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
$\frac{N_{2}}{N_{1}}=\exp \left(-\frac{E_{2}-E_{1}}{k T}\right)$
The middle of the visible range is taken at $\lambda=550 \,nm$
$\Rightarrow E_{2}-E_{1}=\frac{h c}{\lambda}=3.16 \times 10^{-19} J$
$\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(300 k)}\right) $
$\Rightarrow \frac{N_{2}}{N_{1}}=1.1577 \times 10^{-38}$