Q1 / 160 · TS EAMCET 2023 (12 May Shift 2) · Chemistry

The wavelength of second line of Balmer series of hydrogen atom is $\lambda \mathrm{nm}$. What is the wavelength of first line of Lyman series of $\mathrm{He}^{+}$ion (in $\mathrm{nm}$ )?

  • Correct answer
    A
    $\frac{\lambda}{16}$
  • B
    $\frac{16}{\lambda}$
  • C
    $\frac{16}{3 \lambda}$
  • D
    $\frac{3 \lambda}{16}$

Solution

$$ \frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_1{ }^2}-\frac{1}{\mathrm{n}_2{ }^2}\right] $$ For the second line of Balmer series of $\mathrm{H}$-atom:- $$ \frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{2^2}-\frac{1}{4^2}\right]=\frac{3 \mathrm{R}}{16}=\frac{1}{\lambda} \Rightarrow \lambda=\frac{16}{3 \mathrm{R}} $$ For the first line of Lyman series of $\mathrm{He}^{+}$ion:- $$ \begin{aligned} & \frac{1}{\lambda_2}=\mathrm{R}\left[\frac{1}{(1)^2}-\frac{1}{2^2}\right] \mathrm{Z}^2=\mathrm{R}\left[\frac{3}{4}\right](2)^2=3 \mathrm{R} \\ & \Rightarrow \lambda_2=\frac{1}{3 \mathrm{R}} \\ & \Rightarrow \lambda=16 \times \lambda_2 \text { or } \lambda_2=\frac{\lambda}{16} \end{aligned} $$

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

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