Answer
$9\times10^{-7}$
Work Step by Step
The naturally occurring population ratio $\frac{N_x}{N_0}$ of the two states is due to thermal agitation of the gas atoms
$\frac{N_x}{N_0}=e^{-\frac{(E_x-E_0)}{kT}}$
where, $(E_x-E_0)$ is the energy separation between the two states.
Substituting the given values: $(E_x-E_0)=2\times1.2\;eV=2.4\;eV$; $T=2000\;K$
$\frac{N_x}{N_0}=e^{-\frac{2.4}{8.62\times10^{-5}\times 2000}}$
$\frac{N_x}{N_0}\approx9\times10^{-7}$
Therefore, the ratio of the number of atoms in the 13th excited state to the number in the 11th excited state is $9\times10^{-7}$
