Question

If levels $1$ and $2$ are separated by an energy $E_2-E_1$, such that the corresponding transition frequency falls in the middle of the visible range, calculate the ratio of the populations of two levels in the thermal equilibrium at room temperature.

• A. $1.1577\times 10^{-38}$
• B. $29\times 10^{-35}$
• C. $2.168\times 10^{-36}$
• D. $1.96\times 10^{-20}$

Answer 1

Answer: (A)

At thermal equilibrium, the ratio $\frac{N_2}{N_1}$ is given as
$\frac{N_2}{N_1}=\exp \left(-\frac{E_2-E_1}{kT}\right)$
The middle of the visible range is taken at $\lambda =550nm$
$\Rightarrow E_2-E_1=\frac{hc}{\lambda }=3.16\times 10^{-19}J$
$\Rightarrow \frac{N_2}{N_1}=\exp \left(\frac{-3.16\times 10^{-19}J}{\left(1.38\times 10^{-23}1/k\right)\cdot (300k)}\right)$
$\Rightarrow \frac{N_2}{N_1}=1.1577\times 10^{-38}$