Monday, November 21, 2022

H C Verma solutions, PHOTOELECTRIC EFFECT AND WAVE-PARTICLE DUALITY, Chapter-42, EXERCISES, Q1 to Q10, Concepts of Physics, Part-II

Photoelectric Effect and Wave-Particle Duality


EXERCISES, Q1 -Q10


     1.  Visible light has wavelengths in the range of 400 nm to 780 nm. Calculate the range of energy of the photons of visible light.  


ANSWER: The energy of a photon, 

E =hc/λ.

For λ =780 nm =780x10⁻⁹ m

        =7.80x10⁻⁷ m

c =3x10⁸ m/s, h =6.63x⁻³⁴ J-s.

So, E =6.63x10⁻³⁴*3x10⁸/7.8x10⁻⁷ J

       =2.55x10⁻¹⁹ J.

For λ =400 nm

        =4x10⁻⁷ m

E =6.63x10⁻³⁴*3x10⁸/4x10⁻⁷ J

   ≈5.0x10⁻¹⁹ J

So the range of energy of photons of visible light is from 2.55x10⁻¹⁹ J to 5.0x10⁻¹⁹ J.

 

   





     2.  Calculate the momentum of a photon of light of wavelength 500 nm.  


ANSWER: The momentum of a photon of wavelength λ is given as,

p =h/λ

For λ =500 nm =5x10⁻⁷ m

The momentum of the photon,

p =6.63x10⁻³⁴/5x10⁻⁷ kg-m/s

   =1.33x10⁻²⁷ kg-m/s.

    




 

     3.  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.  


ANSWER: The energy of a photon of wavelength 500 nm i.e. 5x10⁻⁷ m is, 

E =hc/λ

  =6.63x10⁻³⁴*3x10⁸/5x10⁻⁷ J

  =3.9x10⁻¹⁹ J.

The energy of a photon of wavelength λ =700 nm =7x10⁻⁷ m.

E' =6.63x10⁻³⁴*3x10⁸/7x10⁻⁷ J

   =2.8x10⁻¹⁹ J.


Hence the net energy absorbed by the atom in this process =E -E'

  =3.9x10⁻¹⁹ -2.8x10⁻¹⁹ J

  =1.1x10⁻¹⁹ J.

    




 

     4.  Calculate the number of photons emitted per second by a 10 W sodium vapor lamp. Assume that 60% of the consumed energy is converted into light. The wavelength of Sodium light =590 nm.  


ANSWER: The energy of a photon of sodium light of wavelength λ =590 nm =5.9x10⁻⁷ m is

E =hc/λ 

   =6.63x10⁻³⁴*3x10⁸/5.9x10⁻⁷ J

   =3.37x10⁻¹⁹ J.

10% of 10 W i.e. 6 W of energy is converted into light. So the energy of 6 J/s is being emitted. Hence the number of photons emitted per second is

=6/3.37x10⁻¹⁹

=1.78x10¹⁹.

    




 

     5.  When the light is directly overhead, the surface of the earth receives 1.4x10³ W/m² of sunlight. Assume that the light is monochromatic with an average wavelength of 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 1.5x10¹¹ m. (a) Calculate the number of photons falling per second on each square meter of the earth's surface directly below the sun. (b) How many photons are there in each cubic meter near the earth's surface at any instant? (c) How many photons does the sun emit per second?   


ANSWER: (a) The light energy falling per m² of earth's surface in a second, 

E' =1.4x10³ J  

The average wavelength of the falling light, 

λ =500 nm =5.0x10⁻⁷ m.

The energy of a photon of this light,

E =hc/λ

  =6.63x10⁻³⁴*3x10⁸/5.0x10⁻⁷ J

  =3.98x10⁻¹⁹ J

The number of photons falling per second on each square meter of the earth's surface directly below the sun,

  =E'/E

  =1.4x10³/3.98x10⁻¹⁹

  =3.5x10²¹.


(b) In a second, a photon travels 3x10⁸ m and 3.5x10²¹ number of photons fall on an m² of area. Thus there are 3.5x10²¹ photons in a volume of,

3.0x10⁸ m*1 m² =3.0x10⁸ m³.

So the number of photons in 1 m³ near the earth's surface,

  =3.5x10²¹/3.0x10⁸

  ≈1.2x10¹³.


(c) The distance between the sun and the earth, 

R =1.5x10¹¹ m.

Keeping the sun at the center, the area of a sphere of radius R is =4πR². The total number of photons P, emitted per second by the sun is falling on this area 4πR².  The number of photons falling per second on each square meter of the area is 3.5x10²¹. Hence the number of photons emitted per second by the sun is,

=3.5x10²¹*4π*(1.5x10¹¹)²

=99x10⁴³

=9.9x10⁴⁴.

    




 

     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 1.0x10¹⁹. Calculate the force exerted by the light beam on the mirror. 


ANSWER: Wavelength of the monochromatic light, 

λ =663 nm =6.63x10⁻⁷ m

Linear momentum of a photon 

p =h/λ

   =6.63x10⁻³⁴/6.63x10⁻⁷ kg-m/s

   =1.0x10⁻²⁷ kg-m/s

The number of photons striking per second =1.0x10¹⁹

Hence the total momentum of the photons falling per second on the mirror,

 p' =1.0x10¹⁹*1.0x10⁻²⁷ kg-m/s

     =1.0x10⁻⁸ kg-m/s

Diagram for Q-6


   The direction of this momentum is at 60° from the normal to the mirror. Since the angle of reflection is equal to the angle of incidence, the reflected photons will also have the same magnitude of momentum but its direction will be 60° from the normal and going away from the mirror. If we resolve the momenta of the incoming and outgoing photons along the normal and along the plane of the mirror, we see that there is no change of direction and hence momentum of the photons along the plane of the mirror. So the light beam does not exert force along the plane of the mirror. The component of the momentum of incoming photons along the normal is equal and opposite to the momentum of outgoing photons along the normal. 

   The component of the total momentum of the incoming photons per second along the normal is

 =p'*cos 60° =p'/2

    The component of the total momentum of the outgoing photons per second along the normal is

 =-p'*cos 60° =-p'/2

Hence the change in momentum of the total photons falling per second along the normal

  =p'/2 -(-p'/2) =p'

But the change in momentum per second is the force exerted. Hence the force exerted by the light beam on the mirror =p'

=1.0x10⁻⁸ N.

 




 

     7.  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.  


ANSWER: The energy of photons falling per second on the plane surface =10 J.

The total momentum of these photons before they strike the plane surface,

p = E/c 

   =(10 J)/(3.0x10⁸ m/s)

   =(10/3)x10⁻⁸ kg-m/s 

70% of the light is absorbed by the surface, so after the strike, 70% of the momentum of photons becomes zero. The remaining 30% of the momentum has the same magnitude but opposite in direction because these photons are reflected back normally. So after the strike, the momentum of the photons falling per second is, 

  =0.70P*0 +(-0.30P) 

  =-0.30P 

Hence the change in momentum per second, 

  =P -(-0.30P) 

  =1.30P

  =1.30*(10/3)x10⁻⁸ kg-m/s 

  =4.3x10⁻⁸ kg-m/s  

The force exerted by the incident beam on the surface is equal to the change in the momentum of the photons per second, i.e.,  

=4.3x10⁻⁸ N.    





 

     8.  A totally reflecting, small plane mirror placed horizontally faces a parallel beam of light as shown in figure (42-E1). The mass of the mirror is 20 g. Assume that there is no 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/s².   
The figure for Q-8


ANSWER: To support the weight of the mirror, the parallel beam of light should apply a force equal to the weight of the mirror, i.e.

 F =mg 

  =(20/1000)*10 N 

  =0.20 N. 

Suppose the momentum of the photons falling per second on the mirror =p. 

Since the mirror is totally reflecting, the momentum of the reflected photons =-p.

So the rate of change of momentum of the incident photons =p -(-p) =2p. 

Since the rate of change of momentum = force applied, 

F =2p 

→p =F/2 

    =0.20/2 kg-m/s 

    =0.10 kg-m/s. 

Let the power of the source =P watt, i.e. P joules per second. The energy of light passing through the lens per second, 

E =0·30P J. 

The relation between energy and momentum of photons is, 

 p =E/c 

→E =pc 

→0.30P =0.10*3.0x10⁸ 

→P =1.0x10⁸ W 

     =100 MW. 


                




 

     9.  A 100 W light bulb is placed at the center of a spherical chamber of radius 20 cm. Assume that 60% of the energy supplied to the bulb is converted into light and that the surface of the chamber is perfectly absorbing. Find the pressure exerted by the light on the surface of the chamber.  


ANSWER: Radius, r =20 cm =0.20 m.

Area of the sphere =4πr² 

→A =4π*(0.20)² m² 

     =0.16π m² 

Since the bulb is placed at the center of the sphere, the emitted photons will fall normally on the absorbing surface. Though the direction of photons will be different, it may be calculated as if falling on a normal surface.       

       The energy of photons emitted per second,  

E =0.60*100 J/s =60 J/s. 

The momentum of photons falling per second, 

p =E/c 

   =60/3x10⁸ kg-m/s

   =2x10⁻⁷ kg-m/s 

Since the final momentum is zero due to absorption, the change in momentum per second, 

  = p =2x10⁻⁷ kg-m/s

Hence the magnitude of the total force on the spherical surface,

F =2x10⁻⁷ N 

 Hence the pressure exerted by the light on the spherical chamber, 

=F/A

=2x10⁻⁷/0.16π N/m² 

=4.0x10⁻⁷ Pa. 





 

     10.  A sphere of radius 1.00 cm is placed in the path of a parallel beam of light of a large aperture. The intensity of the light is 0.50 W/cm². If the sphere completely absorbs the radiation falling on it, find the force exerted by the light beam on the sphere.   


ANSWER: The force on the sphere will be the same as the force on a disc of radius 1.00 cm in the path of light. The area of this disc,

A =π*1² cm² 

   =π cm². 

The intensity of light, I =0.50 W/cm² 

  

So per cm², 0.50 J/s energy of photons are falling. The momentum of photons falling per second on a square cm,

  =0.50/3x10⁸ kg-m/s

  =1.67x10⁻⁹ kg-m/s.  

The momentum of photons falling on the cross-section of the sphere per second, 

p =1.67x10⁻⁹*π kg-m/s 

  =5.2x10⁻⁹ kg-m/s.

Since the sphere absorbs the light falling on it, the momentum of photons after absorption is zero. So the force exerted by the light beam on the sphere is the change of momentum of falling photons per second, which is

=p -0 

=p

=5.2x10⁻⁹ N.        

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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

Click here for → Questions for Short Answer 

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

Click here for → Questions for Short Answer

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)

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

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


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

CHAPTER- 3 - Kinematics - Rest and Motion

Click here for "Questions for short Answers"


Click here for "OBJECTIVE-I"


Click here for EXERCISES (Question number 1 to 10)


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"

Click here for "Questions for Short Answers"


Click here for "OBJECTIVE-II"

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