Class 12 CBSE PHYSICS Dual nature of Radiation and Matter
Questions/Sums on Important Concepts
Hi students, just 9 days left for CBSE Class 12 Physics Board Exam 2025. So as a ready reckoner, I have given the answers to some of the questions here.
For derivations and graphs, please refer your textbook.
1. Why do metals have a large number of electrons?
In metals, the electrons in the outermost shells are loosely bound to the nucleus. Thus there are a large number of free electrons moving in a random manner inside the metals at room temperature.
2. Define work function of a metal.
The minimum amount of energy required by an electron to just escape from the metal surface.
3. Name the different types of electron emission.
Thermionic emission
Field emission
Photoelectric emission
Secondary emission
4. Draw the graph of effect of intensity of light on photoelectric current.
Refer textbook.
5. Define Stopping potential.
The value of the retarding potential at which the photoelectric current becomes zero is called stopping potential or cut off potential. It is denoted by Vₒ.
6. Mention the laws of photoelectric emission.
For a given photosensitive material
1. The photoelectric current is directly proportional to the intensity of light.
2. There exists a minimum cut off frequency below which no photoelectrons are emitted. (threshold frequency).
3. For radiation with higher frequency more than the threshold frequency, the maximum KE of the photoelectrons is directly proportional to the frequency of incident radiation (and it is independent of the intensity).
4. The photoelectric emission is an instantaneous process.
7. Explain why photoelectric effect cannot be explained on the basis of wave nature of light.
It could not explain the main features of photoelectric effect.
It fails to explain the threshold frequency.
The instantaneous nature of photoelectric emission.
The independence of KE on intensity.
8. Explain Einstein's theory of photoelectric effect.
Refer textbook.
9. What is a photon?
It is defined as one quantum of electromagnetic radiation.
10. Write some basic features of photon.
It travels with the speed of light.
Its rest mass is zero
Frequency does not change as it travels from one medium to another.
The speed changes as it travels through different media due to change in wavelength.
They are electrically neutral and are not deflected by electric and magnetic fields.
Energy of a photon E= hc /λ
Momentum of a photon p = h/λ
Mass of a photon = hν/c²
11. How will you justify that the rest mass of photon is zero?
12. Do all the photons have same mass?
No. Different radiations have different frequencies. So accordingly photons have different masses. (Mass of a photon = hν/c²)
13. What happens to the wavelength of a photon after it collides with an electron.
The wavelength of the photon increases. A part of its energy is transferred to the colliding electron and so its energy decreases and wavelength increases.
14. Does each incident photon essentially eject an electron?
No. Only 1 % of incident photons are capable of ejecting out electrons.
15. What is the momentum of a photon from ultraviolet light of wavelength 332 nm?
16. Draw a graph showing the variation of stopping potential with the frequency of incident radiation in relation to photoelectric effect. Deduce an expression for the slope of this graph using Einstein's photoelectric equation.
Refer textbook
17. A monochromatic source emitting light of wavelength 600 nm has a power output of 66 W. Find the number of photons emitted by this source in 2 minutes.
18. The work function of caesium is 2.14 eV. Find (i) the threshold frequency for caesium and (ii) the wavelength of the incident light if the photo current is brought to zero by a stopping potential of 0.60 eV.
19. Derive de-Broglie wave equation for material particles.
Refer textbook
20. Deduce an expression for the de-Broglie wavelength of an electron accelerated through a potential difference of V volts.
Refer textbook
21. The kinetic energy of the electron orbiting in the first excited state of hydrogen atom is 3.4 eV. Find the de-Broglie wavelength associated with it.
h= 6.63 x 10⁻³⁴ Js
K = 3.4 eV = 3.4 x 1.6 x 10⁻¹⁹ J
m= 9.1 x 10^-31 kg
substituting all the values in λ = h/√2mK,
we get 6.63 x 10⁻¹⁰m
22. Find the de-Broglie wavelength associated with an electron moving with a velocity 0.5 c and rest mass equal to 9.1 x 10^-31 kg.
m₀ = 9.1 x 10^-31 kg
v= 0.5 c = 0.5 x 3 x 10⁸ = 1.5 x 10⁸ m/s
h= 6.63 x 10⁻³⁴ Js
Mass of an electron in motion
m=m₀ /(√(1-(v²/c²))
Substituting v=0.5 c,
m=m₀/√0.75
Substituting all the above values in λ = h/mv,
we get λ = 4.2 x 10⁻¹² m
23. The maximum kinetic energy of a photoelectron is 3 eV. What is its stopping potential?
V₀ = Kmax / e = 3 eV/e = 3V
24. What is the effect on the velocity of photoelectrons if the wavelength of incident light is decreased?
25. Two metals A and B have work functions 4 eV and 10 eV respectively. Which metal has higher threshold wavelength?
Wₒ = hc/λ₀ which implies λ₀ ∝ 1/Wₒ
Therefore metal A with lower work function has higher threshold wavelength.
26. How does the maximum kinetic energy of electrons emitted vary with the work function of the metal?
Kmax = hν - Wₒ
Larger the work function smaller is the maximum KE.
27. Is photoelectric emission possible at all frequencies? Give reason for your answer.
No. The frequency of the incident radiation must be greater than the threshold frequency. Photoelectric emission is possible only if the energy of the incident photon is greater than the work function of the metal.
28. Every metal has a definite work function. Why do photoelectrons not come out with same energy if incident radiation is monochromatic? Why is there energy distribution of photoelectrons?
This is because electrons belonging to different energy levels need different energies to come out of the metal surface. Photons transfer different amount of energy to different electrons.
29. Define electron volt.
The energy gained by an electron when accelerated through a potential difference of 1 volt is called electron volt. (eV)
1 eV = 1.6 x 10⁻¹⁹ J
30. How many electron volts make up one joule?
1 J = 1/(1.6 x 10⁻¹⁹) = 6.25 x 10¹⁸ eV
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