= Solution
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 by
$$
\boxed{\Delta S\simeq\frac{hc}{\lambda T}.}
$$
The <Boltzmann entropy> $S=k_B\ln\Omega$ then gives the <microstate multiplier from photon absorption>:
$$
\boxed{\frac{\Omega_f}{\Omega_i}=\exp(\Delta S/k_B)\simeq
\exp\!\left(\frac{hc}{\lambda k_BT}\right).}
$$
For scale, explicitly choosing a representative visible <wavelength> $\lambda=500\ \mathrm{nm}$ and room <temperature> $T=300\ \mathrm K$ gives $\Delta S=1.32\times10^{-21}\ \mathrm{J\,K^{-1}}$, or $\Delta S/k_B=95.9$, and \b[a multiplier approximately $4.5\times10^{41}$]. 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.<Fundamental Physical Constants — Complete Listing by National Institute of Standards and Technology and others>[2022 CODATA adjustment, Planck constant, speed of light in vacuum, and Boltzmann constant entries]{ref}
Back to article page
