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If the series limit wavelength of Lyman series for the hydrogen atom is $912 Å,$ then the series limit wavelength for Balmer series of hydrogen atoms is
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The correct answer is:
$912 \times 4 Å$
$\frac{1}{\lambda}=R\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]$
For limiting wavelength of Lyman series
$$
n_{1}=1, n_{2}=\infty \quad \frac{1}{\lambda_{L}}=R
$$
For limiting wavelength of Balmer series $n_{1}=2, n_{2}=\infty$
$$
\begin{array}{l}
\frac{1}{\lambda_{B}}=R\left(\frac{1}{4}\right) \Rightarrow \lambda_{B}=\frac{4}{R} \\
\therefore \lambda_{B}=4 \lambda_{L}=4 \times 912 \AA
\end{array}
$$
For limiting wavelength of Lyman series
$$
n_{1}=1, n_{2}=\infty \quad \frac{1}{\lambda_{L}}=R
$$
For limiting wavelength of Balmer series $n_{1}=2, n_{2}=\infty$
$$
\begin{array}{l}
\frac{1}{\lambda_{B}}=R\left(\frac{1}{4}\right) \Rightarrow \lambda_{B}=\frac{4}{R} \\
\therefore \lambda_{B}=4 \lambda_{L}=4 \times 912 \AA
\end{array}
$$
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