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Dual Nature of Radiation question

2024 · 30 Jan · Shift 2 · Q78
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  5. /2024 · 30 Jan · Shift 2 · Q78

Dual Nature of Radiation question

2024 · 30 Jan · Shift 2 · Q78

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
If the total energy transferred to a surface in time t\mathrm{t}t is 6.48×105 J6.48 \times 10^5 \mathrm{~J}6.48×105 J, then the magnitude of the total momentum delivered to this surface for complete absorption will be:
  1. A
    2.16×10−3 kg m/s2.16 \times 10^{-3} \mathrm{~kg} \mathrm{~m} / \mathrm{s}2.16×10−3 kg m/s
  2. B
    2.46×10−3 kg m/s2.46 \times 10^{-3} \mathrm{~kg} \mathrm{~m} / \mathrm{s}2.46×10−3 kg m/s
  3. C
    1.58×10−3 kg m/s1.58 \times 10^{-3} \mathrm{~kg} \mathrm{~m} / \mathrm{s}1.58×10−3 kg m/s
  4. D
    4.32×10−3 kg m/s4.32 \times 10^{-3} \mathrm{~kg} \mathrm{~m} / \mathrm{s}4.32×10−3 kg m/s
View written solutionFree

Correct answer: A

  1. For electromagnetic radiation, the relation between energy and momentum is

E=pcE = pcE=pc

So, the total momentum delivered when total energy EEE is completely absorbed is

p=Ecp = \frac{E}{c}p=cE​

  1. Given:

E=6.48×105 JE = 6.48 \times 10^5\ \text{J}E=6.48×105 J

and

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

Therefore,

p=6.48×1053×108p = \frac{6.48 \times 10^5}{3 \times 10^8}p=3×1086.48×105​

  1. Simplify:

p=6.483×105−8p = \frac{6.48}{3} \times 10^{5-8}p=36.48​×105−8

p=2.16×10−3 kg m/sp = 2.16 \times 10^{-3}\ \text{kg m/s}p=2.16×10−3 kg m/s

  1. Compare with the options:
  • A: 2.16×10−3 kg m/s2.16 \times 10^{-3}\ \text{kg m/s}2.16×10−3 kg m/s ✓
  • B: 2.46×10−3 kg m/s2.46 \times 10^{-3}\ \text{kg m/s}2.46×10−3 kg m/s
  • C: 1.58×10−3 kg m/s1.58 \times 10^{-3}\ \text{kg m/s}1.58×10−3 kg m/s
  • D: 4.32×10−3 kg m/s4.32 \times 10^{-3}\ \text{kg m/s}4.32×10−3 kg m/s

Hence, the correct option is A.

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