If the shortest wavelength of $\mathrm{H}$ atom in Lyman series is $\mathrm{x}$, the longest wavelength in Balmer series of $\mathrm{He}^{+}$is

Question

Aicharya · Accepted Answer

$$
\frac{1}{\lambda}=\overline{\mathrm{R}}_{\mathrm{H}} \mathrm{Z}^{2}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}ight]
$$
For Lyman series $\mathrm{n}_{1}=1$
$\lambda$ is shortest if $\mathrm{n}_{2}=\infty$
$\therefore \overline{\mathrm{R}}_{\mathrm{H}}=\frac{1}{\mathrm{x}}$ for H-atom $(\mathrm{Z}=1)$
For Balmer series, $\mathrm{n}_{1}=2$
$\lambda$ is longest if $\mathrm{n}_{2}=3$
Here $\frac{1}{\lambda_{\max }}=\left(\frac{1}{\mathrm{x}}ight)(2)^{2}\left[\frac{1}{2^{2}}-\frac{1}{3^{2}}ight]_{\text {for } \mathrm{He}^{+} \text {ion }}$
$(Z=2)$
$\therefore \lambda_{\max }=\frac{9 x}{5}$ *

If the shortest wavelength of $\mathrm{H}$ atom in Lyman series is $\mathrm{x}$, the longest wavelength in…

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