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MCQEasyJEE 2023Characteristics of EM Waves

JEE Physics 2023 Question with Solution

A small object at rest absorbs a light pulse of power 20 mW20 \, \text{mW} and duration 300 ns300 \, \text{ns}. Assuming speed of light as 3×108 m/s3 \times 10^8 \, \text{m/s}, the momentum of the object becomes equal to:

  • A

    0.5×10−17 kg m/s0.5 \times 10^{-17} \, \text{kg m/s}

  • B

    2×10−17 kg m/s2 \times 10^{-17} \, \text{kg m/s}

  • C

    3×10−17 kg m/s3 \times 10^{-17} \, \text{kg m/s}

  • D

    1×10−17 kg m/s1 \times 10^{-17} \, \text{kg m/s}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Power of the light pulse is P=20×10−3 WP = 20 \times 10^{-3} \, \text{W}, duration is t=300×10−9 st = 300 \times 10^{-9} \, \text{s}, and speed of light is c=3×108 m/sc = 3 \times 10^8 \, \text{m/s}.

Find: Momentum gained by the object after absorbing the light pulse.

For absorbed radiation, the momentum transferred to the object equals the momentum carried by the pulse.

E=P⋅tE = P \cdot tp=Ec=P⋅tcp = \frac{E}{c} = \frac{P \cdot t}{c}

Substituting the given values,

p=20×10−3⋅300×10−93×108p = \frac{20 \times 10^{-3} \cdot 300 \times 10^{-9}}{3 \times 10^8}p=6×10−93×108p = \frac{6 \times 10^{-9}}{3 \times 10^8}p=2×10−17 N sp = 2 \times 10^{-17} \, \text{N s}

Therefore, the momentum of the object is 2×10−17 kg m/s2 \times 10^{-17} \, \text{kg m/s}. The solution concludes that the correct option is A.

Momentum Transfer Explanation

Given: The object is initially at rest and it absorbs the entire light pulse.

Find: The final momentum of the object.

The radiation initially has momentum

pinitial=P⋅tcp_{\text{initial}} = \frac{P \cdot t}{c}

Since the pulse is absorbed, the radiation has zero final momentum, so the entire incident momentum is transferred to the object.

Δp=pinitial=P⋅tc\Delta p = p_{\text{initial}} = \frac{P \cdot t}{c}

Using

P=20×10−3 W,t=300×10−9 s,c=3×108 m/sP = 20 \times 10^{-3} \, \text{W}, \quad t = 300 \times 10^{-9} \, \text{s}, \quad c = 3 \times 10^8 \, \text{m/s}

we get

Δp=20×10−3⋅300×10−93×108=2×10−17 N s\Delta p = \frac{20 \times 10^{-3} \cdot 300 \times 10^{-9}}{3 \times 10^8} = 2 \times 10^{-17} \, \text{N s}

Since 1 N s=1 kg m/s1 \, \text{N s} = 1 \, \text{kg m/s}, the momentum is 2×10−17 kg m/s2 \times 10^{-17} \, \text{kg m/s}. The numerical value matches option B, but the solution labels it as option A. the answer is recorded as A.

Common mistakes

  • Using p=2Ecp = \frac{2E}{c} instead of p=Ecp = \frac{E}{c}. The factor of 22 applies to reflection, not absorption. Here the object absorbs the pulse, so use p=Ecp = \frac{E}{c}.

  • Forgetting to convert 20 mW20 \, \text{mW} and 300 ns300 \, \text{ns} into SI units. Direct substitution without converting milli and nano gives the wrong power of 1010. Convert first to watts and seconds.

  • Treating power itself as energy. Power is energy per unit time, so first calculate energy using E=PtE = Pt, then use p=Ecp = \frac{E}{c}.

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