The ratio of minimum wave length of Balmer series to maximum wavelength in Brackett series in hydrogen…

The ratio of minimum wave length of Balmer series to maximum wavelength in Brackett series in hydrogen spectrum is
  1. 25:16
  2. 4:36
  3. 9:100
  4. 100:9

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, $n_1=4, n_2=5,6,7, \ldots \infty$ $\begin{aligned} & \therefore \frac{1}{\lambda_{\max }}=\mathrm{R}\left(\frac{1}{4^2}-\frac{1}{5^2}\right)=\frac{9 \mathrm{R}}{16 \times 25} \Rightarrow \lambda_{\min }=\frac{16 \times 25}{9 \mathrm{R}} \\ & \therefore \frac{\lambda_{\min }}{\lambda_{\max }}=\frac{\frac{4}{\mathrm{R}}}{\frac{16 \times 25}{9 \mathrm{R}}}=9: 100 \end{aligned}$

Asked in: AP EAMCET 2024 (22 May Shift 1)

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