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In hydrogen spectrum, the shortest wavelengths of Lyman and Balmer series are $\lambda_1$ and $\lambda_2$ respectively. The Rydberg constant of hydrogen is
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Verified Answer
The correct answer is:
$\frac{4\left(\lambda_2-\lambda_1\right)}{3 \lambda_1 \lambda_2}$
For Lymen series
$\begin{aligned} & \frac{1}{\lambda_1}=\mathrm{R}_{\mathrm{H}}\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right] \\ & \mathrm{n}_1=1 \text { and } \mathrm{n}_2=\infty\end{aligned}$
$\frac{1}{\lambda_1}=\mathrm{R}_{\mathrm{H}}$ ...(1)
For balmer series
$\begin{aligned} & \frac{1}{\lambda_2}=\mathrm{R}_{\mathrm{H}}\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right] \\ & \mathrm{n}_1=2 \text { and } \mathrm{n}_2=\infty\end{aligned}$
$\frac{1}{\lambda_2}=\frac{\mathrm{R}_{\mathrm{H}}}{4}$ ...(2)
$\begin{aligned} & \frac{1}{\lambda_2}-\frac{1}{\lambda_1}=\frac{\mathrm{R}_{\mathrm{H}}}{4}-\mathrm{R}_{\mathrm{H}} \\ & \frac{\lambda_1-\lambda_2}{\lambda_1 \lambda_2}=\frac{-3 \mathrm{R}_{\mathrm{H}}}{4}\end{aligned}$
Rydberg constant, $\mathrm{R}_{\mathrm{H}}=\frac{4\left(\lambda_2-\lambda_1\right)}{3 \lambda_1 \lambda_2}$
$\begin{aligned} & \frac{1}{\lambda_1}=\mathrm{R}_{\mathrm{H}}\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right] \\ & \mathrm{n}_1=1 \text { and } \mathrm{n}_2=\infty\end{aligned}$
$\frac{1}{\lambda_1}=\mathrm{R}_{\mathrm{H}}$ ...(1)
For balmer series
$\begin{aligned} & \frac{1}{\lambda_2}=\mathrm{R}_{\mathrm{H}}\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right] \\ & \mathrm{n}_1=2 \text { and } \mathrm{n}_2=\infty\end{aligned}$
$\frac{1}{\lambda_2}=\frac{\mathrm{R}_{\mathrm{H}}}{4}$ ...(2)
$\begin{aligned} & \frac{1}{\lambda_2}-\frac{1}{\lambda_1}=\frac{\mathrm{R}_{\mathrm{H}}}{4}-\mathrm{R}_{\mathrm{H}} \\ & \frac{\lambda_1-\lambda_2}{\lambda_1 \lambda_2}=\frac{-3 \mathrm{R}_{\mathrm{H}}}{4}\end{aligned}$
Rydberg constant, $\mathrm{R}_{\mathrm{H}}=\frac{4\left(\lambda_2-\lambda_1\right)}{3 \lambda_1 \lambda_2}$
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