A strange grass satellite is orbiting Rigel?

The expected number of molecules in the 3rd state is 0.04, if I understand the problem correctly.

The Rydberg is 10^7 m^(-1), which is about 13.6 eV. 4 m^(-1) corresponds to ε = 5*10^(-6) eV (rotational levels, that's why you need diatomic molecules). The Rigel temperature is about T_R = 1 eV. The Solar radius is 0.00465 AU. The Rigel radius is 62 times larger, i.e. 0.29 AU. At the distance of 7 AU, the radiated energy flux density σ (T_R)^4 (assuming that Rigel is a black body) decreases (7/0.29)^2 = 590 times.The grass reflection is about 0.1,so that 0.9 of this flux is absorbed by the satellite. The satellite is heated to the temperature T and radiates the energy back. Assuming that the satellite is a sphere of radius r, the energy balance is: π r^2 (0.9/590) σ(T_R)^4 = 4 π r^2 σ T^4, which gives T = 0.14 T_R = 0.14 eV. The probablitiy to find a molecule in the n-th state is equal to p_n = exp(-n ε/T) / Σ exp(-j ε/T) = exp(-n ε/T)*(1-exp( -ε/T)). For small n, p_n is about 4*10^(-5). Multiply by 1000 to get the answer.

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Al P, I corrected my answer and it is even smaller than originally. Indeed, 400 cm^-1 corresponds to ε = 5E-02 eV. However, 4 m^-1 is equal to 0.04 cm^-1, and the corresponding ε is four orders of magnitude smaller, ε=5E-06. Anyway, if ε= 5E-02 eV (which sounds more like oscillation levels) then, counting from n=0,

p_2 = exp(-2 ε/T)*(1-exp(-ε/T)) = 0.15.

There will we about 150 molecules in this state.