JEE MAIN - Chemistry (2023 - 29th January Morning Shift - No. 7)
The shortest wavelength of hydrogen atom in Lyman series is $$\lambda$$. The longest wavelength is Balmer series of He$$^+$$ is
$$\frac{36\lambda}{5}$$
$$\frac{5}{9\lambda}$$
$$\frac{9\lambda}{5}$$
$$\frac{5\lambda}{9}$$
Explanation
For H: $\frac{1}{\lambda}=\mathrm{R}_{\mathrm{H}} \times 1^{2}\left(\frac{1}{1^{2}}-\frac{1}{\infty^{2}}\right) \quad...(1)$
$$ \frac{1}{\lambda_{\mathrm{He}^{+}}}=\mathrm{R}_{\mathrm{H}} \times 2^{2} \times\left(\frac{1}{4}-\frac{1}{9}\right)\quad...(2) $$
From (1) & (2) $\frac{\lambda_{\mathrm{He}^{+}}}{\lambda}=\frac{9}{5}$
$$ \begin{aligned} & \lambda_{\mathrm{He}^{+}}=\lambda \times \frac{9}{5} \\ & \lambda_{\mathrm{He}^{+}}=\frac{9 \lambda}{5} \end{aligned} $$
$$ \frac{1}{\lambda_{\mathrm{He}^{+}}}=\mathrm{R}_{\mathrm{H}} \times 2^{2} \times\left(\frac{1}{4}-\frac{1}{9}\right)\quad...(2) $$
From (1) & (2) $\frac{\lambda_{\mathrm{He}^{+}}}{\lambda}=\frac{9}{5}$
$$ \begin{aligned} & \lambda_{\mathrm{He}^{+}}=\lambda \times \frac{9}{5} \\ & \lambda_{\mathrm{He}^{+}}=\frac{9 \lambda}{5} \end{aligned} $$
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