Q2 / 160 · TS EAMCET 2024 (09 May Shift 1) · Chemistry

Which one of the following corresponds to the wavelength of line spectrum of H atom in its Balmer series? ( $\mathrm{R}=$ Rydberg constant)

  • A
    $\frac{9}{8 R}$
  • Correct answer
    B
    $\frac{100}{21 R}$
  • C
    $\frac{25}{24 R}$
  • D
    $\frac{16}{15 R}$

Solution

$\overline{\mathrm{v}}=-\mathrm{R}_{\mathrm{H}}\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right]\left(\mathrm{n}_1\gt\mathrm{n}_2\right)$ $\begin{aligned} & =R_H\left[\frac{1}{n_2^2}-\frac{1}{n_1^2}\right]=R_H\left[\frac{n_1^2-n_2^2}{n_1^2 n_2^2}\right] \\ & \Rightarrow \lambda=\frac{1}{\bar{v}}=\frac{n_1^2 n_2^2}{R_H\left(n_1^2-n_2^2\right)} \end{aligned}$ for $\mathrm{n}_1 \equiv 5$ and $\mathrm{n}_2=2$ (for Balmer Series), $\lambda=\frac{(5)^2 \times(2)^2}{\mathrm{R}_{\mathrm{H}}(25-4)}=\frac{100}{21 \mathrm{R}_{\mathrm{H}}}$

Hence, the correct answer is option: 'B'.

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