectron. The larger the number of photons, the larger the volume of photoelectrons, and the larger the photocurrent.


Unit 5 Review 761

20. W = 2.48 eV = (2.48 eV)(1.60  10 –19 J/eV) = 3.968 λ = ?

10 –19 J



The minimum energy of the photons will equal the work function. Thus: EK = Ephoton − W Since E K = 0, E p

= W

hc

= W

λ

hc

λ =

W (6.63 × 10

=

−34

J ⋅ s)(3.00 ×108 m/s)

(3.968

× 10−19 J)

λ = 5.01× 10−7 m The colour of light will liberate photoelectrons from the barium surface are dark green, blue and violet. (All have a –7 wavelength shorter than 5.01  10 m.) 21. W = E K = 5.01 eV (a) f = ? E = hf f =

E h

(5.01 eV ) (1.60 × 10 −19 J/eV )

=

6.63 × 10

−34

J ⋅s

f = 1.21×10 Hz The threshold frequency for the photoelectric effect is 1.21 –7 (b) λ = 1.50  10 m cutoff potential = ? hc E = 15



15

10 Hz.

λ

(6.63 ×10− J ⋅ s )(3.00 ×10 m/s ) 34

=

8

−

1.50 × 10 7 m

E = 1.326 × 10

−18

J

E = 8.29 eV The maximum E K of photoelectrons is 9.29 eV – 5.01 eV = 3.28 eV. Therefore, the cutoff potential is –3.28 eV. 22. W = 2.48 eV –7 λ = 452 nm = 4.52  10 m E K = ? EK = Ephoton − W

=

hc

λ

− W

(6.63 ×10−34 J ⋅ s )(3.00 ×108 m/s ) = − 2.48 eV ( 4.52 ×10−7 m )(1.60 ×10−19 J/eV ) = 2.75 eV − 2.48 eV E K = 0.27 eV The maximum kinetic energy of the ejected photons is 0.27 eV.



23. E K = 1.20 eV W = ? W

= E photon − E K

(6.63 ×10− J ⋅ s )(3.00 ×10 m/s ) = − 1.20 eV − − × × 3.55 10 m 1.60 10 J/eV ( )( ) 34

8

7

19

= 3.50 eV − 1.20 eV W = 2.30 eV The work function of the metal is 2.30 eV. 14 24. f = 8.0  10 Hz W = 1.2 eV v = ? First we must calculate the energy of the photon: E photon = hf

= (6.63 × 10 −34 J ⋅ s )(8.0 × 1014 Hz ) E photon

= 5.304 ×10 −19 J = 2.1 eV

Next, we must calculate the maximum kinetic energy: EK = Ephoton − W

= 3.32 eV − 1.2 eV E K = 2.1 eV But the retarding potential is –1.0 V. Therefore, the maximum energy is 2.1 eV – 1.0 eV = 1.1 eV, or 1.76

–19

10



J.

Finally, we can calculate the maximum speed: 1 EK = mv 2 2 2 E K

v=

m 2 1.76 ×10 −19 J

(

=

9.11 × 10

−31

)

kg

v = 6.2 × 10 m/s 5 The maximum speed of at which the electron reaches the conductor is 6.2  10 m/s. 25. (i) For the photoelectric effect, lower energy photons usually in the visible spectrum initiate the effect. For the Compton effect, high-energy photons in the X-ray region of the electromagnetic spectrum initiate the effect. (ii) For the photoelectric effect, the result of the interaction is that a low-energy photon liberates a photoelectron and gives it kinetic energy. For the Compton effect, it is that an incident high-energy photon emerges as a lower energy photon and scatters an electron. (iii) The photoelectric effect implies that the energy of light is carried as quanta or photons. The Compton effect implies that photons have momentum as well as energy. 26. The emerging photon has lower energy and momentum when compared with the incident photon. It also travels in a different direction, as predicted by the conservation of momentum. 27. λ = 525 nm = 5.25  10 –7 m p = ? h p = 5

λ

=

6.63 × 10

−34

J⋅s

−7

5.25 × 10 m

p = 1.26 × 10

−27

kg ⋅ m/s

The momentum of the photon is 1.26 × 10−27 kg ⋅ m/s .


Unit 5 Review 763

28. f = = 4.5  1015 Hz p = p = ? h c = , Since p = and λ = f λ hf

p =

c

(6.63 ×10− J ⋅ s ) (4.5 ×10 − 34

=

15

Hz

)

3.00 × 10 m/s 8

p = 9.9 ×10

−27

kg ⋅ m/s

The momentum of the photon is 9.9 × 10−27 kg ⋅ m/s . 29. E = = 136 eV p = p = ? h hc = , Since p = and λ = E λ h hc

p =

= =

E E c

(136 eV ) (1.60 × 10−19 J/eV ) 3.00 × 108 m/s

p = 9.86 × 10−24 kg ⋅ m/s The momentum of the photon is 9.86 × 10−24 kg ⋅ m/s . 4

30. E p = 6.0  10 eV 4 E K 5.6  10 eV K = (a) E p´ = ? E p

= Ep′ + E K

E p′ = Ep

− E K

= 6.0 × 104 eV − 5.6 × 10 4 eV E p′ = 0.4 0.4 eV The energy of the scattered proton is 0.4 eV. (b) v = ? 1 2 EK = mv 2 v=

=

2 E K m

(

)( )(

2 5.6 ×10 eV 1.60 ×10 4

−19

J/eV

)

9.11 × 10 −31 kg

v = 1.4 ×108 m/s 8 The speed of the electron is 1.4  10 m/s.



(c) p = p = ? mv

p =

1−

=

v2 c2

(9.11×10−31 kg )(1.4 ×108 m/s ) 2

(1.4 ×108 m/s ) 1− 2 (3.00 ×108 m/s ) p = 1.13 ×10

−22

kg ⋅ m/s, or 1.1× 10

The momentum of the electron is 1.1 6 31. v = 1.0  10 m/s λ = λ = ? h

λ = λ =

=

−22

kg ⋅ m/s

–27

10



kg·m/s.

p h mv 6.63 × 10

(9.11×10−

31

−34

)(

J ⋅s

kg 1.00 × 10 m/s 6

)

λ = 7.28 × 10 −10 m, or 7.3 × 10−10 m –10

The wavelength of the photon is 7.3  10 m. 32. (a) 3.0 eV for an elastic collision is not enough energy to equal an energy level. (b) 8.0 eV − 4.9 eV = 3.1 eV 8.0 eV − 6.7 eV = 1.3 eV The electron would have either 3.1 eV, 1.3 eV, or, for an elastic collision, 8.0 eV of energy after passing through the vapour. 33. An atom can absorb neither a 13.6 eV nor one one greater that 13.6 eV. For a photon to be absorbed, the photon energy must be exactly equal to the energy levels above the ground state. Since this not the case, no absorption is possible. (13.6 eV is the ionization energy, not an energy level for the atom.) (On the other hand, if were an electron colliding with the atom, ionization would be likely in either case. 34. (a) In the second Lyman transition: n = 3 → n =1

−15.1 eV − (−13.6 eV ) = 12.1 eV In the second Balmer transition: n=4→n=2

−0.85 eV − (−3.40 eV ) = 2.55 eV (b) From part (a), the energy the atom must absorb to make make the transition from n = 2 to n = 4 is 2.55 eV. h 35. Although a photon has has zero mass, it does have momentum (see Compton effect p = , page 605). In classical mechanics,

λ

when a particle’s direction is changed there is a change in momentum. For momentum to change, the particle exerts an impulse on the surface given by ∑ F ∆t = m∆ v = ∆p . For a photon, the change in momentum is extremely small compared with that of a nonzero mass of a particle (e.g., an electron). Nevertheless, there is a small impulse on a reflecting surface ∆ p when the momentum of the photon changes. Since ∑ F = , there is a force on the surface that is small, but still ∆t measurable. An example is the “solar wind” created by photons originating from the sun. (Note that Newton made a similar argument in his particle theory to explain the reflection of light; see page 462.)


Unit 5 Review 765

36.

Type

α

Description of Particle helium nucleus

β –

electron

β+

positron

γ

photon

Effect on N and Z N is is decreased by 2 Z is is decreased by 4 N is is increased by 1 Z is is unchanged N is is decreased by 1 Z is is unchanged both N and and Z are are unchanged

Example (examples will vary) 238 92

U→

14 6

He →

15 8

O→

238 92

Th + 42 He He + γ

234 90 14 7

15 7

U→

N+

N+

0

−1 e 0

+1 e

+ γ

Th + 42 He He + γ

234 90

Note that γ decay decay usually occurs in conjunction with either α or or β decay decay as the above examples show. 37. (a) 15 = y + 0 y = 15 8 = x + ( −1) x = 9 From the periodic table, the element is fluorine (F). This is β decay. decay. (b) 226 = y + 4 y = 222 88 = x + 2 x = 90 From the periodic table, the element is thorium (Th). This is α decay. (c) 231 = 227 + y y = 4 91 = 89 + x x = 2 From the periodic table, the element is Helium (He). This is α decay. (d) 214 = y + 0 y = 214 82 = x + ( −1) x = 83 From the periodic table, the element is bismuth (Bi). This is β decay. decay. (e) 239 = y + 0 y = 239 92 = x + ( −1) x = 93 From the periodic table, the element is neptunium (Np). This is β decay. decay. 38. component particles – 29 e = 29 × 0.00055 u = 0.01595 u + 29 p = 29 × 1.00728 u = 29.21112 u 0 36 n = 36 × 1.00867 u = 36.31212 u total mass = 65.53919 u copper 65 mass = 64.92779 u

∆m = 0.61140 u E = 0.61140 u × 931.4 MeV/u = 569.5 MeV



569.3 MeV

= 8.762 MeV/nucleon 65 nucleons The average binding energy per nucleon is 8.762 MeV 214 210 39. (a) 84 Po →82 Pb + 24 He average energy per nucleon =

(b) This energy was some of the binding energy that originally held the component parts of the α particle in the nucleus. (c) Conservation of momentum suggests that both particles would be given some kinetic energy. Initially there is only one object. There are two objects after the reaction, and one of them, the α particle, is likely moving at a considerable speed (around 0.1c). Conservation of momentum suggests that the lead particle would have to recoil in the opposite direction as a result. This means that it also must have some kinetic energy, though not nearly as much as the α particle. 40. (a) The Geiger-Mueller tube detects the particles through the ionizations they produce within the tube. The γ particles produce very little ionization and hence are virtuall y invisible to the device. (b) A scintillation detector uses a crystal that readily absorbs and re-emits the energy from γ rays. 41. The initial activity is 2500 Bq. This activity will be one-half that amount, or 1250 Bq, when the half-life occurs. By inspecting the graph you can see that t = 30 a when A = 1250 Bq. Therefore, the half-life of the sample is 30 a. 12 42. (a) A0 = 2.50 × 10 Bq t ½ = 28.8 a t = one half-life At the half-life the activity is one half the original quantity. Since A0 = 2.50 × 1012 Bq, the activity at the half-life is 12 one-half this amount. Therefore, after one half-life the activity is 1.25 × 10 Bq. 12 (b) A0 = 2.50 × 10 Bq t ½ = 28.8 a t = 5.0 a A = ? t

A = A0 ( 12 )t 1/2 5.0

1  28.8 = ( 2.50 × 1012 )    2  A = 2.22 × 1012 Bq After 5.00 a, the activity is 2.22 × 1012 Bq. 43. N 0 = 2.50 µg t ½ = 1600 a N = 1.00 µg t = ?

 1  t N = N 0    2 

t

1/ 2

t

 1  1600 1.00 = 2.50    2  t

1  1600 =   2.50  2 

1.00

 1.00 

t

 1 1600

 = log  2   2.50     1.00  = t log  1  log      2.50  1600  2   1.00  1600 log   2.50   t =  1  log    2  t = 2.1 ×103 a The amount remaining after 2.1 × 103 a will be 1.00 µ g. log 


Unit 5 Review 767

44. A0 = 2.99 Bq a = 1.93 Bq t ½ = 5730 a t = ?

 1  t A = A0    2 

t

1/ 2

t

 1  5730 1.93 = 2.99    2  t

1  5730 =   2.99  2 

1.93

t

 1.93 

 1  5730

 = log  2   2.99     1.93  = t log  1  log      2.99  5730  2   1.93  5730log   2.99   t =  1  log    2  t = 3.62 × 103 a The age of the sample is 3.62 × 103 a. log 

45. A0 = 450.0 Bq t ½ = 3.7 a t = 10.0 a A = ?

 1  t A = A0    2 

t

1/ 2

10.0

1  3.7 = 450.0    2  A = 69 Bq 46. 47.

48.

49.

After 10.0 a, the activity is 69 Bq. A grand unified theory seeks to unify the electromagnetic, strong nuclear, and weak nuclear forces. The gravitational force is the one left out. With each successive acceleration across an RF gap in a cyclotron, the particle gains kinetic energy. Due to relativistic effects its mass therefore increases. The frequency of rotation in the cyclotron is a function of the mass so the particle will gradually become out of synch with the electric field that is intended to accelerate it across the gap. This sets an upper limit of the amount of energy that can be imparted in a cyclotron. A synchrocyclotron compensates for the relativistic mass increase by varying the frequency of the electric field. In this way much higher energies can be achieved. Both devices accelerate the particles in an RF cavity. The electric field in the cavity is precisely timed to accelerate the particle across the “gap.” The motion in the linac i s linear, as one would expect. The particles move through drift t ubes when not in the RF cavity. One significant linac is located at Stanford University. The motion in the synchrotron is circular. Super-conducting magnets bend the path of the particles when they are not in the RF cavities. Both the frequency of the electric field and the magnetic field strength can be adjusted to compensate for relativistic effects. Significant synchrotrons include the CLS at Saskatoon, the Tevatron at Fermilab, and the various synchrotrons operated by CERN. Charged particles travelling through the wire grids in the tracking detector produce electrical signals that allow their motion to be analyzed. In addition, a magnetic field in this region can be used to help determine the quantity and type of charge. The calorimeters absorb the kinetic energy of most of the particles. By measuring the penetration of the particles one can measure their kinetic energy. The muon chamber is the calorimeter that works with the more penetrating muons. It both detects and measures the kinetic energy of these particles.



50. mneutron = 1.00728 u mantineutron = 1.00728 u mtotal = mneutron + mantineutron = 1.00728 u + 1.00728 u = 2.01456 u E = 2.01456 u × 931.4 MeV/u 3 E = 1.87 × 10 MeV 3 The minimum energy of the photon is 1.87  10 MeV. + 51. The quark composition of a proton consists of uud. In the process of β decay one u quark transmutes to a d quark. In this + process it emits a W boson. The W boson decays, producing a neutrino ( ν) and a positron (e ). 1 52. The quantum numbers were obtained from Table 2, on page 707. Q = 0, and spin = since the particle is a baryon. 2 B = 1 for the same reason, and s = –1. Since the strangeness is –1, one of the components must be an s quark. The charge 1 on the s quark is − . The remaining two particles must have charges that bring the sum to 0. By inspection of the 3 charges on the various quarks you can see that the combination ud provides the correct sum of the charge and baryon number. The quark composition of the lambda–0 particle is uds. 53. Thus far, attempts to reconcile the gravitational force with the other three forces have been unsuccessful. Additionally, only indirect evidence exists for the presence of the graviton, the hypothetical force-carrier for gravity. 2 1 54. (a) u : Q = , B = , s = 0 3 3 1 1 d :Q = − ,B = ,s = 0 3 3 For udd: 1 1 +  −  +  −  = 0 3  3  3

Q=

2

B =

1

1

1

+ + =1

3 3 3 s = 0 + 0 + 0 The particle that has these quantum numbers is the neutron. 1 1 (b) s : Q = − , B = , s = − 1 3 3 1 1 d :Q = − ,B = ,s = 0 3 3 For sdd:

 

Q = − B =

1

1

1 1 +  −  +  −  = − 1  3  3  3 1

1

+ + =1

3 3 3 s = −1 + 0 + 0 = −1 The particle that has these quantum numbers is the ∑ – . 55. According to quantum chromodynamics, the quarks within the nucleus exchange particles called gluons. There are eight types of gluons. The “colour force” is the cause of the strong nuclear force. The theory says that the force between various hadrons in a nucleus is actually due to the force between the quarks from which they are composed. 56. The parent particles have no net charge. Since one of the emitted particles is negative, the unknown particle must be 1 1 positive. The parent particles each have spin ± . Since one of the emitted particles, the electron, has spin ± then the 2 2 1 unknown emitted particle must either have spin ± or 0. One of the parent particles is a lepton and the other is a baryon. 2 Since the electron is a lepton then the unknown particle must be a baryon. This is consistent with the spin number.


Unit 5 Review 769

The neutrino’s mass is negligible compared to that of the neutron. The parents’ combined mass is therefore that of the 1 neutron. The electron’s mass is about that of the neutron so the unknown particle’s mass is slightly less than that of 2000 the neutron. The unknown particle is a positively charged baryon with a mass slightly less than that of the neutron. It is most likely a proton. 57. (a) e + → ν+ γ

e+

Baryon Number

Lepton Number

0

+1

ν

0

+1

γ

0

0

Spin

Charge +1

±

1

±

1

0

2 1

0

2

Charge and spin are not conserved. (b) p + p → p + n + n

p p p

n n

Baryon Number +1 +1 +1 –1 +1

Lepton Number 0 0 0 0 0

Spin

± ± ± ± ±

1

Charge +1

2 1

+1

2 1

+1

2 1

0

2 1

0

2

Baryon number, charge and spin are not conserved. (c) K 0 → µ + νe

K

0

µ νe

Baryon Number B 0 0 0

Lepton Number L 0 +1 –1

Spin s 0

±

1

±

1

Charge Q 0 –1

2

+1

2

All quantum numbers in the table are conserved. 58. Only (c) is theoretically possible since both (a) and (b) do not conserve all the quantum numbers. 59. (a) p + e − → n + ν

p – e n

ν

Mass 938.3 0.511 939.6 small

Charge +1 –1 0 0

Baryon Number +1 0 +1 0

Lepton Number 0 1 0 +1

The small difference in mass can be attributed to the kinetic energy of the particles. The quantum numbers are conserved so the reaction is theoretically possible.



(b) n → π +

+π −

n

π+ π –

Mass 439.6 139.6 135.0

Charge 0 +1 –1

Baryon Number +1 0 0

Lepton Number 0 0 0

The large difference in mass could be attributed to an enormous kinetic energy in the emitted particles, however, because the baryon number is not conserved the reaction is not theoretically possible. 60. (a) The standard model lacks the virtue of simple mathematical elegance. It is also more concerned with describing particle interactions than with explaining them. (b) The chief alternative to the standard model, superstring theory, sees all particles as multi-dimensional closed loops vibrating in all dimensions. The strings are extremely small in size, of the order of about 10 –35 m. The different types of particles are seen as just different harmonics. Owing to the extremely small size of the strings it is highly unlikely that direct evidence will be found in support of the theory in the near future. The current research is therefore purely theoretical.

Applying Inquiry Skills 61. The chip inside the remote control can detect when a key is pressed. A signal is created that modulates the carrier frequency (see Section 10.9). A unique sequence is produced for each key pressed. The chip sends the resultant information out to a transistor, which amplifies it and makes it stronger. At the end of the circuit board there is an infrared light emitting diode (LED). Although LEDs produce visible light, the remote's LED produces infrared light that is invisible to the human eye. However, it is not invisible to a TV, stereo, CD, DVD or VCR. The receptors in these devices are able to detect the infrared light and react appropriately. (You can see the infrared signal being sent by the remote control if you view if through a digital camera since digital cameras can “see” infrared.) 62. It is likely that the film has been “exposed” by the low-level radiation from the luminous dial. The developed photograph should reveal the shape of the luminous features.

Making Connections 63. One implication is time dilation and the twin paradox. Astronauts on long journeys at relativistic speeds will age at a much slower rate than their friends and relatives on Earth. In fact, the friends and relatives on Earth may not be alive when the travellers return, and it is feasible that if the journey is long enough the world may have changed, for example a new ice age. To reach relativistic speeds, the spaceships will require very large amounts of energy, more than can be carried. Some other propulsion system will be necessary, a system the converts mass, picked up along the way, into energy. –7 64. λ = 550 nm = 5.50  10 m number of photons = ? hc E =

λ

(6.63 ×10−34 J ⋅ s )(3.00 ×108 m/s ) = (5.50 ×10−7 m )(1.60 ×10−19 J/eV ) E = 2.26 eV/photons At an efficiency of 100% and a requirement of 60 W or 60 J/s of energy: 60 J/s number of photons = ( 2.25 eV ) 1.60 ×10 −19 J/eV

(

)

number of photons = 1.67 × 10 20 photons Since the efficiency is only 25%, the light bulb requires four times the number of photons to receive adequate power. Therefore, (1.67  1020 photons)(4) = 6.67 1020 photons are required. 65. λ = 5.0  10 –7 m 5 8 m = 4.0  10 t = 4.0  10 kg 8 v = 1.5  10 m/s number of protons = ?


Unit 5 Review 771

First we must calculate the momentum of the photons: h p =

λ

6.63 × 10

=

−34

J ⋅s

−7

5.0 × 10 m

p = 1.33 ×10−27 kg ⋅ m/s Using the unknown x, we can calculate the number of photons: xp photon = pspaceship x =

= =

pspaceship p photon mv p photon

( 4.0 ×108 kg )(1.5 ×108 m/s ) 1.33 × 10 −27 kg ⋅ m/s

x = 4.5 × 10 photons 43 Therefore 4.5  10 photons are required to accelerate the spaceship. 66. Speculation about the movement of the moons of Jupiter added to our knowledge of the speed of light. Jupiter has twelve moons, four of which are easily seen with a simple telescope. All of the moons move relatively quickly around Jupiter. One of them has a period of only 42.5 h. Although the accurate measurement of their periods had been made earlier, it was Olaus Roemer (1644–1710), a Danish astronomer, who attempted to predict the precise moment they would be eclipsed by Jupiter (as viewed from Earth). To his surprise, the eclipses came progressively earlier at the times of year when Earth approached Jupiter, and progressively later when Earth was moving away from Jupiter . Roemer concluded, correctly, that light must travel at a finite speed, and that the eclipses were delayed because the light had to travel a greater distance when Earth and Jupiter were farther apart. He calculated that when Earth and Jupiter were farthest apart, it took 22 min, or 1320 s, for light to cross the diameter of Earth’s orbit. A few years later, Christiaan Huygens, a Dutch mathematician and scientist, calculated the diameter of Earth’s orbit and, using Roemer’s data, calculated the speed of light as follows: 43

3.00  1011 m ∆t = 1320 s ∆d =

v=

∆d ∆t

=

3.00 × 1011 m 1320 s

v = 2.3 × 108 m/s



Modern telescopes and accurate timers have enabled scientists to determine that the time difference is 1.00  103 s; nevertheless, Huygens’ calculation was astonishingly accurate considering the equipment he had. The value Roemer and Huygens gave for the speed of light was so great that their fellow scientists rejected it at first. The work of these two scientists was not accepted until after both were dead. In 1905, Albert Michelson made very accurate measurements of the speed of light. His work was recognized, and he was awarded the Nobel Prize for Physics in 1907. Michelson’s method involved an ingenious arrangement of mirrors (see the diagram that follows). Light from a very bright source was reflected from surface A, on an eight-sided, rotatable mirror, to a mirror located about 35 km away. The distant mirror reflected the light back to surface G where it was observed in a telescope. The octagonal mirror rotated, and there were only certain positions in which it reflected light that could be seen in the telescope. These positions occurred every one-eighth of a rotation. If the mirror moved by this amount in the time interval taken by a light pulse making the round trip, then the telescope would detect the pulse reflected from face A via face F, in its new position. For this to happen, the octagonal mirror had to rotate very quickly—approximately 32 000 times/min. The period of rotation of the mirror was accurately determined, and the time for the light to make the round trip was measured. The speed of light could then be calculated, using this time and the distance travelled by the light on its round trip.

In 1933, Michelson supervised the remeasurement of the speed of light with a long, evacuated tube in which light rays were repeatedly reflected for a total distance of over 16 km. Although he died before the final results were calculated, his experiment produced an even more accurate value for the speed of light. Today we have very accurate measurements made with lasers. The current accepted value for the speed of light in a 8 8 vacuum, c, is 2.997 924 562 6 0.000 000 011  10 m/s. In air, the speed of light is only about 0.000 87  10 m/s less than in a vacuum. 69. l π − = 8.0 cm

= 0.3 cm l Σ− = 0.6 cm mπ − = 140.0 MeV/c 2 mΣ− = 1197.0 MeV/c 2 θ = 135° (a) m Χ = ? sπ −

0

r π −

=

l − 2 π

8 sπ −

+

sπ − 2 2

=

(8.0 cm )

8(0.3 cm)

+

0.3 cm 2

r = 26.8 cm pπ − pπ −


= (6.86 MeV/c ⋅ cm )r π − = (6.86 MeV/c ⋅ cm)(26.8 cm) = 184 MeV/c

Unit 5 Review 773

Physics-12-Ch 11 & 12 & 13-Solns

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