Q156 / 160 · AP EAMCET 2024 (22 May Shift 1) · Physics
The ratio of minimum wave length of Balmer series to maximum wavelength in Brackett series in hydrogen spectrum is
- A25:16
- B4:36
- Correct answerC9:100
- D100:9
Solution
For hydrogen atom,
$\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)$
For Balmer series, $\mathrm{n}_1=2, \mathrm{n}_2=3,4,5, \ldots . \infty$ $\frac{1}{\lambda_{\min }}=\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{\infty^2}\right)=\frac{\mathrm{R}}{4} \Rightarrow \lambda_{\min }=\frac{4}{\mathrm{R}}$
For Bracket series, $n_1=4, n_2=5,6,7, \ldots \infty$ $\begin{aligned} & \therefore \frac{1}{\lambda_{\max }}=\mathrm{R}\left(\frac{1}{4^2}-\frac{1}{5^2}\right)=\frac{9 \mathrm{R}}{16 \times 25} \Rightarrow \lambda_{\min }=\frac{16 \times 25}{9 \mathrm{R}} \\ & \therefore \frac{\lambda_{\min }}{\lambda_{\max }}=\frac{\frac{4}{\mathrm{R}}}{\frac{16 \times 25}{9 \mathrm{R}}}=9: 100 \end{aligned}$
For Balmer series, $\mathrm{n}_1=2, \mathrm{n}_2=3,4,5, \ldots . \infty$ $\frac{1}{\lambda_{\min }}=\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{\infty^2}\right)=\frac{\mathrm{R}}{4} \Rightarrow \lambda_{\min }=\frac{4}{\mathrm{R}}$
For Bracket series, $n_1=4, n_2=5,6,7, \ldots \infty$ $\begin{aligned} & \therefore \frac{1}{\lambda_{\max }}=\mathrm{R}\left(\frac{1}{4^2}-\frac{1}{5^2}\right)=\frac{9 \mathrm{R}}{16 \times 25} \Rightarrow \lambda_{\min }=\frac{16 \times 25}{9 \mathrm{R}} \\ & \therefore \frac{\lambda_{\min }}{\lambda_{\max }}=\frac{\frac{4}{\mathrm{R}}}{\frac{16 \times 25}{9 \mathrm{R}}}=9: 100 \end{aligned}$
Hence, the correct answer is option: 'C'.