Q1 / 160 · TS EAMCET 2024 (10 May Shift 1) · Chemistry

The wavenumber of first spectral line of Lyman series of $\mathrm{He}^{+}$ion is $x \mathrm{~m}^{-1}$. What is the wavenumber (in $\mathrm{m}^{-1}$ ) of second spectral line of Balmer series of $\mathrm{Li}^{2+}$ ion?

  • Correct answer
    A
    $\frac{9 x}{16}$
  • B
    $\frac{16 x}{9}$
  • C
    $\frac{6 x}{27}$
  • D
    $\frac{27 x}{8}$

Solution

For Lyman $\begin{aligned} & \overline{\mathrm{v}}=\mathrm{R}_{\mathrm{H}} \times \mathrm{Z}^2\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right] \\ & =\mathrm{R}_{\mathrm{H}} \times 2^2\left[\frac{1}{1^2}-\frac{1}{2^2}\right] \\ & =\mathrm{R}_{\mathrm{H}} \times 4 \times \frac{3}{4} \\ & \overline{\mathrm{v}}=3 \mathrm{R}_{\mathrm{H}}=\mathrm{x} \\ & \mathrm{R}_{\mathrm{H}}=\frac{\mathrm{x}}{3}...(i) \end{aligned}$
For Balmer series $\begin{aligned} & \overline{\mathrm{v}}=\mathrm{R}_{\mathrm{H}} \times 3^2 \times\left[\frac{1}{2^2}-\frac{1}{4^2}\right] \\ & =\mathrm{R}_{\mathrm{H}} 9 \times \frac{3}{16}. \end{aligned}$
Putting the value of $R_H$ from eq. (i) $\bar{v}=\frac{x}{3} \times 9 \times \frac{3}{16}=\frac{9 x}{16}$

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

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