The shortest wavelength of hydrogen atom in Lyman series is $\lambda$. The longest wavelength is Balmer series of He$^+$ is
Solution
For H: $\frac{1}{\lambda}=\mathrm{R}_{\mathrm{H}} \times 1^{2}\left(\frac{1}{1^{2}}-\frac{1}{\infty^{2}}\right) \quad...(1)$
<br/><br/>
$$
\frac{1}{\lambda_{\mathrm{He}^{+}}}=\mathrm{R}_{\mathrm{H}} \times 2^{2} \times\left(\frac{1}{4}-\frac{1}{9}\right)\quad...(2)
$$
<br/><br/>
From (1) & (2) $\frac{\lambda_{\mathrm{He}^{+}}}{\lambda}=\frac{9}{5}$
<br/><br/>
$$
\begin{aligned}
& \lambda_{\mathrm{He}^{+}}=\lambda \times \frac{9}{5} \\
& \lambda_{\mathrm{He}^{+}}=\frac{9 \lambda}{5}
\end{aligned}
$$
About this question
Subject: Chemistry · Chapter: Atomic Structure · Topic: Bohr's Model
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