Q3 / 160 · AP EAMCET 2021 (23 Aug Shift 2) · Chemistry

Among the following, which transition in the hydrogen spectrum would have the same wavelength as Balmer transition, $n=4$ to $n=2$ in $\mathrm{He}^{+}$spectrum?

  • A
    $n=3 \longrightarrow n=1$
  • B
    $n=3 \longrightarrow n=2$
  • C
    $n=4 \longrightarrow n=1$
  • Correct answer
    D
    $n=2 \longrightarrow n=1$

Solution

We know that, $\frac{1}{\lambda}=R_{\mathrm{H}} Z^2\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$ For $\mathrm{He},(Z=2)$ $\frac{1}{\lambda_{\mathrm{He}}}=R_{\mathrm{H}}(2)^2\left[\frac{1}{2^2}-\frac{1}{4^2}\right]$ $=R_{\mathrm{H}} 4\left[\frac{1}{4}-\frac{1}{16}\right]=R_{\mathrm{H}} 4 \times \frac{3}{16}$ $\frac{1}{\lambda_{\mathrm{He}}}=\frac{3}{4} R_{\mathrm{H}}$ For option (d), $n=1$ and $n_2=2$ $\frac{1}{\lambda_{\mathrm{H}}}=\frac{R_{\mathrm{H}}}{1}\left[\frac{1}{1^2}-\frac{1}{2^2}\right]=\frac{3}{4} R_{\mathrm{H}}$ So, option (d) is correct.

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

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