Monday, November 28, 2022

H C Verma solutions, PHOTOELECTRIC EFFECT AND WAVE-PARTICLE DUALITY, Chapter-42, EXERCISES, Q11 to Q20, Concepts of Physics, Part-II

Photoelectric Effect and Wave-Particle Duality


EXERCISES, Q11 -Q20


     11.  Consider the situation described in the previous problems. 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.  


ANSWER: Let us take the radius of the sphere = r and the intensity of the light = I. In the figure below, OX is the axis of the sphere parallel to the falling beam. P is a point on the sphere such that radius OP makes an angle θ from OX. If OP is moved through a very small angle dθ to point Q, then PQ =rdθ. 
Diagram for Q-11

    Suppose PQ is moved on the surface of the sphere at the same inclination around OX, we get a ring with width rdθ. The area of this ring,

 dA = rdθ*2πrSinθ =2πr²Sinθdθ. 

    The rays make an angle θ with radius OP, hence even after reflection, the rays will make an angle θ with the OP because the radius is perpendicular to the surface at P.  

  Consider a small area 'S' on this ring. Light energy falling on this area per second,

=S cosθ*I =SI cosθ.

The momentum of light falling on area S

 =SI cosθ/c.

Since the light is reflected through the same angle from OP, the change of momentum of light energy per second is,

=2(SI cosθ/c)*cosθ

=2SI cos²θ/c, along PO.

Hence the force on this small area =2SIcos²θ/c.

Component of this force along the parallel beam (which is the force on the sphere for a small area S)

=(2SI cos²θ/c)*cosθ

=2SI cos³θ/c.

Hence the force on the ring of which part is S, {Substituting S with the area of the ring dA}

dF =(2dAIcos³θ/c) 

    =(4πr²I/c) cos³θSinθdθ.

For the total force on the sphere, we integrate it for the area of the exposed hemisphere, i.e. from θ =0 to θ =π/2.

F = ∫dF

  =(4πr²I/c)∫cos³θ.sinθdθ

Let y =cosθ, dy =-sinθdθ. The limit of integration will be from, cos0 =1 to cosπ/2 =0. Hence 

F =-(4πr²I/c)∫y³dy

  =-(4πr²I/c)[y⁴/4]

  =-(4πr²I/c)[0 -1/4]

  =πr²I/c.  ---------------- (i)

In the previous problem, if the intensity is I', the force on the sphere is 

=(I'/c)*πr²

=πr²I'/c.   ---------------- (ii)


(i) is derived for a fully reflecting sphere while (ii) is for a fully absorbing sphere. Our given sphere is partly absorbing. Suppose out of 0.50 W/cm² intensity (in the previous problem) I watt/cm² is fully reflected and the rest I' watt/cm² is fully absorbed. i.e. I +I' =0.50 W/cm², So the total force on the sphere 

F =πr²I/c +πr²I'/c

   =πr²(I+I')/c

   =π*1²*0.50/3x10⁸ N

   =5.2x10⁻⁹ N.

So the force on the sphere is the same even if the sphere is not fully absorbing. 







     12.  Show that it is not possible for a photon to be completely absorbed by a free electron.  


ANSWER: A free electron in a metal is attracted by the other molecules and nucleus of the atom. To bring it out of the metal a minimum amount of energy is needed which is called the work function φ of the metal. When a photon of energy hv is used to bring out the free electron from a metal surface, a part of photon energy hv is used as a work function. So the electron can only absorb the rest of the energy hv-φ and this energy shows up as the kinetic energy of the electron. Thus it is not possible for a photon to be completely absorbed by a free electron.   






     13.  Two neutral particles are kept 1 m apart. Suppose by some mechanism some charge is transferred from one particle to the other and the electric potential energy lost is completely converted into a photon. Calculate the longest and the next smaller wavelength of the photon possible.  


ANSWER: Since both the particles are neutral, if the -q charge is transferred from one particle to another, the first particle will have +q charge while the other will have -q charge. For a separation r, the electric potential energy between these two charges,

U =kq²/r. 

For the same energy of a photon let its wavelength be equal to λ, then 

U =hc/λ.

Equating the two energies, 

kq²/r = hc/λ.

→λ =hcr/kq².

  In this expression h, c, r, and k are constants. For the wavelength λ to be maximum, q must be minimum. And the minimum possible value of a charge is the charge of an electron.

So λₘₐₓ =hcr/ke²

  =6.63x10⁻³⁴*3x10⁸*1/{9x10⁹*(1.6x10⁻¹⁹)²} m

   =0.86x10³ m

   =860 m.


The next greater charge will give the next smaller wavelength. And the next greater charge will be q =2e. Hence, 

λ' =hcr/{k(2e)²}.

 =6.63x10⁻³⁴*3x10⁸*1/{9x10⁹*4*2.56x10⁻³⁸} m

=0.215x10³ m 

=215 m.          





 

     14.  Find the maximum kinetic energy of the photoelectrons ejected when the light of wavelength 350 nm is incident on a cesium surface. The work function of Cesium =1.9 eV.  


ANSWER: The energy of a photon of the light of wavelength λ =350 nm 

E =hc/λ

 =4.14x10⁻¹⁵x3x10⁸/350x10⁻⁹ eV 

 =3.5 eV

The work function of Cesium,

 φ =1.9 eV

     Hence the maximum kinetic energy of the emitted photoelectron, 

= E-φ

=3.5 -1.9 eV

=1.6 eV.

 





 

     15.  The work function of a metal is 2.5x10⁻¹⁹ J. (a) Find the threshold frequency for photoelectric emission. (b) If the metal is exposed to a light beam of frequency 6.0x10¹⁴ Hz. What will be the stopping potential?


ANSWER: The work function, 

 φ =2.5x10⁻¹⁹ J.  

(a) Threshold frequency 𝜈 of photons is the minimum frequency for which photoelectrons are just able to come out of the metal. The energy of this photon is equal to the work function. So,

h𝜈 =φ

→𝜈 =φ/h

   =2.5x10⁻¹⁹/6.63x10⁻³⁴ Hz 

  =3.8x10⁻¹⁴ Hz.

  

(b) Frequency of light beam,

𝜈 =6.0x10¹⁴ Hz 

If the stopping potential is Vₒ, then

eVₒ =Kₘₐₓ =h𝜈 -φ

→Vₒ =h𝜈/e -φ/e

=6.63x10⁻³⁴*6.0x10¹⁴/1.6x10⁻¹⁹ -2.5x10⁻¹⁹/1.6x10⁻¹⁹ volts

=2.48 -1.56 volts

=0.92 volts.  





 

     16.  The work function of a photoelectric material is 4.0 eV. (a) What is the threshold wavelength? (b) Find the wavelength of light for which the stopping potential is 2.5 V.  


ANSWER: (a) Given φ =4.0 eV,

→φ =4.0 eV.

Let the threshold wavelength =λ,

Then hc/λ =φ

→λ =hc/φ

  =4.14x10⁻¹⁵*3x10⁸/4.0 m

  =3.10x10⁻⁷ m

  =310x10⁻⁹ m

  =310 nm.


(b) Vₒ =2.5 V, 

Let the corresponding wavelength of light for this stopping potential be λ. Then,

eVₒ =hc/λ -φ

→2.5=4.14x10⁻¹⁵*3x10⁸/λ-4.0 eV

→6.5 =12.42x10⁻⁷/λ

→λ =1.90x10⁻⁷ m

      =190x10⁻⁹ m

      =190 nm. 





 

     17.  Find the maximum magnitude of the linear momentum of a photoelectron emitted when the light of wavelength 400 nm falls on a metal having work function 2.5 eV.


ANSWER: λ =400 nm

→λ =4x10⁻⁷ m

φ =2.5 eV =2.5x1.6x10⁻¹⁹ J

The maximum kinetic energy of the photoelectron,

E =hc/λ -φ

=6.63x10⁻³⁴*3x10⁸/4x10⁻⁷ -2.5x1.6x10⁻¹⁹

=5.0x10⁻¹⁹ -4.0x10⁻¹⁹ J

=1.0x10⁻¹⁹ J

If p is the linear momentum of the photoelectron then,

E = p²/2m

→p² =2mE

  =2x9.1x10⁻³¹*1.0x10⁻¹⁹

  =18.2x10⁻⁵⁰

→p =4.2x10⁻²⁵ kg-m/s.

 





 

     18.  When a metal plate is exposed to a monochromatic beam of light of wavelength 400 nm, a negative potential of 1.1 V is needed to stop the photocurrent. Find the threshold wavelength for the metal.


ANSWER: The stopping potential,

Vₒ =1.1 volts

Wavelength, λ =400 nm =4x10⁻⁷ m

If φ is the work function then. 

eVₒ =hc/λ -φ

→eVₒ =hc/λ -hc/λ'

{Where λ' is the threshold wavelength.}

→hc/λ' =hc/λ -eVₒ

→1/λ' =1/λ -eVₒ/hc

    =1/4x10⁻⁷ -1.6x10⁻¹⁹*1.1/{6.63x10⁻³⁴*3x10⁸}

  =2.5x10⁶ -0.88x10⁶

→1/λ' =1.62x10⁶

→λ' =6.20x10⁻⁷ m

      =620x10⁻⁹ m

      =620 nm.

 





 

     19.  In an experiment on the photoelectric effect, the stopping potential is measured for monochromatic light beams corresponding to different wavelengths. 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 the inverse of wavelength (1/λ) on graph paper and find (a) the Plank constant, (b) the work function of the emitter, and (c) the threshold wavelength.    


ANSWER: The relation between stopping potential and the wavelength of the incident light is, 

eVₒ +φ =hc/λ

→1/λ =eVₒ/hc +φ/hc

This is in the form of y =mx+C, which is an equation of a straight line.

Here, y = 1/λ, x =Vₒ, 

m =e/hc, {Slope of the line},

C =φ/hc, {Intercption of the line on the Y-axis}. 

Let us plot the graph. We have,

1/λ (*10⁶)=2.86  2.50  2.22  2.00  1.82

Vₒ (V) =      1.45  1.00  0.66  0.38  0.16

Graph for Q-19


From the graph, C =φ/hc =1.72x10⁶ m⁻¹

→φ =h*3x10⁸*1.72x10⁶ J

      =5.16x10¹⁴*h J  -------------- (i)


(a) Slope of the graph m =tanθ =1.14x10⁶/1.45

→e/hc =0.8x10⁶

→h =ex10⁻⁶/(0.8c) J-s

    =1.0x10⁻⁶/(0.8c) eV-s

   =1.0x10⁻⁶/(0.8x3x10⁸) eV-s

  =4.2x10⁻¹⁵ eV-s.


(b) From (i), work function

φ = 5.16x10¹⁴ h

   =5.16x10¹⁴*4.2x10⁻¹⁵ eV 

   =2.16 eV.


(c) Threshold wavelength λ is given as

hc/λ =φ

→λ =hc/φ

  =4.2x10⁻¹⁵*3x10⁸/2.16 m

  =5.83x10⁻⁷ m

  =583x10⁻⁹ m

  =583 nm.

 

{Note: Accuracy of the result depends upon accuracy of plotting and eye estimation}




 

     20.  The electric field associated with a monochromatic beam becomes zero 1.2x10¹⁵ times per second. Find the maximum kinetic energy of the photoelectrons when this light falls on a metal surface whose work function is 2.0 eV.


ANSWER: The electric field of an electromagnetic wave becomes zero two times in a cycle. So in a second, it becomes zero twice its frequency. Thus the frequency of the given monochromatic beam is half of the given number. 

Frequency 𝜈 =6.0x10¹⁴

      The maximum kinetic energy of the photoelectrons, when this beam falls on a metal surface of work function φ =2.0 eV, is

  =h𝜈 -φ

  =4.14x10⁻¹⁵*6x10¹⁴-2.0 eV 

  =2.48 -2.0 eV 

  =0.48 eV.

 

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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

Click here for → Question for Short Answers

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → Exercises (1-10)

Click here for → Exercises (11-20)

Click here for → Exercises (21-30)

Click here for → Exercises (31-42)

Click here for → Exercise(43-54)

CHAPTER- 7 - Circular Motion

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Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → EXERCISES (11-20)

Click here for → EXERCISES (21-30)

CHAPTER- 6 - Friction

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Click here for → OBJECTIVE-I

Click here for → Friction - OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → Exercises (11-20)

Click here for → EXERCISES (21-31)

For more practice on problems on friction solve these- "New Questions on Friction".

---------------------------------------------------------------------------------

CHAPTER- 5 - Newton's Laws of Motion


Click here for → QUESTIONS FOR SHORT ANSWER

Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)

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CHAPTER- 4 - The Forces

The Forces-

"Questions for short Answers"    


Click here for "The Forces" - OBJECTIVE-I


Click here for "The Forces" - OBJECTIVE-II


Click here for "The Forces" - Exercises


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CHAPTER- 3 - Kinematics - Rest and Motion

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Click here for "OBJECTIVE-I"


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Click here for EXERCISES (Question number 11 to 20)


Click here for EXERCISES (Question number 21 to 30)


Click here for EXERCISES (Question number 31 to 40)


Click here for EXERCISES (Question number 41 to 52)


CHAPTER- 2 - "Physics and Mathematics"

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