Tuesday, December 20, 2022

H C Verma solutions, PHOTOELECTRIC EFFECT AND WAVE-PARTICLE DUALITY, Chapter-42, EXERCISES, Q21 to Q35, Concepts of Physics, Part-II

Photoelectric Effect and Wave-Particle Duality


EXERCISES, Q21 -Q35


     21.  The electric field associated with a light wave is given by

E =Eβ‚’sin[(1.57x10⁷ m⁻¹)(x -ct)].

Find the stopping potential when the light is used in an experiment on the photoelectric effect with the emitter having work function 1.9 eV.   


ANSWER: The equation for the electric field in a wave is written as, 

E =Eβ‚’Sin ⍵(x/c -t)

   =Eβ‚’ Sin (⍵/c)(x -ct)

Comparing to the given equation, we get,

⍵/c =1.57x10⁷ rad/s

So the frequency 𝜈 is given as,

2Ο€πœˆ/c = 1.57x10⁷ rad/s

→𝜈 =1.57x10⁷*3x10⁸/2Ο€ Hz

     =7.5x10¹⁴ Hz

The stopping potential Vβ‚’ is given as

eVβ‚’ =h𝜈 -Ο†

→Vβ‚’ =(6.63x10⁻³⁴*7.5x10¹⁴-1.9*1.60x10⁻¹⁹)/1.6x10⁻¹⁹ volt

  =3·1 -1·9 volt 

  =1·2 V.      




 

     22.  The electric field at a point associated with a light wave is

E =(100 V/m) sin[(3x10¹⁵ s⁻¹)t] sin[(6x10¹⁵ 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?

ANSWER: From,

Cos(A-B)-Cos(A+B) =2SinA.SinB,

The given equation is, 

E =(100 V/m)*½[(Cos 3x10¹⁵ s⁻¹t)-(Cos 9x10¹⁵ s⁻¹t)]

So there are two waves mixed in the incident light having angular frequencies ⍵ =3x10¹⁵ s⁻¹ and 9x10¹⁵ s⁻¹.

     The maximum kinetic energy of a photoelectron will be related to the maximum frequency of the incident light. So we take ⍵ =9x10¹⁵ s⁻¹. 

→2Ο€πœˆ =9x10¹⁵ s⁻¹

→𝜈 =1.43x10¹⁵ Hz.

So the maximum kinetic energy of a photoelectron,

=h𝜈 -Ο† 

If we take the unit as eV, 

Ο† =2.0 eV, h =4.14x10⁻¹⁵ eV-s

So, Kβ‚˜β‚β‚“ =4.14x10⁻¹⁵*1.43x10¹⁵-2.0 eV

        =5·92 -2·0 eV =3·92 eV.

     



 

     23.  A monochromatic light source of intensity 5 mW emits 8x10¹⁵ photons per second. This light ejects photoelectrons from a metal surface. The stopping potential for this setup is 2.0 eV Calculate the work function of the metal.    


ANSWER: Energy emitted by light per second is the intensity of light. So in the given problem, 8x10¹⁵ photons have energy =5 mW =5x10⁻³ J.

So the energy of each photon,

h𝜈 =5x10⁻³/8x10¹⁵ J 

   =6.25x10⁻¹⁹ J

   =3.9 eV

Stopping potential in terms of eV,

Vβ‚’ =h𝜈 -Ο†

→2.0 =3.9 -Ο†

→Ο† =3.9 -2.0 eV

     =1·9 eV.

    



 

     24.  Figure (42-E2) is the plot of the stopping potential versus the frequency of the light used in an experiment on the photoelectric effect. Find (a) the ratio h/e and (b) the work function.   
Figure for Q-24


ANSWER: (a) The stopping potential is given as, 

eV。=hv -Ο†

→V。=hv/e -Ο†/e  ------ (i)

It is an equation of a straight line in the form of 

y =mx +C

The slope of the line, 

m =tan ΞΈ =h/e,, and the intercept on the y-axis, C =Ο†/e.

From the graph,

 m =1.656/4x10¹⁴ V-s

→h/e =4·14x10⁻¹⁵ V-s.


(b) From equation (i),

V。=(h/e)v -Ο†/e
Taking V。=1.656 V and v =5x10¹⁴ Hz
from the graph,
1.656 =4.14x10⁻¹⁵*5x10¹⁴ -Ο†/e
→Ο†/e =2.07 -1.656
→Ο† =0·414*e J
      =0·414 eV.




 

     25.  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.    


ANSWER: From the problem, we conclude that the minimum energy of a photon needed to dissociate a silver bromide is 0.6 eV. It will correspond to the maximum wavelength πœ† of the light. So, 

  hc/πœ† =0.6 eV

→6.63x10⁻³⁴*3x10⁸/πœ† =0.6*1.6x10⁻¹⁹ J

→1.99x10⁻²⁵/πœ† =9.6x10⁻²⁰

→πœ† =2.070x10⁻⁶ m =2070 nm.

  



 

     26.  In an experiment on the 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 reaches its saturation value. Assuming that on average one out of every 10⁶ photons is able to eject a photoelectron, find the photocurrent in the circuit.   


ANSWER: The energy of each photon of the incident light,

E =hc/πœ† 

  =6.63x10⁻³⁴*3x10⁸/(400x10⁻⁹) J

  ≈5x10⁻¹⁹ J. 

The total energy of the light =5.0 J/s. 

Number of photons moving in the light beam, 

  =5.0/5x10⁻¹⁹ per/s

  =1x10¹⁹ per/s.

Since one out of 10⁶ photon is able to eject a photoelectron, the number of photoelectrons emitting out per second,

 N =1x10¹⁹/10⁶ 

  =1x10¹³.

Hence the current in the circuit,

 i =N*e A 

  =1x10¹³*1.6x10⁻¹⁹ A

  =1.6x10⁻⁶ A 

  =1.6 Β΅A.            





 

     27.  A silver ball of radius 4.8 cm is suspended by a thread in a vacuum chamber. Ultraviolet light of wavelength 200 nm is incident on the ball for some time during which a total light energy of 1.0x10⁻⁷ J falls on the surface. Assuming that on average one photon out of every ten thousand is able to eject a photoelectron, find the electric potential at the surface of the ball assuming zero potential at infinity. What is the potential at the center of the ball?   


ANSWER: The wavelength of the incident ultraviolet light,

 Ξ» =200 nm =2x10⁻⁷ m.

The energy of each photon,

E =hc/Ξ»

  =6.63x10⁻³⁴*3x10⁸/2x10⁻⁷ J

  ≈1.0x10⁻¹⁸ J.

Total number of photons that fell on the ball, 

N =1.0x10⁻⁷ J/1.0x10⁻¹⁸ J

   =1.0x10¹¹.

The number of ejected photoelectrons,

 N'=1.0x10¹¹/10⁴

    =1.0x10⁷.

The positive charge on the ball due to ejected photoelectrons,

q =1.0x10⁷*1.6x10⁻¹⁹ C

   =1.6x10⁻¹² C.

The electric potential on the surface of the ball,

V =Kq/r

K =9x10⁹ N-m²/C², 

r =4.8 cm =0.048 m

Hence V =9x10⁹*1.6x10⁻¹²/0.048 volt

     =0·30 volts.


          Since the electric potential inside a sphere remains constant and equal to the surface potential, hence the electric potential at the center of the ball =0·30 volts.


 





 

     28.  In an experiment on the 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 (figure 42-E3). A magnetic field B exists parallel to the plates. The work function of the emitter is 2.39 eV and the light incident on it has wavelengths between 400 nm and 600 nm. Find the minimum value of B for which the current registered by the ammeter is zero. Neglect any effect of space charge.   
Figure for Q-28


ANSWER: The value of B will be minimum when the emitted electrons having the maximum kinetic energy fail to reach the collector plate. Electrons having maximum kinetic energy will be emitted by the light having minimum wavelength i.e. Ξ» =400 nm. 

The maximum kinetic energy of the photoelectron,

K =hc/Ξ» -Ο†

 =4.14x10⁻¹⁵*3x10⁸/400x10⁻⁹ -2.39 eV

 =3.105 -2.39 eV

 =0.715 eV.

    Since the emitted electrons encounter a magnetic field B that is perpendicular to its direction of motion, they experience a force perpendicular to both B and v and it results in a circular motion of the photoelectrons. If the radius of this circular path is just equal to the separation of plates, there will be no current in the circuit. The radius is given as,

r =mv/qB

Here, r =d and mv =√(2mK)

So, d =√(2mK)/qB

→B =√(2mK)/qd

=√(2*9.1x10⁻³¹*0.715*1.6x10⁻¹⁹)/(1.6x10⁻¹⁹*0.10)

=2.85x10⁻⁵ T.




 

     29.  In the arrangement shown in figure (42-E4). y =1.0 mm, d =0.24 mm and D =1.2 m. The work function of the material of the emitter is 2.2 eV. Find the stopping potential V needed to stop the photocurrent.   
Figure for Q-29


ANSWER: The fringe width,

w=2y=2*1.0 mm=2.0 mm=2.0x10⁻³ m.

D =1.2 m, d =0.24 mm =2.4x10⁻⁴ m.  

The wavelength of light, πœ† =wd/D.

→πœ† =2x10⁻³*2.4x10⁻⁴/1.2 m 

     =4x10⁻⁷ m.

The energy of a photon, E =hc/πœ† 

→E =4.14x10⁻¹⁵*3x10⁸/4x10⁻⁷ eV

    =3.1 eV.  

The work function of the emitter, Ο† =2.2 eV. 

Stopping potential V' needed to stop the photocurrent is given as, 

eV' =E -Ο† =3.1 -2.2 =0.9 eV 

→V' ={0.9*e/e} V = 0·9 V.






 

     30.  In a photoelectric experiment, the collector plate is at 2.0 V with respect to the emitter plate made of copper (Ο† =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.  


ANSWER: When photons are incident on the emitter surface, the electrons in the metal absorbing the photons are detached from the molecules but they may either lose some or all their kinetic energy with collisions with other molecules in the metal. So the electrons coming out of the emitter surface may have kinetic energies varying from zero to a maximum K =hc/Ξ»-Ο†.

→K =4.14x10⁻¹⁵*3x10⁸/(200x10⁻⁹) -4.5 eV

→K =6.21 -4.5 eV =1.71 eV.


Since here the collector plate is at 2.0 V with respect to the emitter plate, the kinetic energy of 2.0 eV will be added to the electrons coming out of the emitter plate. The electrons coming out of the emitter plate have already kinetic energy from zero to 1.71 eV. Thus the electrons reaching the collector plate will have minimum kinetic energy =0+2.0 eV =2.0 eV and the maximum kinetic energy =1.71+2.0 eV =3.71 eV.

 




 

     31.  A small piece of cesium metal (Ο† =1.9 eV) is kept at a distance of 20 cm from a large metal plate having a charge density of 1.0x10⁻⁹ C/m² on the surface facing the cesium piece. A monochromatic light of wavelength 400 nm is incident on the cesium piece. Find the minimum and maximum kinetic energy of the photoelectrons reaching the large metal plate. Neglect any change in the electric field due to the small piece of cesium present. 


ANSWER: Given Ξ» =400 nm and Ο† =1.9 eV. Hence without any external force, the emitted photoelectrons will have kinetic energy varying from zero to K =hc/Ξ»-Ο†,

→K=4.14x10⁻¹⁵*3x10⁸/400x10⁻⁹-1.9 eV

    =3.1 -1.9 eV =1.2 eV.


   Here we have an accelerating force for the emitted photoelectrons due to the large charged plate. The electric field near a charged plate is E =𝜎/Ξ΅β‚’, where 𝜎 is the charge density. Given,

 πœŽ =1.0x10⁻⁹ C/m² and we know, 

Ξ΅β‚’ =8.85x10⁻¹² C²/N-m².

So, E =1.0x10⁻⁹/8.85x10⁻¹² N/C

     =113 N/C.

The electric potential at the cesium metal due to the charged plate,

V =E*d

   =113*0.20 V

   =22.6 V.

This will impart a kinetic energy of 22.6 eV to the emitted photoelectrons near the cesium plate. So the minimum kinetic energy of the photoelectrons reaching the large metal plate is =0+22.6 eV =22·6 eV and the maximum kinetic energy =1.2 +22.6 eV =23·8 eV.

  



 

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


ANSWER: The fastest electron emitted parallel to the large metal plate will have velocity v parallel to the plate and its kinetic energy K =1.2 eV, (as calculated in the previous problem). But its velocity perpendicular to the plates is zero. 

So v =√(2K/m) 

→v =√(2*1.2*1.6x10⁻¹⁹/9.1x⁻³¹) m/s

     =6.5x10⁵ m/s. 

     Now it is a problem like a stone thrown parallel to the ground. 
Diagram for Q-32

     Force on the electron perpendicular to the plate due to the electric field, E =113 N/C, 

F =eE, and the acceleration along this force, 

a =F/m =eE/m

  =1.6x10⁻¹⁹*113/9.1x10⁻³¹ m/s 

  =2x10¹³ m/s². 

If t is time to reach the large plate, then 

d =½at² 

→t =√(2d/a) s

  =√(2*0.20/2x10¹³) s 

  =1.41x10⁻⁷ s


Hence the displacement of this electron parallel to its initial velocity before it strikes the large metal plate,

S =vt

   =6.5x10⁵*1.41x10⁻⁷ m 

   =0.092 m

   =9·2 cm.              




 

     33.  A horizontal cesium plate (Ο† =1.9 eV) is moved vertically downward at a constant speed v in a room full of radiation of wavelength 250 nm and above. What should be the minimum value of v so that the vertically upward component of velocity is nonpositive for each photoelectron?  


ANSWER: When the plate is unmoved, the maximum upward velocity v' of an emitted photoelectron will be corresponding to the maximum possible kinetic energy K of the emitted photoelectron. From Einstien's photoelectric equation,

K =hc/Ξ» -Ο†

  =4.14x10⁻¹⁵*3x10⁸/250x10⁻⁹ -1.9 eV

  =5.0 -1.9 eV

  =3.1 eV.

Velocity v' =√(2K/m)

→v'=√(2*3.1*1.6x10⁻¹⁹/9.1x10⁻³¹) m/s

   =1.04x10⁶ m/s.


   When the plate is moved vertically downward with a velocity v, each emitted photoelectron gets additional velocity v. Net velocity of the fastest vertically upward moving electron,

=v' -v.

For the vertically upward component of the velocity of emitted photoelectrons to be nonpositive,

v' -v =0

→v =v' =1·04x10⁶ m/s.

 



 

     34.  A small metal plate (work function Ο†) 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 around the ion along a circle.  


ANSWER: The photoelectrons going around the ion must have a force of attraction as centripetal force. Thus the charge on the ion is positive e so that the electrostatic force between the ion and the electron provides this force. The centripetal electrostatic force,

F =Ke²/d²

When the photoelectrons having the maximum kinetic energy have a velocity parallel to the emitter plate and have centrifugal force just equal to F, they will go around the ion. The maximum kinetic energy of a photoelectron, 

E =hc/Ξ» -Ο†

Its velocity, v =√(2E/m)

The centrifugal force =mv²/d

  =m*2E/md

  =2E/d 

  =2(hc/Ξ» -Ο†)/d 


Equating the two forces,

 Ke²/d² =2(hc/Ξ» -Ο†)/d

Ke²/d =2(hc/Ξ» -Ο†)

→hc/Ξ» =Ke²/2d +Ο† =(Ke²+2Ο†d)/2d

→Ξ» =2dhc/(Ke²+2Ο†d)

     =2dhc/{e²/4πΡₒ +2Ο†d}

    =2dhc/{(e² +8πΡₒφd)/4πΡₒ}

    =8πΡₒdhc/(e² +8πΡₒφd).

 



 

     35.  A light beam of wavelength 400 nm is incident on a metal plate of work function 2.2 eV. (a) A particular electron 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 electron when it comes out of the metal. (b) Under the same assumptions, find the maximum number of collisions the electrons can suffer before they become unable to come out of the metal.  


ANSWER: (a) Ξ» =400 nm, Ο† =2.2 eV.

The energy of the photon is,

E = hc/Ξ»

  =4.14x10⁻¹⁵*3x10⁸/400x10⁻⁹ eV

  =3.1 eV.

  This energy of the photon is absorbed by the electron and gains kinetic energy. The energy lost by the absorbing electron in one collision is 10%, hence remaining energy after two collisions,

  =(0.9)²*3.1 eV =2.51 eV


The electron will lose energy equal to the work function in coming out of the metal. So the kinetic energy of this electron after coming out =2.51 -Ο†

  =2.51 -2.2 eV =0.31 eV.


(b) Remaining kinetic energy after two collisions =2.51 eV. 

After the third collision remaining energy =0.9*2.51 eV =2.26 eV.

After the fourth collision remaining energy =0.9*2.26 eV =2.03 eV.

 The remaining energy is now less than the work function and it will not be able to come out of the metal. 

   The electron will have 4 number of collisions before it is unable to come out of the metal. 

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Links to the Chapters




CHAPTER- 42- Photoelectric Effect and Wave-Particle Duality



CHAPTER- 34- Magnetic Field

CHAPTER- 29- Electric Field and Potential











CHAPTER- 28- Heat Transfer

OBJECTIVE -I







EXERCISES - Q51 to Q55


CHAPTER- 27-Specific Heat Capacities of Gases

CHAPTER- 26-Laws of Thermodynamics


CHAPTER- 25-CALORIMETRY

Questions for Short Answer

OBJECTIVE-I

OBJECTIVE-II


EXERCISES - Q-11 to Q-18


CHAPTER- 24-Kinetic Theory of Gases







CHAPTER- 23 - Heat and Temperature






CHAPTER- 17 - Light Waves




CHAPTER- 14 - Fluid Mechanics




CHAPTER- 13 - Fluid Mechanics


CHAPTER- 12 - Simple Harmonic Motion









CHAPTER- 11 - Gravitation





CHAPTER- 10 - Rotational Mechanics






CHAPTER- 9 - Center of Mass, Linear Momentum, Collision


CHAPTER- 8 - Work and Energy

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CHAPTER- 5 - Newton's Laws of Motion


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CHAPTER- 4 - The Forces

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CHAPTER- 3 - Kinematics - Rest and Motion

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