Discussion
If the total energy transferred to a completely absorbing surface by an EM wave in unit time is 3.6 J, then the radiation pressure exerted by the wave on the surface is
Answer: E
The key gives \(1.2 \times 10^{-8}\) Nm-2.
For a completely absorbing surface, the momentum delivered per second is the energy delivered per second divided by \(c\). Reading the 3.6 J per unit time as falling on unit area (an intensity of 3.6 W m-2), the pressure is
\[ \begin{aligned} P &= \frac{I}{c} \\ &= \frac{3.6}{3 \times 10^{8}} \\ &= 1.2 \times 10^{-8}\ \text{Nm}^{-2} \end{aligned} \]
The other options multiply by \(c\) instead of dividing, which gives enormous pressures. A perfectly reflecting surface would receive twice this pressure.
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