Photoelectric Effect

Chapter # 42

Photoelectric Effect & Wave Particle duality

Solved Examples 42.1

Sol.

Consider a parallel beam of light of wavelength 600 nm and intensity 100 W/m2. (a) Find the energy and linear momentum of each photon. (b) How many photons cross 1 cm2 area perpendicular to the beam in one second ? (a) The energy of each photon E = hc/  =

( 4.14  10 15 eV  s)  (3  10 8 m / s)

600  10  9 m The linear momentum is

 2.07 eV.

E 2.07eV  = 0.69 × 10-8 eV –s/m. c 3  10 8 m / s (b) The energy crossing 1 cm2 in one second = (100 W/m2) × (1 cm2) × (1 s) = 1.0 × 10-2 J. The number of photons making up this amount of energy is p

1.0  10 2 1.0  10 2 J = = 3.0 × 1016. 2.07eV 2.07  1.6  10 19 For a given wavelength , the energy of light is an integer times hg/ Thus, the energy of light can be varied n=

only in quantums (steps) of 42.2 Sol.

hc . The photon theory is, therefore, also called the quantum theory of light. 

Find the maximum wavelength of light that can cause photoelectric effect in lithium. From table (42.1), the work function of lithium is 2.5eV. The threshold wavelength is  = hc/  =

( 4.14  10 16 eV  s)  (3  10 8 m / s) 2.5 eV

This is the required maximum wavelength. 42.3

Sol.

A point source of monochromatic light of 1.0 mW is placed at a distance of 5.0 m from a metal surface. Light falls perpendicularly on the surface. Assume wave theory of light to hold and also that all the light falling on the circular area with radius = 1.0 × 10-9 m (which is few times the diameter of an atom) is absorbed by a single electron on the surface. Calculate the time required by the electron to receive sufficient energy to come out of the metal if the work function of the metal is 2.0 eV. The energy radiated by the light source per second is 1.0 mJ. This energy is spread over the total solid angle 4. The solid angle subtended at the source by the circular area mentioned is

d 

dA r2



  (1.0  10 9 m)2 (5.0m)2



  10 18 sr. 25

Hence the energy heading towards the circular area per second is t=

2.0  1.6  10 19 J 10  20 mJ / s

= 3.2 × 104 s = 8.8 hours.

Worked out Solved Examples 1. Sol.

How many photons are emitted per second by a 5 mW laser source operating at 632.8 nm ? The energy of each photon is E

=

hc 

(6.63  10 34 J  s)  (3  10 8 m / s) (632.8  10 9 m)

= 3.14 × 10–19 J. manishkumarphysics.in

Page # 1

Chapter # 42

Photoelectric Effect & Wave Particle duality

The energy of the laser emitted per second is 5 × 10-3 J. Thus the number of photons emitted per second 5  10 3 J

2.

= 1.6 × 1016. 3.14  10 19 J A monochromatic source of light operating at 200 W emits 4 × 1020 photons per second. Find the wavelength of light.

Sol.

The energy of each photon =

=

200 J / s

4  10 20 / s = 5 × 10-19 J.

Wavelength =  

=

hc E

(6.63  10 34 J  s)  (3  10 8 m / s) (5  10 19 J)

= 4.0 × 10 -7 m = 400 nm. 3. Sol.

A hydrogen atom moving at a speed v absorbs a photon of wavelength 122 nm and stops. Find the value of  Mass of a hydrogen atom = 1.67 × 10-27 kg. The linear momentum of the photon h 6.63  10 34 J  s  = 5.43 × 10-27 kg–m/s.  122  10 9 m As the photon is absorbed and the atom stops, the total final momentum is zero. From conservation of linear momentum, the initial momentum must be zero. The atom should move opposite to the direction of motion of the photon and they should have the same magnitudes of linear momentum. Thus, (1.67 × 10 -27 kg) = 5.43 × 10-27 kg–m/s

=

or, 4.

Sol.



5.43  10 27 1.67  10 27

m/s = 3.25 m/s.

A parallel beam of monochromatic light of wavelength 500 nm is incident normally on a perfectly absorbing surface. The power through any cross-section of the beam is 10 W. Find (a) the number of photons absorbed per second by surface and (b) the force exerted by the light beam on the surface. (a) The energy of each of photon is

E

hc ( 4.14  10 15 eV  s)  (3  10 8 m / s)   500nm

1242 eV  nm = 2.48 eV.. 500nm In one second, 10 J of energy passes through any cross-section of the beam. Thus, the number of photons crossing a cross-section is

=

10 J n = 2.48 eV = 2.52 × 1019. This is also the number of photons falling on the surface per second and being absorbed. (b) the linear momentum of each photon is h hv  .  c The total momentum of all the photons falling per second on the surface is

p=

nhv 10 J 10 J   = 3.33 × 10 - 8 N-s. c c 3  10 8 m / s As the photons are completely absorbed by the surface, this much momentum is transferred to the surface per second. The rate of change of the momentum of the surface, i.e., the force on it is =

F=

dp 3.33  10  8 N  s  = 3.33 × 10 - 8 N. dt 1s manishkumarphysics.in

Page # 2

Chapter # 42 5.

Sol.

Photoelectric Effect & Wave Particle duality

Figure shows a small, plane strip suspended from a fixed support through a string of length l. A continuous beam of monochromatic light is incident horizontally on the strip and is completely absorbed. The energy falling on the strip per unit time is W. (a) Find the deflection of the string from the vertical if the mirror stays in equilibrium. (b) If the strip is deflected slightly from its equilibrium position in the plane of the figure, what will be the time period of the resulting oscillations ? (a) The linear momentum of the light falling per unit time on the strip is W/c. As the light in incident on the strip, its momentum imparted to the strip per unit time is thus W/c. This is equal to the force on the strip by the light beam. In equilibrium, the force by the light beam, the weight of the strip and the force due to tension add to zero. If the string makes an angle with the vertical,  l light

and

T cos  = mg T sin  = W/c.

Thus,

tan  

W mgc .

(b) In equilibrium, the tension is 2  W  T  (mg)2       c  

1/ 2

1/ 2

2 T  2 W    g     or, m   mc    This plays the role of effective g. The time period of small oscillations is

t  2

   2 1/ 4 T/m 2  W   2 g      mc   

6.

A point source of light is placed at the centre of curvature of a hemispherical surface. The radius of curvature is r and the inner surface is completely reflecting. Find the force on the hemisphere due to the light falling on it if the source emits a power W.

Sol.

The energy emitted by the source per unit time, i.e., W falls on an area 4r 2 at a distance r in unit time. Thus, the energy falling per unit area per unit time is W/( 4r 2 ). Consider a small area dA at the point P of the hemisphere (figure). The energy falling per unit time on it is

W dA 4r 2

. The corresponding momentum incident

dA cos 0



p dA X on this area per unit time is

W dA 4r 2c

. As the light is reflected back, the change in momentum per unit time,

i.e., the force on dA is

dF 

2W dA

. 4r 2c Suppose the radius OP through the area dA makes an angle  with the symmetry axis OX. The force on dA is along this radius. By symmetry, the resultant force on the hemisphere is along OX. The component of dF manishkumarphysics.in

Page # 3

Chapter # 42

Photoelectric Effect & Wave Particle duality

along OX is

dF cos  

2W dA

dF cos  

2W

cos  4r 2c If we project the area dA on the plane containing the rim, the projection is dA cos. Thus, the component of dF along OX is, 4r 2c The net force along OX is F

2W 4r 2c

(projection of dA).

  projection of dA 

2 When all the small areas dA are projected, we get the area enclosed by the rim which is r . Thus,

F

7.

2W 4r c 2

 r 2 

W . 2c

A perfectly reflecting solid sphere of radius r is kept in the path of a parallel beam of light of large aperture. If the beam carries an intensity , find the force exerted by the beam on the sphere.

Sol.

Let O be the centre of the sphere and OZ be the line opposite to the incident beam (figure). Consider a radius OP of the sphere making an angle  with OZ. Rotate this radius about OZ to get a circle on the sphere. Change  to  + d and rotate the radius about OZ to get another circle on the sphere. The part of the sphere between these circles is a ring of area 2r2 sin d. Consider a small part A of this ring at P. Energy of the light falling on this part in time t is U = t(A cos ) The momentum of this light falling on A is U/c along QP. The light is reflected by the sphere along PR. The change in momentum is U 2 cos = t (A cos2 ) c c along the inward normal. The force on A due to the light falling on it, is

p = 2

p 2 =  A cos2 . t c This force is along PO. The resultant force on the ring as well as on the sphere is along ZO by symmetry. The component of the force on A along ZO p 2 cos  = A cos3 . t c The force acting on the ring is 2 (2r2 sin d)cos3 . c The force on the entire sphere is

dF =

/2

F=

 0

4r 2  cos3  sin  d c

/2

= 

 0

4 r 2  cos3  d(cos ) c

manishkumarphysics.in

Page # 4

Chapter # 42

Photoelectric Effect & Wave Particle duality /2

= 



0

4 r 2  c

/2

 cos 4      4  0

=

r 2  c

Note that integration is done only for the hemisphere that faces the incident beam. 8. Sol.

Find the threshold wavelengths for photoelectric effect from a copper surface, a sodium surface and a cesium surface. The work functions of these metals are 4.5 eV, 2.3eV and 1.9eV respectively. If 0 be the threshold wavelength and  be the work function,

0 

hc  =

9. Sol.

1242 eV  nm . 

For copper,  0 

1242 eV  nm = 276 nm. 4.5 eV

For sodium,  0 

1242 eV  nm = 540 nm. 2.3 eV

For cesium,  0 

1242 eV  nm = 654 nm. 1.9 eV

Ultraviolet light of wavelength 280 nm is used in an experiment on photoelectric effect with lithium (= 2.5 eV) cathode. Find (a) the maximum kinetic energy of the photoelectrons and (b) the stopping potential. (a) The maximum kinetic energy is Kmax=

hc  

=

1242 eV  nm – 2.5 eV 280 nm

= 4.4 eV – 2.5 eV = 1.9 eV. (b) Stopping potential V is given by eV = Kmax or, 10.

Sol.

V=

K max 1.9 eV  = 1.9 V.. e e

In a photoelectric experiment, it was found that the stopping potential decreases from 1.85 V to 0.82 V as the wavelength of the incident light is varied from 300 nm to 400 nm. Calculate the value of the planck constant from these data. The maximum kinetic energy of a photoelectron is hc   and the stopping potential K max 

K max hc    . e e e If V1, V2 are the stopping potential at wavelengths 1 and 2 respectively. V

V1 

hc   1e e

and

V2 

hc   .  2e e

This gives,

V1  V2 

hc  1 1     e  1  2  manishkumarphysics.in

Page # 5

Chapter # 42

Photoelectric Effect & Wave Particle duality e( V1  V2 )  1 1 c    1  2 

h

or,

e(1.85 V  0.82 V ) 1 1   = c    9  9  300  10 m 400  10 m 

1.03eV = (3  10 8 m / s) 1  10 7 m 1  12  = 4.12 × 10-15 eV-s. 11.

Sol.

A beam of 450 nm light is incident on a metal having work function 2.0 eV and placed in a magnetic field B. The most energetic electrons emitted perpendicular to the field are bent in circular arcs of radius 20 cm. Find the value of B. The kinetic energy of the most energetic electrons is

K

=

hc  

1242 eV  nm – 2.0 eV 450 nm

= 0.76 eV = 1.2 × 10 -19 J. The linear momentum = m   2mK When a charged particle is sent perpendicular to a magnetic field, it goes along a circle of radius

r

Thus, 12.

Sol.

0.20 m =

m qB

4.67  10  25 kg  m / s (1.6  10 19 C)  (0.20 m)

= 1.46 × 10-5 T..

A monochromatic light of wavelength  is incident on an isolated metallic sphere of radius a. The threshold wavelength is 0 which is larger than . Find the number of photoelectrons emitted before the emission of photoelectrons will stop. As the metallic sphere is isolated, it becomes positively charged when electrons are ejected from it. There is an extra attractive force on the photoelectrons. If the potential of the sphere is raised to V, the electron should have a minimum energy  + eV to be able to come out. Thus, emission of photoelectrons will stop when

hc    eV  hc =   eV 0 or,

hc  1 1  V = e      0

The charge on the sphere needed to take its potential to V is

Q  ( 4 0 a)V. The number of electrons emitted is, therefore, n=

=

13.

Q 4 0 aV  e e

4 0 ahc  1 1     e 2   0 

Light described at a place by the equation E = (100 V/m) [sin (5 × 1015 s–1) t + sin (8 × 1015 s–1)t] manishkumarphysics.in

Page # 6

Chapter # 42

Sol.

Photoelectric Effect & Wave Particle duality

falls on a metal surface having work function 2.0 eV. Calculate the maximum kinetic energy of the photoelectrons. The light contains two different frequencies. The one with larger frequency will cause photoelectrons with largest kinetic energy. This larger frequency is

 8  1015 s 1  2 2 The maximum kinetic energy of the photoelectrons is Kmax = hv –  v= v 

 8  1015 1    = (4.14 × 10 eV-s) ×  2 s  – 2.0 eV   = 5.27 eV – 2.0 eV = 3.27 eV. –15

QUESTION FOR SHORT ANSWER 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

11. 12. 13.

Can we find the mass of a photon by the definition p = mv ? Is it always true that for two sources of equal intensity, the number of photons emitted in a given time are equal ? What is the speed of a photon with respect to another photon if (a) the two photons are going in the same direction and (b) they are going in opposite directions ? Can a photon be deflected by an electric field ? By a magnetic field ? A hot body is placed in a closed room maintained at a lower temperature. Is the number of photons in the room increasing ? Should the energy of a photon be called its kinetic energy or its internal energy ? In an experiment on photoelectric effect, a photon is incident on an electron from one direction and the photoelectron is emitted almost in the opposite direction. Does this violate conservation of momentum ? It is found that yellow light does not eject photoelectrons from a metal. Is it advisable to try with orange light ? With green light ? It is found that photosynthesis starts in certain plants when exposed to the sunlight but it does not start if the plant is exposed only to infrared light. Explain The threshold wavelength of a metal is 0. Light of wavelength slightly less than 0 is incident on an insulated plate made of this metal. It is found that photoelectrons are emitted for sometime and after that the emission stops. Explain. Is p = E/c valid for electrons ? Consider the de Broglie wavelength of an electron and a proton. Which wavelength is smaller if the two particles have (a) the same speed (b) the same momentum (c) the same energy ? If an electron has a wavelength, does it also have a colour ?

OBJECTIVE - I 1.

2.

Planck constant has the same dimension as (A) force x time (B) force x distance (C) force x speed Iykad fu;rkad dh foek,¡] fuEu ds leku gksrh gS (A) cy x le; (B) cy x foLFkkiu (C) cy x pky

(D*) force x distance x time (D*) cy x foLFkkiu x le;

Two photons having (A) equal wavelengths have equal linear momenta (B) equal energines have equal linear momenta (B) equal frequencies have equal linear momenta (D*) equal linear momenta have equal wavelength nks QksVkWu ftuds fy;s leku (A) rjaxnS/;Z gksrh gS] jSf[kd laosx leku gksrs gSaA (B) ÅtkZ,¡ gksrh gS] jSf[kd laosx leku gksrs gSA (B) vko`fÙk;k¡ gksrh gS] jSf[kd laosx leku gksrs gSAa (D*) jSf[kd laosx gksrs gSa] rjaxnS/;Z leku gksrh gSA manishkumarphysics.in

Page # 7

Chapter # 42

Photoelectric Effect & Wave Particle duality

3.

Let p and E denote the linear momentum and energy of a photon. If the wavelength is decreased (A*) both p and E increase (B) p increases and E decreases (C) p decreases and E increases (D) both p and E decrease ekukfd p rFkk E fdlh QksVksu dk jSf[kd laosx ,oa ÅtkZ O;Dr djrs gSAa ;fn rjaxnS/;Z de dh tk;s rks (A*) p rFkk E nksuksa gh c<+sxsa (B) p c<+sxk o E de gksxhA (C) p de gksxk o E c<+sxk (D) p o E nksuksa gh de gksxsaA

4.

Let nr and nb be reprectively the number of photons emitted by a red bulb and a blue bulb of equal power in a given time. (A) nr = nb (B) nr < nb (C*) nr > nb (D) the information is insufficient to get a relation between nr and nb ekukfd fdlh fuf'pr le; esa leku 'kfDr okys yky cYc rFkk uhys cYc ls mRlftZr QksVkWuksa dh la[;k Øe'k% nr o nb gS] rks (A) nr = nb (B) nr < nb (C*) nr > nb (D) nr o nb esa laca/k izkIr djus ds fy;s nh x;h lwpuk vi;kZIr gSA

5.

The equation E = pc is valid (A) for an electron as well as for a photon (C*) for a photon but nor for an electron lehdj.k E = pc lR; gS (A) bysDVªkWu ds fy;s o lkFk gh QksVkWu ds fy;s (C*) QksVkWu ds fy;s fdUrq bysDVªkWu ds fy;s ugha

(B) for an electron but not for a photon (D) neither for an electron nor for a photon (B) bysDVªkWu ds fy;s fdUrq (D) u rks bysDVªkWu ds fy;s

QksVkWu ds fy;s ugha vkSj u gh QksVkWu ds fy;sA

6.

The work function of a metal is hv0. Light of frequency v falls on this metal. The photoelectric effect will take place only if fdlh /kkrq ds fy;s dk;Z Qyu hv0 gSA bl /kkrq ij v vko`fÙk dk izdk'k vkifrr gksrk gSA izdk'k fo|qr izHkko mRiUu gksxk] ;fn (A*) v  v0 (B) v  2v0 (C) v  v0 (D) v  v0 /2

7.

Leigth of wavelength  falls on a metal having work function hc/0. Photoelectric effect will take place only if  rjaxnS/;Z dk izdk'k ,d /kkrq ij vkifrr gS] ftldk dk;ZQyu hc/0 gSA izdk'k fo|qr izHkko dsoy rc gh mRiUu gksxk] ;fn (A) 0 (B) 0 (C*) 0 (D) 0 /2.

8.

When stopping potential is applied in an experiment on photoelectric effect, no photocurrent is observed. This means that (A) the emission of photoelectrons is stopped (B*) the photoelectrons are emitted but are reabsorbed by the emitter metal (C) the photoelectrons are accumulated near the collector plate (D) the photoelectrons are dispersed from the sides of the apparatus.

izdk'k fo|qr izHkko iz;ksx esa tc fujks/kd foHko vkjksfir fd;k tkrk gS] dksbZ izdk'k fo|qr /kkjk izfs {kr ugha gksrh gSA bldk rkRi;Z gS (A) QksVks bysDVªkWuksa dk mRiknu can gks x;k gSA (B*) QksVks bysDVªkWuksa mRiUu gks jgs gSa] fdUrq mRltZd /kkrq }kjk iqu% vo'kksf"kr fd;s tk jgs gSaA (C) QksVks bysDVªkWuksa laxkz gd IysV ds lehi ,df=kr gks jgs gSAa (D) QksVks bysDVªkWu midj.k ds fdukjksa ls ckgj foyksfir gks jgs gSaA 9.

If the frequency of light in a photoelectric experiment is double, the stopping potential will (A) be doubled (B) be halved (C*) become more than double (D) become less than double

izdk'k fo|qr izHkko iz;ksx esa ;fn vkifrr izdk'k dh vko`fÙk nqxuh dj nh tk;s] rks fujks/kd foHko gks tk;sxk (A) nqxuk (B) vk/kk (C*) nqxus ls vf/kd (D) nqxus ls de 10.

The frequency and intensity of a light source are both doubled, the stopping potential statements. (a) The saturation photocurrent remains almost the same. (b) The maximum kinetic energy of the photoelectrons is doubled (A) Both (a) and (b) are true. (C) (a) is false but (b) is true

(B*) (a) is true but (b) is false (D) Both (a) and (b) are false manishkumarphysics.in

Page # 8

Chapter # 42

Photoelectric Effect & Wave Particle duality

izdk'k lzksr dh vko`fÙk ,oa rhozrk nksuksa nqxuh dj nh tkrh gSA fuEu dFkuksa ij fopkj dhft;s (a) larI` r izdk'k fo|qr /kkjk yxHkx leku jgrh gSA (b) QksVks bysDVªkWuksa dh xfrt ÅtkZ nqxuh gks tkrh gSA (A) (a) vkSj (b) nksuksa gh lR; gSA (C) (a) vlR; gS] fdUrq (b) lR; gSA 11.

(B*) (a) lR; fdUrq (b) vlR; (D) (a) o (b) nksuksa gh vlR;

gSA gSA

A point source of ligth is used in a photoelectric effect. If the source is removed farther from the emitting metal, the stopping potential (A) will increase (B) will decrease (C*) will remain constant (D) will either increase or decrease

izdk'k fo|qr izHkko esa izdk'k dk ,d fcUnq lzkrs iz;D q r fd;k x;k gSA ;fn lzkrs dks mRltZd /kkrq ls nwjh ys tk;k tk;s rks fujks/kd foHko (A) c<+ tk;sxk (B) de gks tk;sxk (C*) fu;r jgsxk (D) c<+ tk;sxk vFkok de gks tk;sxk 12.

A point source causes photoelectric effect from a small metal plate. Which of the following curves may represent the saturation photocurrent as a function of the distance between the source and the metal ?

,d fcUnq lzkrs ] /kkrq dh ,d NksVh IysV ls /kkrq dh lrg ds chp nwjh ds Qyu ds :i esa larI` r izdk'k fo|qr /kkjk dks O;Dr djus okyk Qyu gSA

(A) a

(B) b

(C) c

(D*) d

13.

A nonmonochromatid light is used in an experiment on photoelectric effect. The stopping potential (A) is related to the mean wavelength (B) is related to the longest wavelength (C*) is related to the shortest wavelenth (D) is not related to the wavelength ,d izdk'k fo|qr izHkko iz;ksx esa cgqo.khZ izdk'k iz;D q r fd;k tkrk gSA fujks/kd foHko (A) ek/; rjaxnS/;Z ls lacaf/kr gSA (B) lcls yEch rjaxnS/;Z ls lacfa /kr gSA (C*) lcls NksVh rjaxnS/;Z ls lacaf/kr gSA (D) rjaxnS/;Z ls lacaf/kr ugha gSA

14.

A proton and an electron are acceleration by the same potential diffenrence. Let e and p denote the de Broglie wavelength of the electron and the proton respectively (A) e = p (B) e < p (C*) e > p (D) The relation between e and p depends on the acceleration potential difference

,d izkVs kWu rFkk ,d bysDVªkWu leku foHkokarj ls Rofjr gSA ekuk fd bysDVªkWu rFkk izksVkWu dh Mh&czkxs yh rjaxnS/;Z Øe'k% e rFkk p gS (A) e = p (D) e rFkk p esa

(B) e < p

lac/a k Rojd foHkokarj ij fuHkZj djrk gSA

(C*) e > p

OBJECTIVE - II 1.

When the intensity of a light source is increased, (A*) the number of photons emitted by the source in unit time increases (B*) the total energy of the photons emitted per unit time increases (C) more energetic photons are emitted (D) faster photons are emitted tc izdk'k lzkrs dh rhozrk c<+k;h tkrh gS (A*) ,dkad le; esa lzksr ls mRlftZr QksVkWuksa dh la[;k c<+rh gSA (B*) ,dkda le; esa mRlftZr QksVkWu dh dqy ÅtkZ c<+ tkrh gSA (C) vf/kd ÅtkZ ds QksVkWu mRlftZr gksrs gSaA (D) vf/kd rst xfr'khy QksVkWu mRlftZr gksrs gSaA

2.

Photoelectric effect supports quantum nature of light because (A*) there is a minimum frequency below which no photoelectrons are emitted manishkumarphysics.in

Page # 9

Chapter # 42

Photoelectric Effect & Wave Particle duality

(B*) the maximum kinetic energy of photoelectrons depends only on the frequency of light and not on its intensity (C*) even when the metal surface is faintly illuminated the photoelectrons leave the surface imme diately (D) electric charge of the photoelectrons is quantized izdk'k fo|qr izHkko izdk'k dh Dok.Ve izÑfr dh iqf"V djrk gS] D;ksfa d (A*) ,d fuf'pr vko`fÙk ls de vko`fÙk ij dksbZ QksVkW bysDVªkWu mRlftZr ugha gksrk gSA (B*) QksVks bysDVªkuW ksa dh vf/kdre xfrt ÅtkZ dsoy vkifrr izdk'k dh vko`fÙk ij gh fuHkZj djrh gS] bldh rhozrk (C*) ;|fi vkifrr izdk'k cgqr /kq/a kyk gks rks Hkh QksVks bysDVªkWu lrg ls rqjr a mRlftZr gksus yxrs gSaA (D) QksVks bysDVªkWuksa dk fo|qr vkos'k Dok.VhÑr gksrk gSA 3.

4.

5.

6.

A photon of energy hv is absorbed by a free electron of a metal having work function < hv (A) The electron is sure to come out (B) The electron is sure to come out with a kineitc energy hv -  (C) Either the electrons does not come out or it comes out with a kinetic energy hv -  (D*) It may come out wiht a kinetic energy less than hv -  hv ÅtkZ okyk ,d QksVkWu] ,d /kkrq ds eqDr bysDVªkWu }kjk vo'kksf"kr fd;k tkrk gS] ftldk dk;ZQyu < hv (A) bysDVªkWu fuf'pr :i ls mRlftZr gksxkA (B) bysDVªkWu fuf'pr :i ls mRlftZr :i ls mRlftZr gksxkA ftldk xfrt ÅtkZ hv - gksxhA (C) ;k rks bysDVªkWu mRlftZr gh ugha gksxk ;k fQj ;g hv - ÅtkZ ds lkFk mRlftZr gksxkA (D*) ;g hv -  ls de ÅtkZ ds lkFk mRlftZr gks ldrk gSA If the wavelength of light in an experiment on photoelectric effect is doubled, (A) the photoelectric emission will not take place (B*) the exposure time is increased (C) the intensity of the source is decreased (D*) the exposure time is decreased ;fn fdlh izdk'k fo|qr iz;ksx esa izdk'k dh rjaxnS/;Z nqxuh dj nh tk;s (A) izdk'k fo|qr mRltZu ugha gksxk (B*) izdk'k fo|qr mRltZu gks Hkh ldrk gS vkSj (C) fujks/kd foHko c<+ tk;sxk (D*) fujks/kd foHko de gks tk;sxkA The photocurrent in an experiment on photoelectric effect increases if (A*) the intensity of the source is increased (B) the exposure time is increased (C) the intensity of the source is decreased (D) the exposure time is decreased izdk'k&fo|qr izHkko iz;ksx esa QksVks increases if (A*) lzksr dh rhozrk c<+ tkrh gS (B) mn~Hkklu le; c<+ tkrk gSA (C) lzkr s dh rhozrk de gks tkrh gSA (D) mn~Hkklu le; de gks tkrk

ij ughaA

gS &

ugha HkhA

gSA

The collector plate in an experiment on photoelectric effect is kept vertically above the emitter plate. Light source is put on and a saturation photocurrent is recorded. An electric field is switched on which has vertically downward direction (A) The photocurrent will increase (B*) The kinetic energy of the electrons will increase (C) The stopping potential will decrease (D) The threshold wavelength will increase

izdk'k fo|qr izHkko iz;ksx esa laxkz gd IysV] mRltZd IysV ij Bhd m/okZ/kj Åij j[kh gqbZ gSA ,d izdk'k lzkrs pkyw j[kk x;k gS rFkk QksVks /kkjk vafdr dh tk jgh gSA ,d fo|qr {ks=k m/okZ/kj uhps dh vksj fLop vkWu fd;k tkrk gS (A) QksVks /kkjk esa o`f) gksxh (B*) bysDVªkWuksa dh xfrt ÅtkZ c<+ tk;sxh (C) fujks/kd foHko de gks tk;sxk (D) nsgyh rjaxnS/;Z c<+ tk;sxh 7.

In which of the following situations the heavier of the two particles has smaller be Broglie wavelength? The two particles (A*) move with the same speed (B) move with the same linear momentum (C*) move with the same kinetic energy (D*) have fallen through the same height fuEu esa ls fdu ifjfLFkfr;ksa esa nks d.kksa esa ls Hkkjh d.k dh Mh&czkxs yh rjaxnS/;Z de gksxh\ nksuksa d.k (A*) ,d leku pky ls xfr'khy gSA (B) ,d leku jSf[kd laosx ds lkFk xfr'khy gSA (C*) ,d gh xfrt ÅtkZ ds lkFk xfr'khy gSA (D*) ,d gh Å¡pkbZ ls fxjk;s x;s gSaA

EXERCISE 1.

Visible light has wavelengths inthe range of 400 nm to 780 nm. Calculate the range of energy of the photons of visible light (in joules). n`'; izdk'k dh rjaxnS/;Z ijkl 400 nm ls 780 nm gSA n`'; izdk'k ds QksVkWu dh ÅtkZ dh ijkl Kkr djksA manishkumarphysics.in

Page # 10

Chapter # 42

Photoelectric Effect & Wave Particle duality HCV_Ch.42_1

Ans. 2.56 × 10–19 J to 5.00 × 10–19 J 2.

3.

Calculate the momentum of a photon of light of wavelength 500 nm. 500 nm rjaxnS/;Z okys çdk'k ds QksVkWu dh rjaxnS/;Z dh x.kuk dhft,A Ans. 1.33 × 10–27 kg-m/s

HCV_Ch.42_2

An atom absorbs a photon of wavelength 500 nm and emits another photon of wavelength 700 nm. Find the net energy absorbed by the atom in the process. ,d ijek.kq 500 nm dk QksVkWu vo'kksf"kr djrk gS rFkk 700 nm rjaxnS/;Z dk nwljk QksVkWu mRlftZr djrk gSA bl

çfØ;k esa ijek.kq }kjk dqy vo'kksf"kr Å"ek D;k gksxh \

HCV_Ch.42_3

Ans. 1.1 × 10–19 J 4.

Calculate the number of photons emitted per second by a 10 W sodium vapour lamp. Assume that 60% of the consumed energy is converted into light. Wavelength of sodium light = 590 nm. HCV_Ch.42_4 10 W ds lksfM;e ok"i ySEi }kjk mRlftZr QksVkWu dh la[;k dh x.kuk dhft,A ;g ekfu, fd vo'kksf"kr Å"ek dk 60% Hkkx izdk'k esa cny tkrk gSA lksfM;e izdk'k dh rjaxnS/;Z = 590 nm. Ans. 1.77 × 1019

5.

When the sun is directly overhead, the surface of the earth receives 1.4 × 103 W/m 2 of sunlight. Assume that the light is monochromatic with average wavelength 500 nm and that no light is absorbed in between the sun and the earth’s surface. The distance between the sun and the earth is 1.5 × 1011 m. (a) Calculate the number of photons falling per second on each square metre of earth’s surface directly below the sun. (b) How many photons are there in each cubic metre near the earth’s surface at any instant ? (c) How many photons does the sun emit per second ? tc lw;Z flj ds ,dne Åij gksrk gS rks i`Foh dh lrg 1.4 × 103 okWV@eh2 lkSj izdk'k xzg.k djrh gSA eku yhft;s fd izdk'k ,do.khZ gS] ftldh vkSlr rjaxnS/;Z 500 nm gS rFkk lw;Z ,oa i`Foh dh lrg ds chp esa izdk'k vo'kksf"kr ugha gksrk gSA lw;Z ,oa i`Foh dh nwjh 1.5 × 1011 eh- gSA (a) lw;Z ds Bhd uhps i`Foh ds ,dkad {ks=kQy ij izfr lsd.M vkifrr gksus okys QksVkWuksa dh la[;k dh x.kuk dhft;sA (b) bl {k.k ij i`Foh dh lrg ds lehi izR;sd ?kuehVj esa fdrus QksVkWu gSa\ (c) lw;Z izfr lsd.M fdrus QksVkWu mRlftZr djrk gS\ Ans. (a) 3.35 × 1021 (b) 1.2 × 1013 (c) 9.9 × 1044

6.

A parallel beam of monochromatic light of wavelength 663 nm is incident on a totally reflecting plane mirror. The angle of incidence is 60° and the number of photons striking the mirror per second is 1.0 × 1019. Calculate the force exerted by the light beam on the mirror. iw.kZ ijkorZd lery niZ.k ij 663 nm rjaxnS/;Z okys ,d o.khZ; çdk'k dk lekukUrj iqt a vkifrr fd;k tkrk gSA vkiru dks.k 60° gS rFkk çfr lSd.M niZ.k ls Vdjkus okys QkWVksu dh la[;k 1.0 × 1019 gSA niZ.k ij fxjus okys çdk'k iqat ds

dkj.k yxus okys cy dh x.kuk dhft,A

HCV II_Ch-42_6

Ans. 1.0 × 10–8 N 7.

8.

A beam of white light is incident normally on a plane surface absorbing 70% of the light and reflecting the rest. If the incident beam carries 10 W of power, find the force exerted by it on the surface. lQsn çdk'k dk iqt a tc yEcor~ :i ls ,d lery lrg ij fxjrk gS rks ;g lrg 70% çdk'k dks vo'kksf"kr djrh gS rFkk 'ks"k dks ijkofrZr dj nsrh gSA vxj lrg ij vkifrr çdk'k dh 'kfDr 10 W gS rks blds }kjk lrg ij yxk, x, cy dk eku crkb, \ HCV_Ch.42_Ex._7 Ans. 7.43 × 10–8 N A totally reflecting, small plane mirror placed horizontally faces a parallel beam oflight as shown in figure. The mass ofthe mirror is 20g. Assume that there isno absorption in the lens and that 30% of the light emitted by the source goes through the lens. Find the power of the source needed to support the weight of the mirror. Take g = 10 m/s 2 .

,d iw.kZr;k ijkorZd NksVk niZ.k {kSfrt :i ls izdk'k ds lekUrj iqt a ds lkeus fp=kkuqlkj j[kk x;k gSA niZ.k dk nzO;eku manishkumarphysics.in

Page # 11

Chapter # 42

Photoelectric Effect & Wave Particle duality

20 xzke

gSA ;g eku yhft;s fd ySl a }kjk dksbZ vo'kks"k.k ugha gksrk gS rFkk lzkrs ls mRlftZr izdk'k dk 30% ySl a ls xqtjrk gSA niZ.k ds Hkkj dks larfq yr djus ds fy;s vko';d izdk'k lzkrs dh 'kfDr Kkr dhft;sA g = 10 m/s2 gSA

9.

A 100 W light bulb is placed at the centre of a spherical chamber of radius 20cm. Assume that 60% of the energy supplied to the bulb is converted into light and that the surface of the chamber is absorbing. Find the pressure exerted by the light on the surface of the chamber. 20 lseh f=kT;k okys xksykdkj d{k ds dsUnz ij 100 okWV dk izdk'k cYc j[kk x;k gSA ;g eku yhft;s fd cYc dks nh x;h 60% ÅtkZ izdk'k esa ifjofrZr gks tkrh gS rFkk d{k dh lrg iw.kZr;k vo'kks"kd gSA d{k dh lrg ij izdk'k }kjk yxk;k x;k

nkc Kkr dhft;sA Ans: 10.

100 MW A sphere of radius 1.00 cm is placed in the path of a parallel beam of large aperture. The intensity of the light is 0.50 W/cm If the sphere completely absorbs the radiation falling on it Find the force exerted by the light beam on the sphere. cgqr cM+s }kjd okys izdk'k ds lekUrj iqt a ds ekxZ esa 1.00 lseh f=kT;k dk ,d xksyk j[k x;k gSA izdk'k dh rhozrk 0.50

okWV@lseh2 gSA ;fn xksyk blds Åij vkifrr fofdj.k dks iw.kZr;k vo'kksf"kr dj ysrk gS] rks izdk'k iqt a }kjk xksys ij vkjksfir cy Kkr dhft;sA Ans: 11.

4.0 × 10 – 7 Pa Consider the situation described in the previous problem. Show that the force on the sphere due to the light falling on it is the same even if the sphere is not perfectly absorbing.

fiNys iz'u esa of.kZr fLFkfr ij fopkj dhft;sA iznf'kZr dhft;s fd xksyk iw.kZr;k vo'kks"kd ugha gksxk rks Hkh vkifrr izdk'k ds dkj.k xksys ij cy leku gh gksxkA Ans : 12.

5.2 × 10 – 9 N Show that it is not possible for a photon to be completely absorbed by a free electron.

iznf'kZr dhft;s fd ,d eqDr bysDVªkWu ds }kjk QksVkWu dk iw.kZr;k vo'kks"k.k laHko ugha gSA Ans : 13

Two neutral particles are kept 1m apart .Suppose by some mechanism some charge is transferred from one particle to the other and the electronic potential energy lost is completely converted into a photon .Calculate the longest and the next smaller wavelength of the photon possible . nks mnklhu d.k 1 eh- nwj j[ks x;s gSAa ;g eku yhft;s fd fdlh izdkj ls ,d d.k ls nwljs d.k ij dqN vkos'k LFkkukarfjr

fd;k tkrk gS rFkk [kks;h gqbZ lEiw.kZ oS|rq fLFkfrt ÅtkZ] QksVkWu esa ifjofrZr gks tkrh gSA laHko QksVkWu dh lcls yEch rFkk mlls NksVh vxyh rjaxnS/;Z dh x.kuk dhft;sA Ans: 14.

Ans: 15.

Ans: 16.

17.

860 m, 215 m Find the maximum kinetic energy of the photoelectrons ejected when light of wavelength 350 nm is incident on a cesium surface .Work function of cesium = 1.9 ev lhft;e dh lrg ij 350 nm rjaxnS/;Z dk izdk'k vkifrr gksus ij mRlftZr gksus okys QksVkWu bysDVªkWuksa dh vf/kdre xfrt ÅtkZ Kkr dhft;sA lhft;e dk dk;Z Qyu = 1.9 ev 1.6 V The work function of a metal is 2.5 × 10–19 J (a) Find the threshold frequency for photoelectric emission (b) If The metal is exposed to a light beam of frequency 6.0 × 1014 Hz what will be the stopping potential is ,d /kkrq dk dk;ZQyu 2.5 × 10–19 twy gSA (a) izdk'k oS|rq mRltZu gsrq nsgyh vko`fÙk Kkr dhft;sA (b) ;fn /kkrq dks 6.0 × 1014 gV~t Z vko`fÙk ds izdk'k iqat ls mn~Hkkflr fd;k tk;s] fujks/kd foHko fdruk gksxk\ (a) 3.8 × 10 14 Hz (b) 0.91 V The work function of a photoelectric material is 4.0 eV. (a) What is the threshold wavelength ? (b) Find the wavlength of light for which the stopping potential is 2.5 V. ,d izdk'k oS|qr /kkrq dk dk;Z Qyu 4.0 eV gSA (a) nsgyh rjaxnS/;Z fdruh gS\ (b) ml izdk'k dh rjaxnS/;Z Kkr dhft;s] ftlds fy;s fujks/kd foHko 2.5 oksYV gSA Ans (a) 310 nm (b) 190 nm. Find the maximum magnitude of the linear momentum of a photoelectron emitted when light of wavelength manishkumarphysics.in

Page # 12

Chapter # 42

Photoelectric Effect & Wave Particle duality

400 nm falls on a metal having work function 2.5 eV 2.5 eV dk;ZQyu okyh /kkrq ij 400 nm rjaxnS/;Z dk izdk'k

dk vf/kdre ifjek.k Kkr dhft;sA Ans: 18.

Ans: 19.

vkifrr gksus ij mRlftZr QksVks bysDVªkWu ds jSf[kd laosx

4.2 ×10 – 25 kg/ ms When a metal plate is exposed to a monochromatic beam of light of wavelength 400 nm a negative potential of 1.1 is needed to stop the photocurrent .Find the threshold wavelength for the metal. tc ,d /kkrq ifV~Vdk] 400 nm rjaxnS/;Z okys ,d o.khZ izdk'k iat q ls mn~Hkkflr dh tkrh gS] izdk'k fo|qr /kkjkdks jksdus ds fy;s 1.1 oksYV _.kkRed foHko dh vko';drk gksrh gSA /kkrq ds fy;s nsgyh rjaxnS/;Z Kkr dhft;sA 620 nm n an experiment on potential of 1.1 V is needed to stop the photocurrent the data collected are as follows : wavelength (nm) : 350 400 450 500 550 stopping potential (v) 1.45 1.00 0.66 0.38 0.16 Plot the stopping potential against inverse of wavelength (1/ l ) on a graph paper and find (a) then planck constant, (b) the work function of the emitter and (c) the threshold wavelength.

izdk'k fo|qr izHkko ds ,d iz;ksx es]a fHkUu&fHkUu rjaxnS/;Z okyh ,d o.khZ izdk'k iqt a ksa ds fy;s fujks/kd foHko ekis x;s gSAa vkadM+s bl izdkj gSA rjaxnS/;Z (nm) : 350 400 450 500 550 fujks/kd foHko (V) : 1.45 1.00 0.66 0.38 0.16 ,d xzkQ isij ij rjaxnS/;Z ds O;qRØe ds lkis{k] fujks/kd foHko dk vkjs[k cukb;s rFkk Kkr dhft;s % (a) Iykad fu;rkad, (b) mRltZd dk dk;Z Qyu rFkk (c) nsgyh rjaxnS/;Z Ans : 20.

(a) 4.2 × 10 – 15 eV –s (b) 2.15 eV (c) 585 nm The electric field associated with a monochromatic beam becomes zero 1.2 × 1015 times per second Find the maximum kinetic energy of the photoelectrons when this light falls on a metal surface whose work functions is 2.0 eV ,d o.khZ iqt a ls lac/a k fo|qr {ks=k izfr lsd.M 1.2 × 1015 ckj 'kwU; gks tkrk gSA tc ;g izdk'k] 2.0 eV dk;ZQyu okyh

/kkrq dh lrg ij vkifrr gksrk gS rks QksVks bysDVªkWuksa dh vf/kdre xfrt ÅtkZ Kkr dhft;sA Ans : 21.

0.48 eV The electric field associated with a light wave is give by E = Eo sin [(1.57 × 107 m – 1 ) (x – ct).] Find the stopping potential when this lights is used in an experiment on photoelectric effect with the emitter having work function 1.9 eV. ,d izdk'k rjax ls laca) fo|qr {ks=k fuEu lw=k }kjk O;Dr fd;k tkrk gS - E = Eo sin [(1.57 × 107 m – 1 ) (x – ct)] tc bl izdk'k dk mi;ksx 1.9 eV dk;ZQyu okys mRltZd ds lkFk izdk'k fo|qr izHkko iz;ksx gsrq fd;k tkrk gS rks

fujks/kh foHko dk eku Kkr dhft;sA Ans: 22.

1.2 V The electric field at a point associated with a light wave is E = (100 V/m ) sin [3.0 × 1015 S – 1)t] sin [(6.0 × 10 15 s – 1)t] If this light falls on a metal surface having a work function of 2.0 eV , what will be the maximum kinetic energy of the photoelectrons ? fdlh fcUnq ij ,d izdk'k rjax ds lkFk laca) fo|qr {ks=k E = (100 oksYV/eh) sin [3.0 × 1015 S – 1)t] sin [(6.0 × 10 15 s – 1)t] gSA ;fn ;g 2.0 eV dk;ZQyu okyh /kkrq dh lrg ij vkifrr gksrk gS rks QksVks bysDVªkWuksa o

vf/kdre xfrt ÅtkZ fdruh gksxh\ Ans :

3.93 eV

23.

A monochromatic light source of intensity 5m W emits 8 × 1015 photons per second .This light ejects photoelectrons from a metal surface. The stopping potential for this setup is 2.0 V. Calculate the work function of the metal. 5 feyh okWV rhozrk dk ,d o.khZ izdk'k lzkr s izfr lsd.M 8 × 1015 QksVkWu mRlftZr djrk gSA ;g izdk'k ,d /kkrq dh lrg ls QksVks bysDVªkuW mRlftZr djrk gSA bl O;oLFkk ds fy;s fujks/kd foHko 2.0 oksYV gSA /kkrq ds dk;ZQyu dh x.kuk dhft;sA 1.9 eV Show figure is the plot of the stopping potential versus the frequency of the light used in an experiment on photoelectric effect .Find (a) ratio h/e and (b) the work function.

Ans : 24

,d izdk'k fo|qr iz;ksx esa iz;D q r izdk'k dh vko`fÙk ds lkis{k fujks/kd foHko dk xzkQ fp=k esa n'kkZ;k x;k gSA Kkr dhft;s% (a) vuqikr h/e rFkk (b) dk;Z Qyu

manishkumarphysics.in

Page # 13

Chapter # 42

Ans : 25.

Photoelectric Effect & Wave Particle duality

(a) 4.14 × 10 – 15 Vs (b) 0.414 eV A photographic film is coated with a silver bromide layer. When light falls on this film silver bromide molecules dissociate and the film records the light there. A minimum of 0.6 eV is needed to dissociate a silver bromide molecule .Find the maximum wavelength of light that can be recorded by the film

,d QksVksxkz fQd fQYe ij flYoj czkes kbM dh ijr p<+k;h x;h gSA tc bl fQYe ij izdk'k vkifrr gksrk gS rks flYoj czkes kbM ds v.kq fo;ksftr gks tkrs gSa rFkk fQYe ds ml Hkkx ij izdk'k lalfw pr gks tkrk gSA ,d flYoj czkes kbM v.kq dks fo;ksftr djus ds fy;s U;wure 0.6 eV dh vko';drk gksrh gSA og vf/kdre rjaxnS/;Z Kkr dhft;s tks bl fQYe }kjk lalfw pr dh tk ldrh gSA Ans: 26.

2070 nm In an experiment on photoelectric effect, light of wavelength 400 nm is incident on a cesium plate at the rate of 5.0 W .The potential of the collector plate is made sufficiently positive with respect to the emitter so that the current teaches its saturation value .Assuming that on the average one of every 106 photons is able to eject a photoelectron, find the photocurrent in the circuit . izdk'k fo|qr izHkko ds ,d iz;ksx es]a lhft;e IysV ij 400 nm rjaxnS/;Z dk izdk'k 5.0 okWV dh nj ls vkifrr gSA mRltZd ds lkis{k laxkz gd dk foHko bruk i;kZIr j[kk tkrk gS fd /kkjk larI` r eku izkIr dj ysrh gSA ;g ekurs gq, fd vkSlru 106

QksVkWu ij ,d QksVks bysDVªkWu dk mRltZu laHko gksrk gS] ifjiFk esa QksVks&/kkjk Kkr dhft;sA Ans : 27.

1.6 µ A A silver ball of radius 4.8 cm is suspended by a thread in a vacuum chamber. Ultraviolet lights of wavelength 200 nm is incident on the ball for some time during which a total light energy of 1.0 × 10 – 7 J falls on the surface. Assuming that on the average one photoelectron fine the electric potential at the surface of the ball assuming zero potential at infinity. What is the potential at the centre of the ball? ,d fuokZfrr d{k esa /kkxs dh lgk;rk ls 4.8 lseh f=kT;k okyh pkanh dh xsna yVdk;h x;h gSA dqN le; ds fy;s xsna ij 200 nm rjaxnS/;Z okyk ijkcSx a uh izdk'k vkifrr gksrk gSA ftlls lrg ij dqy 1.0 × 10–7 twy izdk'k ÅtkZ vkifrr gksrh

gSA ;g eku yhft;s fd izR;sd nl gtkj QksVkWu ij ,d QksVks bysDVªkuW dk mRltZu laHko gksrk gS ,oa vuar ij fo|qr foHko 'kwU; gksrk gS] rks xsna dh lrg ij foHko Kkr dhft;sA xsna ds dsUnz ij foHko fdruk gS\ Ans : 28.

0.3 V in each case In an experiment on photoelectric effect , the emitter and the collector plates are placed at a separation of 10 cm and are connected through an ammeter without any cell ,d izdk'k fo|qr iz;ksx esa mRltZd ,oa laxkz gd IysVas 10 lseh nwjh ij j[kh x;h gS rFkk fcuk fdlh lsy ds ,d vehVjls tksM+h x;h gS ¼fp=k½A connected through an ammeter without any cell k

figure (42-E3) A, Magnetic field B exists parallel to the plates. The work function of the emitter is 2.39eV and the light incident on it has wavelengths between 400 nm and 600 nm Find minium value of B for which the current registered by the ammeter is zero Neglect any effect of space charge. IysVksa ds lekukarj ,d pqEcdh; {ks=k mifLFkr gSA mRltZd dk dk;ZQyu 2.39 eV gS rFkk vkifrr izdk'k dh rjaxnS/;Z 400 nm ls 600 nm ds chp esa gSA B dk og U;wure eku Kkr dhft;s] ftlds fy;s vehVj 'kwU; /kkjk n'kkZ;sxkA fdlh

vUrjdk'kh vkos'k dk izHkko ux.; eku yhft;sA Ans: 29.

2.85 × 10 – 5 T In the arrangement shown in figure (42 E4) , y = 10 nm d = 0.24mm and D = 1.2 m. The work function of the material of the emitter of 2.2 eV stop the photocurrent. fp=k esa iznf'kZr O;oLFkk esa] y = 1.0 mm ] d = 0.24 mm rFkk D = 1.2 m gSA mRltZd ds inkFkZ dk dk;ZQyu 2.2 eV manishkumarphysics.in

Page # 14

Chapter # 42

Photoelectric Effect & Wave Particle duality

gSA QksVks&/kkjk dks jksdus ds fy;s vko';d fujks/kd foHko V Kkr dhft;sA

30.

In a photoelectric experiment the collector plate is at 2.0 V with respect to the emitter plate made of copper ( = 4.5 eV.). The emitter is illuminated by a source of monochromatic light of wavelength 200 nm Find the minimum and maximum kinetic energy of the photoelectrons reaching the collector. ,d izdk'k fo|qr iz;ksx esa] rkacs ( = 4.5 eV.) dh cuh gqbZ mRltZd IysV ds lkis{k laxzkgd IysV 2.0 V ij gSA mRltZd dks 200 nm rjaxnS/;Z okys ,d o.khZ izdk'k ls izdkf'kr fd;k tkrk gSA laxkz gd ij igqp a us okys QksVks bysDVªkuW ksa dh U;wure

,oa vf/kdre xfrt&ÅtkZ Kkr dhft;sA Ans: 31.

2.0 eV, 3.7 eV A small piece of cesium metal ( = 1.9 eV) is kept at a distance of 20cm from a large metal plate having a charge density of 1.0 10 – 9 C/m on the surface facing the cesium piece. A monochromatic light of wavelength 400 nm is incident on the cesium piece. Find the minimum and the maximum kinetic energy of the photoelectrons reaching the large metal plate. Neglect any change in electric field due to the small piece of cesium present. lhft;e /kkrq ( = 1.9 eV) dk ,d NksVk VqdM+k] ,d cM+h /kkrq dh IysV ls 20 lseh nwj j[kk x;k gSA ftlds lhft;e VqdM+s ds lkeus okyk lrg ij 1.0 10– 9 dwyke@eh2 vkos'k ?kuRo gSA lhft;e ds VqdM+s ij 400 nm rjaxnS/;Z okyk ,d o.khZ izdk'k

vkifrr gSA /kkrq dh cM+h IysV ij igqp a us okys QksVks bysDVªkuW ksa dh U;wure rFkk vf/kdre xfrt ÅtkZ Kkr dhft;sA lhft;e ds NksVs VqdM+s dh mifLFkfr ds dkj.k fo|qr {ks=k esa ifjorZu dks ux.; eku yhft;sA Ans: 32.

22.6 eV , 3.7 eV Consider the situation of the previous problem. Consider the fastest electron emitted parallel to the large metal plate. Find the displacement of this electrons parallel to its initial velocity before it strikes the large metal plate

fiNys iz'u esa of.kZr fLFkfrij fopkj dhft;sA eku yhft;s fd mRlftZr rhozre bysDVªkWu /kkrq dh cM+h IysV ds lekukarj mRlftZr gksrk gSA /kkrq dh cM+h IysV ls Vdjkus ls iwoZ bl bysDVªkWu dk izkjfEHkd osx ds lekUrj foLFkkiu Kkr dhft;sA Ans: 33.

9.2 cm A horizontal cesium plate ( = 1.9 eV) is moved vertically downward at a constant speed  in a room full of radiation of wavelength 250 nm and above .What should be the minimum value of  so that the vertically upward component of velocity is non positive for each photoelectron? lhft;e ( = 1.9 eV) dh ,d {kSfrt IysV] ,d ,sls dejs esa tgk¡ 200 nm rjaxnS/;Z dk fofdj.k IysV ds Åij Hkjk gqvk gS] m/okZ/kj uhps dh vksj fu;r pky  ls xfr'khy djok;h tkrh gSA  dk U;wure eku fdruk gks fd izR;sd QksVks&bysDVªkWu

ds fy;s osx dk m/okZ/kj Åij dh vksj ?kVd v/kukRed gks\

Ans : 34.

1.04 10 6 m/s A small metal plate (work function  ) is kept at a distance d from a singly ionized. Fixed ion. A monochromatic light beam is incident on the metal plate and photoelectrons are emitted. Find the maximum wavelength of the light beam so that some of the photoelectrons may go round the ion along a circle. /kkrq ¼dk;ZQyu  ) dh ,d NksVh IysV] ,d ,d/kk vk;fur vdsys fLFkj vk;u ls d nwjh ij j[kh gqbZ gSA /kkrq dh lrg ij

,d o.khZ izdk'k iqt a vkifrr gksrk gS rFkk QksVks&bysDVªkuW mRlftZr gksrs gSAa izdk'k iqt a dh og vf/kdre rjaxnS/;Z Kkr dhft;s ftlds dkj.k mRlftZr dqN QksVks bysDVªkWu vk;u ds pkjksa vksj o`Ùkkdkj Hkze.k dj ldsA 8  o dhc

Ans:

e  8 o d

35.

A light beam of wavelength 400 nm is incident on a metal plate of work function 2.2 eV (a) A particular electrons absorbs a photon and makes two collisions before coming out of the metal Assuming that 10% of the extra energy is lost to the metal in each collision find the kinetic energy of this electrons as it comes out of the metal (b) Under the same assumptions find the maximum number of collisions the electrons can suffer before it becomes unable to come out of metal. 2.2 eV dk;ZQyu okyh /kkrq dh IysV ij 400 nm rjaxnS/;Z dk izdk'k iqat vkifrr gSA (a) ,d fof'k"V bysDVªkWu QksVks bysDVªkWu dk vo'kks"k.k djus ds i'pkr~ /kkrq ls ckgj vkus ls iwoZ nks VDdj esa vfrfjDr ÅtkZ dk 10% /kkrq esa [kks tkrk gS] /kkrq ls ckgj vkus ij bysDVªkuW dh xfrt ÅtkZ Kkr dhft;sA (b) leku /kkj.kk ds lkFk] /kkrq ls ckgj vkus dh lkeF;Z

2

[kksus ls iwoZ bysDVªkWu }kjk dh x;h dqy VDdjksa dh vf/kdre la[;k Kkr dhft;sA Ans

(a) 0.31 eV (b) 4 manishkumarphysics.in

Page # 15