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Dual Nature of Radiation question

2023 · 30 Jan · Shift 1 · Q42
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  5. /2023 · 30 Jan · Shift 1 · Q42

Dual Nature of Radiation question

2023 · 30 Jan · Shift 1 · Q42

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
A small object at rest, absorbs a light pulse of power 20 mW20 \mathrm{~mW}20 mW and duration 300 ns300 \mathrm{~ns}300 ns. Assuming speed of light as 3×108 m/s3 \times 10^{8} \mathrm{~m} / \mathrm{s}3×108 m/s, the momentum of the object becomes equal to :
  1. A
    1×10−17 kg m/s1 \times 10^{-17} \mathrm{~kg} \mathrm{~m} / \mathrm{s}1×10−17 kg m/s
  2. B
    0.5×10−17 kg m/s0.5 \times 10^{-17} \mathrm{~kg} \mathrm{~m} / \mathrm{s}0.5×10−17 kg m/s
  3. C
    3×10−17 kg m/s3 \times 10^{-17} \mathrm{~kg} \mathrm{~m} / \mathrm{s}3×10−17 kg m/s
  4. D
    2×10−17 kg m/s2 \times 10^{-17} \mathrm{~kg} \mathrm{~m} / \mathrm{s}2×10−17 kg m/s
View written solutionFree

Correct answer: D

  1. Use momentum of absorbed light

When a body completely absorbs light of energy EEE, the momentum transferred is

p=Ecp = \frac{E}{c}p=cE​

where c=3×108 m/sc = 3 \times 10^8\ \text{m/s}c=3×108 m/s.

  1. Find the energy of the light pulse

Given:

  • Power, P=20 mW=20×10−3 WP = 20\ \text{mW} = 20 \times 10^{-3}\ \text{W}P=20 mW=20×10−3 W
  • Duration, t=300 ns=300×10−9 st = 300\ \text{ns} = 300 \times 10^{-9}\ \text{s}t=300 ns=300×10−9 s

Energy of pulse:

E=PtE = PtE=Pt

E=(20×10−3)(300×10−9)E = (20 \times 10^{-3})(300 \times 10^{-9})E=(20×10−3)(300×10−9)

E=6000×10−12E = 6000 \times 10^{-12}E=6000×10−12

E=6×10−9 JE = 6 \times 10^{-9}\ \text{J}E=6×10−9 J

  1. Calculate momentum transferred

p=Ec=6×10−93×108p = \frac{E}{c} = \frac{6 \times 10^{-9}}{3 \times 10^8}p=cE​=3×1086×10−9​

p=2×10−17 kg m/sp = 2 \times 10^{-17}\ \text{kg m/s}p=2×10−17 kg m/s

  1. Match with the options

This corresponds to:

  • Option D: 2×10−17 kg m/s2 \times 10^{-17}\ \text{kg m/s}2×10−17 kg m/s

Therefore, the correct answer is D.

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