Soluciones 1

Modern Physics: PHY 104

Spring Semester 2013

Second Collaborative homework Solution Note: Note: Deadli Deadline ne is 15 March 2013, 10 am. Place Place your homework homework in the orange box on physics physics floor. Mention Mention names of all collaborators collaborators in the grou group. p. Work ork alon alone e or in grou groups ps upto five. five. Solut Solutio ion n will will be posted online at 10 am. 1. An electron in chromium makes a transition from the n = 2 state to the n = 1 statez without emitting a photon. Instead, the excess energy is transferred to an outer electron (in the n = 4 state), state), which which is ejected ejected by the atom. (This (This is called called an Auger process, process, and the ejected ejected electron electron is referred referred to as an Auger Auger electron.) electron.) Use the Bohr theory to find the kinetic energy of the Auger electron.

Coulomb interaction Auger electron is ejected n=1 +Ze

n=2 n=3

n=4

Answer 1:

Chromium is making a transition from n = 2 to n = 1 state state.. In order order to calcul calculat atee the kinetic energy of the Auger electron, first calculate the energy of the three levels n = 1, n = 2 and n = 3. According to Bohr’s atomic model, 2

E n = −13. 13.6

Date: 11, March 2013

� Z � n2

eV, eV,

n = 1, 2, 3.....

1



Hence E 1 = =

(24)2

−13.6 ×

1 −7.83 ×

103 eV.

−13.6 ×

(24)2

Similarly setting n = 2 yields, E 2 = =

22 −1.96 ×

103 eV.

−13.6 ×

(24)2

and setting n = 4 gives, E 4 =

42 = −489.6 × 103 eV.

The energy available from the n = 2 to n = 1 transition is, ∆E = E 2 − E 1 =

�

−1.96 − (−7.83)

= 5.87 × 103 eV.

�

×

103 eV

This energy is not directly emitted, rather it is transferred to the Auger electron. Energy required to completely remove an electron from its orbit is equal to the binding energy of that electron. Since the energy of n = 4 orbit state is −489.8 eV, this is also the binding energy for n = 4 electron. Hence the energy available ∆E is providing the binding energy and the excess energy appears as the K.E of the Auger electron, K = ∆E − E 4 K = 5.87 × 103 eV − 489.8 eV = 5.38 × 103 eV = 5.4 keV. 2. An electron initially in the n = 3 state of a one-electron atom of mass M at rest undergoes a transition to the n = 1 ground state. (a) Show that the recoil speed of the atom from emission of a photon is given approximately by, v = Date: 11, March 2013

8hR . 9M 2



(b) Calculate the percent of the 3

→ 1

transition energy that is carried off by the

recoiling atom if the atom is deuterium. Answer 2:

(a) Energy carried off by the electron as the transition from n = 3 to n = 1 takes place is,

� 1�

∆E = E 3 − E 1 = hcR 1 −

9

=

8hcR , 9

where R is the Rydberg constant. This emitted photon has momentum, P = ∆E/c = 8hR/9. Since P = M v,

⇒

v = 8hR/9M as required. Photon is emitted with momentum P

Atom takes a kick in the opposite direction and moves with a speed v, to conserve momentum

(b) If the atom is 2 H , M ≃ mass of 1 proton + mass of 1 neutron = m p + mn (as mass of electron is negligible).

� �

1 1 8hR Energy of the recoiling atom = M v 2 = M 2 2 9M

2

.

This is a fraction of the energy of the emitted photon. The fraction is,

� �

1 8hR M 2 9M

2

·

9 4hR = 8hcR 9Mc 4 × 6.67 × 10 = 9 × (1.67 × 10 = 3.2 × 10 9

−34

−27

Js × 1.097 × 107 m 1 kg × 2) × 3 × 108 m/s −

−

−7

= 3.2 × 10

%.

3. Apply classical mechanics to an electron in a stationary state of hydrogen to show that L2 = m e ke 2r and L3 = m e k2 e4 /ω. Here k is the Coulomb constant, L is the magnitude Date: 11, March 2013

3



of the orbital angular momentum of the electron, and me , e, r, and ω are the mass, charge, orbit radius, and orbital angular frequency of the electron, respectively. Answer 3:

Magnitude of angular momentum is given by, L = rP = r(me v) = rme v ⇒

L2 = r 2m2e v 2 .

(1)

Coulomb’s force between electron and nucleus is, e2 F c = k 2 . r This force must be equal to the centripetal force of the electron, e2 v2 k 2 = me × r r 2 ke v2 = . me r

(2)

Use value of v 2 from equation (2) in equation (1), 2

2

2

2

L = r me

� ke �

me r = me ke2 r, as required.

Now L3 can be written as, L3 = L2 × L = me ke 2r × rme v = m 2e ke2 r2 v m2e ke2 r2 2 = ×v v 2 me ke2 r2 ke2 × = from equation (2) v me r r 1 = me k2 e4 × = m e k2 e4 × v ω 2 4 me k e L3 = , as required. ω 4. Wavelengths of spectral lines depend to some extent on the nuclear mass. This occurs because the nucleus is not an infinitely heavy stationary mass and both the electron and nucleus actually revolve around their common center of mass. It can be shown Date: 11, March 2013

4



that a system of this type is entirely equivalent to a single particle of reduced mass µ that revolves around the position of the heavier particle at a distance equal to the electronnucleus separation. See Figure below. (a)

(b)

ω M ω

ω

+ CM

r

Nucleus

r

(at rest)

me Electron

Moving particle of reduced mass µ

(a) Both the electron and the nucleus actually revolve around the center of mass. (b) To calculate the effect of nuclear motion, the nucleus can be considered to be at rest and m e is replaced by the reduced mass µ .

Here, µ = m e M/(me + M ), where m e is the electron mass and M is the nuclear mass. To take the moving nucleus into account in the Bohr theory we replace me with µ. Thus equation for allowed energy levels becomes, E n =

−µke

2

2me a0

�1� n2

,

and equation for emitted wavelength becomes, 1 µke2 1 = λ 2me a0 hc n2f

�

−

� �µ� �1

1 = n2i

me

R

n2f

−

�

1 . n2i

Determine the corrected values of wavelength for the first Balmer line (n = 3 to n = 2 transition) taking nuclear motion into account for (a) hydrogen, 1 H , (b) deuterium, 2

H , and (c) tritium, 3 H . (Deuterium, was actually discovered in 1932 by Harold Urey,

who measured the small wavelength difference between 1 H and 2 H .) Answer 4: For Hydrogen:

First calculate the reduced mass. µ = Date: 11, March 2013

me M , me + M 5



where me = mass of electron = 9.11 × 10 −27

10

−31

kg and M = mass of hydrogen = 1.67 ×

kg. 9.11 × 10 31 kg × 1.67 × 10 ⇒ µ = 9.11 × 10 31 kg + 1.67 × 10 1.52 × 10 57 kg2 = 1.67 × 10 27 kg = 9.09 × 10 31 kg. −

−27

−

−27

kg kg

−

−

−

Wavwlength of the first Balmer line will be,

� µ � � 1

�

� �

1 5 µ R 2 − 2 = R me 2 3 36 me 5 9.09 × 10 31 kg 7 = × × 1.1 × 10 m 31 36 9.11 × 10 kg 1 = 1524423.71 m

1 = λ

−

−1

−

−

λ = 6.56 × 10

−7

m

= 656 nm. For Deuterium:

For deuterium M = 2 × 1.67 × 10 ⇒

µ =

−27

kg = 3.34 × 10

−27

kg.

9.11 × 10 31 kg × 3.34 × 10 27 kg 9.11 × 10 31 kg + 3.34 × 10 27 kg 3.04 × 10 57 kg2 3.34 × 10 27 kg 9.10 × 10 31 kg 5 µ R 36 me 5 9.10 × 10 31 kg 7 × × 1.1 × 10 m 36 9.11 × 10 31 kg 1526100.744 m 1 −

−

−

−

−

= = ⇒

1 = λ

−

−

� �

−

= =

λ = 6.55 × 10

−1

−

−

−7

m

= 655 nm. For Tritium:


6



For tritium M = 3 × 1.67 × 10 ⇒

µ =

−27

kg = 5.01 × 10

−27

kg.

9.11 × 10 31 kg × 5.01 × 10 27 kg 9.11 × 10 31 kg + 5.01 × 10 27 kg 4.564 × 10 57 kg2 5.01 × 10 27 kg 9.11 × 10 31 kg 5 µ R 36 me 5 9.11 × 10 31 kg 7 × × 1.1 × 10 m 31 36 9.11 × 10 kg 1527777.778 m 1 −

−

−

−

−

= = ⇒

1 = λ

−

−

� �

−

= =

−1

−

−

λ = 6.54 × 10

−7

m

= 654 nm. The small wavelength difference between 1 H , 2 H and 3 H in fact is a signature of the nuclear mass and can be used as technique for isotopic identification. 5. An electron with kinetic energy less than 100 eV collides head-on in an elastic collision with a massive mercury atom at rest. (a) If the electron reverses direction in the collision (like a ball hitting a wall), show that the electron loses only a tiny fraction of its initial kinetic energy, given by, ∆K 4M = , K me (1 + M/m e )2 where me is the electron mass and M is the mercury atom mass. (b) Using the accepted values for me and M , show that, ∆K 4me = , K M and calculate the numerical value of ∆K/K . Answer 5: Before collision m ev

After collision -m ev

MV


7



(a) We are given that, Initial kinetic energy of electron = K = 100 eV Initial kinetic energy of mercury atom = 0 Let energy loss by electron = ∆K Final kinetic energy of electron = K − ∆K Final kinetic energy of mercury atom = ∆K. Since collision between mercury atom and electron is elastic, both the energy and momentum will be conserved. Let v and v be the speed of electron before and after ′

collision respectively, and let V be the speed of mercury atom after collision, then according to law of conservation of energy, 1 1 1 2 me v 2 + 0 = me v + MV 2 2 2 2 M 2 2 v2 = v + V . me ′

′

(3)

According to law of conservation of momentum, ′

me v + 0 = (−me v ) + MV M v = −v + V. me ′

(4)

Using the value of v from equation (4) in equation (3), we obtain,

� v

′2

M V −v + me

� M � + V m � M �

′

2

me

2 ′2

= v +

M 2 V me

−

2

M M 2 2 v V = v + V me me

−

2

M M 2 v V = V me me

e

V

�

2

′

′

′

M V − 2v = V me 1 M v = 2 me ′

′


�

−

�

1 V.

(5)

8



Inserting value of v from equation (5) to equation (3), we obtain, ′

2

v

2

� 1 � M � � M = 1 V + V 2 m m �� M � � M 1 M = +1 2 V + V 4 m m m �� M � � 1 M M = +1 2 +4 V 4 m m m �� M � � 1 � M � 1 M = +1+2 + 1 V V = 4 m m 4 m � � 1 M 2

−

e

e

2

2

−

e

2

e

e

2

2

−

e

e

e

2

2

2

e

v = V =

2 me 2v

�

M m

e

e

2

e

+ 1 V

+1

�.

The equation of energy conservation will become, v2 = v

′2

=

2 M 2v 2 v + me M/m e + 1 M 4v 2 2 v − me (M/m e + 1)2 M (M/m e + 1)2 − 4 m v2 2 (M/m e + 1) (M/m e − 1)2 2 v . (M/m e + 1) 2 ′

� �

� = � =

�

�

e

�

�

Energy transferred to the mercury atom will be, 1 1 2 me v 2 − me v 2 2 1 1 (M/m e − 1)2 2 2 = me v − me v 2 2 (M/m e + 1) 2 1 (M/m e − 1)2 2 = me v 1 − 2 (M/m e + 1)2 1 (M/m e + 1) 2 − (M/m e − 1)2 2 = me v . 2 (M/m e + 1)2 ′

∆K =

�

� � �

� ��

�

Use identity, (a + b)2 − (a − b)2 = 4ab,

�

�

1 4M/m e ∆K = me v 2 . 2 (M/m e + 1)2


9



Since K = mv 2 /2, therefore,

�

� �

4M/m e ∆K = K (M/m e + 1) 2 ∆K 4M = , K me (M/m e + 1)2

�

as required.

(b) We know that, Mass of electron = me = 9.11 × 10 Mass of mercury atom = M = 3.21 × 10 Since M ≫ me ,

⇒

1 + M/m e

≈

−31

−25

kg kg.

M/m e . Therefore,

∆K = K

�

4M me (M/m e )2 4me = . M

�

Numerical value of ∆K/K will be, ∆K 4 × 9.11 × 10 31 kg = K 3.21 × 10 25 kg = 1.1 × 10 5 . −

−

−


10

Soluciones 1

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