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' $\lambda_1$ ' is the wavelength of series limit of Lyman series, ' $\lambda_2$ ' is the wavelength of the first line of Lyman series and ' $\lambda_3$ ' is the series limit of the Balmer series. Then the relation between ' $\lambda_1$ ' $\lambda_2$ and $\lambda_3$ is
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Verified Answer
The correct answer is:
$\frac{1}{\lambda_1}-\frac{1}{\lambda_2}=\frac{1}{\lambda_3}$
Series limit of Lyman series is given by
$$
\frac{1}{\lambda_1}=\mathrm{R}\left(\frac{1}{1}-\frac{1}{\infty}\right)=\mathrm{R}
$$
Series limit of Balmer series given by
$$
\frac{1}{\lambda_3}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{\infty}\right)=\frac{\mathrm{R}}{4}
$$
First line of Lyman series is given by
$$
\begin{aligned}
& \frac{1}{\lambda_2}=\mathrm{R}\left(\frac{1}{1}-\frac{1}{4}\right)=\mathrm{R}-\frac{\mathrm{R}}{4}=\frac{1}{\lambda_1}-\frac{1}{\lambda_3} \\
& \therefore \frac{1}{\lambda_1}-\frac{1}{\lambda_2}=\frac{1}{\lambda_3}
\end{aligned}
$$
$$
\frac{1}{\lambda_1}=\mathrm{R}\left(\frac{1}{1}-\frac{1}{\infty}\right)=\mathrm{R}
$$
Series limit of Balmer series given by
$$
\frac{1}{\lambda_3}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{\infty}\right)=\frac{\mathrm{R}}{4}
$$
First line of Lyman series is given by
$$
\begin{aligned}
& \frac{1}{\lambda_2}=\mathrm{R}\left(\frac{1}{1}-\frac{1}{4}\right)=\mathrm{R}-\frac{\mathrm{R}}{4}=\frac{1}{\lambda_1}-\frac{1}{\lambda_3} \\
& \therefore \frac{1}{\lambda_1}-\frac{1}{\lambda_2}=\frac{1}{\lambda_3}
\end{aligned}
$$
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