If the total energy transferred to a surface in time $t$ is $6.48 \times 10^{5} \mathrm{~J}$, then the magnitude of the total momentum delivered to this surface for complete absorption will be:

(B) $2.16 \times 10^{-3} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$

(A) $2.46 \times 10^{-3} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$

(C) $1.58 \times 10^{-3} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$

(D) $4.32 \times 10^{-3} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$

JEE Main · 2024 · 30 Jan 2024 · Shift 2 · EM Waves

If the total energy transferred to a surface in time

t

6.48 \times 1 0^{5} J

, then the magnitude of the total momentum delivered to this surface for complete absorption will be:

(A) $2.46 \times 1 0^{- 3} kg m / s$
(B) $2.16 \times 1 0^{- 3} kg m / s$ ✓ Correct
(C) $1.58 \times 1 0^{- 3} kg m / s$
(D) $4.32 \times 1 0^{- 3} kg m / s$

Answer: (B)

Solution

For electromagnetic radiation, energy and momentum are related by

p = \frac{E}{c}

For complete absorption, the total momentum delivered equals

p = \frac{6.48 \times 1 0 ^{5}}{3 \times 1 0 ^{8}} = 2.16 \times 1 0^{- 3} kg m/s

JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Single correct option

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