Q155 / 160 · TS EAMCET 2024 (09 May Shift 2) · Physics
The ratio of the wavelengths of radiation emitted when an electron in the hydrogen atom jumps from 4th orbit to 2 nd orbit and from 3 rd orbit to 2 nd orbit is
- A$27: 25$
- B$20: 27$
- Correct answerC$20: 25$
- D$25: 27$
Solution
Wavelength of radiation emitted from hydrogen atom is given by
$\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right]$
In first case $\frac{1}{\lambda_1}=\mathrm{R}\left[\frac{1}{2^2}-\frac{1}{4^2}\right]=\mathrm{R}\left[\frac{1}{4}-\frac{1}{16}\right]=\frac{3 \mathrm{R}}{16}$
In second case $\begin{aligned} & \frac{1}{\lambda_2}=\mathrm{R}\left[\frac{1}{2^2}-\frac{1}{3^2}\right]=\mathrm{R}\left[\frac{1}{4}-\frac{1}{9}\right]=\frac{5 \mathrm{R}}{36} \\ & \therefore \frac{\frac{1}{\lambda_1}}{\frac{1}{\Lambda_2}}=\frac{\frac{3 R}{16}}{\frac{5 R}{36}}=\frac{3 \mathrm{R}}{16} \times \frac{36}{5 \mathrm{R}}=\frac{27}{20} \\ & \Rightarrow \frac{\lambda_1}{\lambda_2}=\frac{20}{27} \end{aligned}$
In first case $\frac{1}{\lambda_1}=\mathrm{R}\left[\frac{1}{2^2}-\frac{1}{4^2}\right]=\mathrm{R}\left[\frac{1}{4}-\frac{1}{16}\right]=\frac{3 \mathrm{R}}{16}$
In second case $\begin{aligned} & \frac{1}{\lambda_2}=\mathrm{R}\left[\frac{1}{2^2}-\frac{1}{3^2}\right]=\mathrm{R}\left[\frac{1}{4}-\frac{1}{9}\right]=\frac{5 \mathrm{R}}{36} \\ & \therefore \frac{\frac{1}{\lambda_1}}{\frac{1}{\Lambda_2}}=\frac{\frac{3 R}{16}}{\frac{5 R}{36}}=\frac{3 \mathrm{R}}{16} \times \frac{36}{5 \mathrm{R}}=\frac{27}{20} \\ & \Rightarrow \frac{\lambda_1}{\lambda_2}=\frac{20}{27} \end{aligned}$
Hence, the correct answer is option: 'C'.