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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?
Options:
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Verified Answer
The correct answer is:
$n=2 \longrightarrow n=1$
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.
$\frac{1}{\lambda}=R_{\mathrm{H}} Z^2\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$
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.
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