CBSE Class 12 Physics Dual Nature of Radiation and Matter Worksheet

Question. Which of the following metals is not sensitive to visible light?
a. Cesium
b. Sodium
c. Rubidium
d. Cadmium
Answer. D

Question. Photons are deflected by
a. electric field only
b. magnetic field only
c. electromagnetic field
d. None of these
Answer. D

Question. The momentum of a photon of wavelength λ is
a. hλ
b. h/λ
c. λ/h
d. h/c λ
Answer. B

Question. Maximum KE of photo electrons is 4 e V Then the stopping potential is
a. 4 V
b. 1.6 V
c. 4 J
d. 4 e V
Answer. A

ASSERTION- REASON

Directions: In each of the following questions, a statement of Assertion (A) is given followed by a corresponding statement of Reason (R) just below it. Of the statements, mark the correct answer as:
(A)If both assertion and reason are true and reason is the correct explanation of assertion
(B)If both assertion and reason are true but reason is not the correct explanation of assertion
(C)If assertion is true and reason is false
(D)If both assertion and reason are false

Question. Assertion: A photon has no rest mass , yet it carries definite momentum.
Reason: Momentum of photon is due to its energy and hence its equivalent mass.
(a) A
(b) B
(c) C
(d) D
Answer. A

Question. Assertion: Photoelectric effect demonstrates the wave nature of light.
Reason. The number of photoelectrons is proportional to the frequency of light.
(a) A
(b) B
(c) C
(d) D
Answer. D

Question. Assertion: When light of certain wavelength falls on a metal surface it ejects electron.
Reason: Light has wave nature.
(a) A
(b) B
(c) C
(d) D
Answer. B

Question. Assertion: As work function of a material increases by some mechanism, it requires greater energy to excite the electrons from its surface.
Reason. A plot of stopping potential (V) versus frequency (v) for different materials, has greater slope for metals with greater work functions
(a) A
(b) B
(c) C
(d) D
Answer. C

Question. Assertion: Light of frequency 1.5 times the threshold frequency is incident on photosensitive material. If the frequency is halved and intensity is doubled the photo current remains unchanged.
Reason:The photo electric current varies directly with the intensity of light and frequency of light.
(a) A
(b) B
(c) C
(d) D
Answer. D

Question. Assertion: The de-Broglie wavelength of a neutron when its kinetic energy is k is λ. Its wavelength is 2 λ when its kinetic energy is 4k.
Reason. The de - Broglie wavelength λ is proportional to square root of the kinetic energy.
(a) A
(b) B
(c) C
(d) D
Answer. A

1 Marks Questions

Question. A proton and an electron have same kinetic energy. Which one has greater de-Broglie wavelength and why?
Answer. λ = h /𝑝 = ℎ/ √2𝑚𝐾, λe > λp

Question. The de-Broglie wavelength associated with an electron accelerated through a potential difference V is λ. What will be its wavelength when the accelerating potential is increased to 4V?
Answer. 𝜆 = 12.27/ √𝑉 𝐴̇⁰
for V = 4 V, 𝜆 = 12.27/ √4 𝐴̇ = 6.135𝐴̇0

Question. The maximum kinetic energy of a photoelectron is 3 eV. What is its stopping potential ?
Answer. KEmax = 3 eV => 3eV = eV₀ => V₀ = 3V

Question. It is difficult to eject out an electron from copper than sodium, which of the two metals has greater work function and which has greater threshold wavelength?
Answer. Since electron ejection is difficult from copper than sodium, so copper has greater work function than sodium. As threshold wavelength is inversely related with work function, so sodium has higher threshold wavelength than copper.

2 Marks Questions

Question. An alpha particle and a proton are accelerated through same potential difference. Find the ratio (vᾳ/vp) of velocities acquired by two particles.
Answer. Gain in K.E = qV
½ m_pv_p² =qp V, ½ m_αv_α² =q_α V
½ m_αvα² /       ½ m_pvp² = q_α /q_α
V_α² / V_p² = q_α /q_α    x m_p /m_α = 2/1 x 1/4 =1/2
V_α / V_p = 1/√2

Question. (i). The stopping potential in an experiment on photoelectric effect is 1.5 V. What is the maximum kinetic energy of the photoelectrons emitted?
(ii) State one factor which determines the intensity of light in the photon picture of light.
Answer. (i) 1.5eV
(ii) Number of photons emitted per second

Question. Monochromatic light of frequency 6x10¹⁴ Hz is produced by a laser. The power emitted is 2x10³ W i) What is the energy of photon in the light? ii) How many photons per second on the average are emitted by the source
Answer. ν = 6 x 10¹⁴ Hz, P = 2 x 10^-3 W
i) energy of photon = E = hν = 6.6x10^-34 x 6 x10¹⁴ = 3.98 x10^-19 J
ii) If N is the number of photons emitted per second by the source then Power = N x energy of photon
P = NE N = P/E = 2x10^-3/3.98 x 10^-19 = 5 x 10¹⁵ photons per second

Question. The work function of a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?
Answer. E = hc/λ = 6.6×10⁻³⁴×3×10⁸ / 330 x10^-9 = 3.767eV   ϕo = 4.2eV
Since E < W0 no photoelectric emission.

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