Q1 / 160 · AP EAMCET 2024 (22 May Shift 1) · Chemistry

If the longest wavelength of spectral line of Paschen series of $\mathrm{Li}^{2+}$ ion spectrum is $x Å$, then the longest wavelength (in $Å$ ) of Lyman series of hydrogen spectrum is

  • A
    $\frac{12}{7 x}$
  • Correct answer
    B
    $\frac{7 x}{12}$
  • C
    $\frac{20 x}{27}$
  • D
    $\frac{27 x}{20}$

Solution

$\bar{v}=R_H \times Z^2 \times\left(\frac{1}{n_1^2}-\frac{1}{\mathrm{n}_2^2}\right)$
For paschen series longest wavelength $\begin{aligned} & \mathrm{n}_1=3 \\ & \mathrm{n}_2=4 \\ & \bar{v}=\mathrm{R}_{\mathrm{H}} \times(3)^2 \times\left(\frac{1}{(3)^2}-\frac{1}{(4)^2}\right) \\ & \bar{v}=R_H \times \frac{9 \times 7}{9 \times 16} \\ & \bar{v}=R_H \times \frac{7}{16} \end{aligned}$
$\frac{1}{\lambda}=\mathrm{R}_{\mathrm{H}} \times \frac{7}{16}$ or, $\lambda=\frac{16}{7 \times \mathrm{R}_{\mathrm{H}}}=\mathrm{x}$...(i) for Lyman series of hydrogen spectrum $\begin{aligned} & \bar{v}=R_{\mathrm{H}} \times \mathrm{Z}^2 \times\left[\frac{1}{\mathrm{~N}_1^2}-\frac{1}{\mathrm{~N}_2^2}\right] \\ & \bar{v}=\mathrm{R}_{\mathrm{H}} \times(1)^2 \times\left[\frac{1}{1^2}-\frac{1}{2^2}\right] \\ & \bar{v}=\mathrm{R}_{\mathrm{H}} \times \frac{3}{4} \\ & \lambda=\frac{4}{3 \mathrm{R}_{\mathrm{H}}}...(ii) \end{aligned}$
By equation (i) and (ii) we get $\lambda=\frac{7 x}{12}$

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

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