Momentum Delivered By Radiation

Electromagnetic radiation carries momentum equal to the energy it delivers divided by the speed of light, p = U/c, for a perfectly absorbing surface. First find the total energy absorbed: \cup = I \times A \times t, where the area collects the flux over the exposure time. With I = 1.2 \times 10^{3}\ \text{W/m}^2, A = 0.50\ \text{m}^2, and t = 600\ \text{s}, the energy is \cup = (1.2 \times 10^{3})(0.50)(600) = 3.6 \times 10^{5}\ \text{J}. The momentum is then p = U/c = (3.6 \times 10^{5})/(3.0 \times 10^{8}) = 1.2 \times 10^{-3}\ \text{kg·m/s}. The 2.4 \times 10^{-3}\ \text{kg·m/s} option wrongly applies the reflector factor of two. The 6.0 \times 10^{-4}\ \text{kg·m/s} value uses only half the exposure time. The 3.6 \times 10^{-3}\ \text{kg·m/s} answer forgets to divide by c correctly. This two-step combination of energy collection and the p = U/c relation is characteristic of JEE Advanced radiation problems, where the difficulty lies less in the physics than in correctly assembling intensity, area, and time before applying the momentum relation. A common error is to forget converting ten minutes into 600 seconds, which would scale the answer down by a large factor. As a check, the units work out to joules over metres-per-second giving kilogram-metres-per-second, confirming a valid momentum, and the tiny magnitude is expected given how small radiation momentum is.