A population inversion for two energy levels is often described by assigning a negative Kelvin temperature to the system. What negative temperature would describe a system in which the population of the upper energy level exceeds that of the lower level by and the energy difference between the two levels is 2.26 eV?

The population of the upper energy level in comparison to the lower energy level, ${\mathrm{N}}_2=1.1{\mathrm{N}}_1$

The energy difference between the two levels,$∆\mathrm{E}=2.26\mathrm{eV}$

The law states that the chances of a molecule gaining energy decreases with increasing energy equal to the Boltzmann factor $\mathbf{exp}\left(-\epsilon /\mathrm{kT}\right)$.

Using the Boltzmann-distribution equation which is the expression for the probability for stimulated emission of radiation to the probability for spontaneous emission of radiation under thermal equilibrium, get the negative temperature to describe the system under the given conditions.

Formula:

The Boltzmann energy distribution equation,

$\frac{{\mathrm{N}}_1}{{\mathrm{N}}_2}={\mathrm{e}}^{-\left({\mathrm{E}}_2-{\mathrm{E}}_1\right)/\mathrm{kT}}$ ….. (1)

Here, the Boltzmann constant is,$\mathrm{k}=1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}$

Let us consider two levels with energies of the upper level as${\mathrm{E}}_2$ and energy of the lower level as ${\mathrm{E}}_1$ where ${\mathrm{E}}_2>{\mathrm{E}}_1$.

Rewrite equation (1) as below.

$$\begin{gathered} \frac{{\mathrm{N}}_1}{{\mathrm{N}}_2}={\mathrm{e}}^{-\left({\mathrm{E}}_2-{\mathrm{E}}_1\right)/-\left(\mathrm{kT}\right)} \\ ={\mathrm{e}}^{-\left({\mathrm{E}}_2-{\mathrm{E}}_1\right)/\left(\mathrm{kT}\right)} \\ >1 \\ {\mathrm{N}}_2>{\mathrm{N}}_1 \end{gathered}$$

Now, for the case of population inversion, we are given a negative temperature that is applied to the system, which can be calculated using the given data in equation (1) as follows:

$$\begin{gathered} T=-\left|T\right| \\ =-\frac{E_2-E_1}{k\ln \left(N_2/N_1\right)} \\ =-\frac{2.26eV\left(1.6\times {10}^{-19}J/eV\right)}{\left(1.38\times {10}^{-23}J/K\right)\ln \left(1.1\right)} \\ =-2.75\times {10}^{5}K \end{gathered}$$

Hence, the value of the negative temperature is $-2.75\times {10}^{5}K$.