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In Lyman series, series limit of wavelength is $\lambda_1$. The wavelength of first line of Lyman series is $\lambda_2$ and in Balmer series, the series limit of wavelength is $\lambda_3$. Then the relation between $\lambda_1$, $\lambda_2$ and $\lambda_3$ is
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The correct answer is:
$\frac{1}{\lambda_1}-\frac{1}{\lambda_2}=\frac{1}{\lambda_3}$
According to Rydberg's formula,
$$
\frac{1}{\lambda}=R\left(\frac{1}{\mathrm{n}^2}-\frac{1}{\mathrm{~m}^2}\right)
$$
For series limit of Lyman series, $\mathrm{n}=1, \mathrm{~m}=\infty, \lambda=\lambda_1$
$$
\therefore \quad \frac{1}{\lambda_1}=\mathrm{R}
$$
For $1^{\text {st }}$ line of Lyman series,
$$
\mathrm{n}=1, \mathrm{~m}=2, \lambda=\lambda_2
$$
$$
\therefore \quad \frac{1}{\lambda_2}=\frac{3 R}{4}
$$
For series limit of Balmer series,
$$
\begin{array}{ll}
& \mathrm{n}=2, \mathrm{~m}=\infty, \lambda=\lambda_3 \\
\therefore \quad & \frac{1}{\lambda_3}=\frac{\mathbb{R}}{4} \\
& \text { Now, } \frac{1}{\lambda_1}-\frac{1}{\lambda_2}=\mathbb{R}-\frac{3 \mathrm{R}}{4}=\frac{\mathbb{R}}{4} \\
\therefore \quad & \frac{1}{\lambda_1}-\frac{1}{\lambda_2}=\frac{1}{\lambda_3}
\end{array}
$$
$$
\frac{1}{\lambda}=R\left(\frac{1}{\mathrm{n}^2}-\frac{1}{\mathrm{~m}^2}\right)
$$
For series limit of Lyman series, $\mathrm{n}=1, \mathrm{~m}=\infty, \lambda=\lambda_1$
$$
\therefore \quad \frac{1}{\lambda_1}=\mathrm{R}
$$
For $1^{\text {st }}$ line of Lyman series,
$$
\mathrm{n}=1, \mathrm{~m}=2, \lambda=\lambda_2
$$
$$
\therefore \quad \frac{1}{\lambda_2}=\frac{3 R}{4}
$$
For series limit of Balmer series,
$$
\begin{array}{ll}
& \mathrm{n}=2, \mathrm{~m}=\infty, \lambda=\lambda_3 \\
\therefore \quad & \frac{1}{\lambda_3}=\frac{\mathbb{R}}{4} \\
& \text { Now, } \frac{1}{\lambda_1}-\frac{1}{\lambda_2}=\mathbb{R}-\frac{3 \mathrm{R}}{4}=\frac{\mathbb{R}}{4} \\
\therefore \quad & \frac{1}{\lambda_1}-\frac{1}{\lambda_2}=\frac{1}{\lambda_3}
\end{array}
$$
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