Cold interstellar molecular clouds often contain the molecule cyanogen (CN), whose first rotational excited states have an energy of 4.7x 10^-4 eV (above the ground state). There are actually three such excited states, all with the same energy. In 1941, studies of the absorption spectrum of starlight that passes | through these molecular clouds showed that for every ten CN molecules that are in the ground state, approximately three others are in the three first excited states (that is, an average of one in each of these states). To account for this data, astronomers suggested that the molecules might be in thermal equilibrium with some "reservoir" with a well-defined temperature. What is that temperature?

Cold interstellar molecular clouds often contain the molecule cyanogen (CN), whose first rotational excited states have an energy of 4.7x 10^-4 eV (above the ground state). There are actually three such excited states, all with the same energy. In 1941, studies of the absorption spectrum of starlight that passes | through these molecular clouds showed that for every ten CN molecules that are in the ground state, approximately three others are in the three first excited states (that is, an average of one in each of these states). To account for this data, astronomers suggested that the molecules might be in thermal equilibrium with some "reservoir" with a well-defined temperature.

The probability is provided by: Let E1 be the ground state energy and E2 be the first excited state energy.

$P\left(s\right)=\frac{1}{Z}{e}^{-E/kT}$

So,

$P\left(E_2\right)=\frac{1}{Z}{e}^{-E_2/kT}P\left(E_1\right)=\frac{1}{Z}{e}^{-E_1/kT}$

The ratio of these probabilities is:

$\frac{P\left(E_2\right)}{P\left(E_1\right)}=\frac{e^{-E_2/kT}}{e^{-E_1/kT}}=e^{-\left(E_2-E_1\right)/kT}$

There are typically three CN molecules in the initial excited states for every ten CN molecules in the ground state, implying an average of one in each of these states, therefore the probability ratio is 1/10, thus:

$e^{-\left(E_2-E_1\right)/kT}=\frac{1}{10}$

Solving for T:

$-\frac{E_2-E_1}{kT}=\ln \left(\frac{1}{10}\right)=-2.3026$

$T=\frac{E_2-E_1}{(2.3026)k}$

The energy of the first excited states have an energy of 4.7 x 10^-4 eV, which means E2- E1 = 4.7 x 10^-4 eV, substitute with Boltzmann constant in eV, k = 8.617 x 10^-5 eV/K, so:

$$\begin{gathered} T=\frac{4.7\times {10}^{-4}\mathrm{eV}}{(2.3026)\left(8.617\times {10}^{-5}\mathrm{eV}/\mathrm{K}\right)} \\ T=2.369K \end{gathered}$$

The temperature is closer to 2.7 K, which is the background temperature.