Past exam of the physics B course of the University of Cambridge / 2015 / Paper 1 / Section A / A5

After the photon has thermalized, its energy $E_{\gamma }=hc/\lambda$ belongs to the macroscopic body's internal energy. The body's heat capacity makes the resulting temperature change negligible. Its equilibrium entropy therefore increases byThe Boltzmann entropy $S=k_B\ln \Omega$ then gives the microstate multiplier from photon absorption:For scale, explicitly choosing a representative visible wavelength $\lambda =500\mathrm{nm}$ and room temperature $T=300\mathrm{K}$ gives $\Delta S=1.32\times 10^{-21}\mathrm{J}{\mathrm{K}}^{-1}$, or $\Delta S/k_B=95.9$, and a multiplier approximately $4.5\times 10^{41}$. These are illustrative choices, not additional supplied data: the exact multiplier depends exponentially on the particular visible wavelength and temperature. The exact Planck constant, speed of light, and Boltzmann constant fix the conversion.^[ref]