QJEEM23Ch40df: JEE Main 2023 Chemistry - Class 11 JEEM23Ch40df Solution: The shortest wavelength of hydrogen atom in Lyman series is λ . …

Question

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The shortest wavelength of hydrogen atom in Lyman series is $λ$ . The longest wavelength is Balmer series of He $^{+}$ is

Accepted Answer

For H:

\frac{1}{\lambda}=\mathrm{R}_{\mathrm{H}} \times 1^{2}\left(\frac{1}{1^{2}}-\frac{1}{\infty^{2}}\right) \quad...(1)

\frac{1}{λ_{{He}^{+}}} = R_{H} \times 2^{2} \times (\frac{1}{4} - \frac{1}{9}) . . . (2)

From (1) & (2)

\frac{\lambda_{\mathrm{He}^{+}}}{\lambda}=\frac{9}{5}

\begin{aligned} λ_{{He}^{+}} = λ \times \frac{9}{5} \\ λ_{{He}^{+}} = \frac{9 λ}{5} \end{aligned}

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