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Q.
If the series limit wavelength of Lyman series for the hydrogen atom is $912 \,\mathring{A} $, then the series limit wavelength for Balmer series of hydrogen atoms is
$\frac{1}{\lambda}=R\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]$
For limiting wavelength of Lyman series
$n_{1}=1, n_{2}=\infty$
$ \frac{1}{\lambda_{L}}=R$
For limiting wavelength of Balmer series
$n_{1}=2, n_{2}=\infty $
$\frac{1}{\lambda_{L}}=R$
For limiting wavelength of Balmer series
$n_{1}=2, n_{2}=\infty$
$ \frac{1}{\lambda_{L}}=R\left(\frac{1}{4}\right) $
$\Rightarrow \lambda_{B}=\frac{4}{R}$
$\therefore \lambda_{B}=4 \lambda_{L}=4 \lambda_{L}=4 \times 912 \,\mathring{A} $