Q122 / 170 · Manipal (MET) 2015 · Physics
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.
- Correct answerA$1.1577 \times 10^{-38}$
- B$2.9 \times 10^{-35}$
- C$2.168 \times 10^{-36}$
- D$1.96 \times 10^{-20}$
Solution
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 $\begin{array}{ll} \lambda=550 \mathrm{~nm} \\ \Rightarrow & E_2-E_1=\frac{h c}{\lambda}=3.16 \times 10^{-19} \mathrm{~J} \\ \Rightarrow & \frac{N_2}{N_1}=\exp \left(\frac{-3.16 \times 10^{-19} \mathrm{~J}}{\left(1.38 \times 10^{-23} 1 / \mathrm{k}\right) \cdot(300 \mathrm{k})}\right) \\ \Rightarrow & \frac{N_2}{N_1}=1.1577 \times 10^{-38} \end{array}$
The middle of the visible range is taken at $\begin{array}{ll} \lambda=550 \mathrm{~nm} \\ \Rightarrow & E_2-E_1=\frac{h c}{\lambda}=3.16 \times 10^{-19} \mathrm{~J} \\ \Rightarrow & \frac{N_2}{N_1}=\exp \left(\frac{-3.16 \times 10^{-19} \mathrm{~J}}{\left(1.38 \times 10^{-23} 1 / \mathrm{k}\right) \cdot(300 \mathrm{k})}\right) \\ \Rightarrow & \frac{N_2}{N_1}=1.1577 \times 10^{-38} \end{array}$
Hence, the correct answer is option: 'A'.