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Q.
In hydrogen spectrum if the shortest wavelength in Balmer series is $\lambda$, the shortest wavelength in Bracket series is
TS EAMCET 2020
Solution:
For Balmer series,
$\frac{1}{\lambda_{\text {Balmer }}}=R\left(\frac{1}{2^{2}}-\frac{1}{n^{2}}\right)$,
where $n=3,4,5 \ldots$
For shortest wavelength, $n=\infty$
$\Rightarrow \frac{1}{\lambda}=R\left(\frac{1}{2^{2}}\right)=\frac{R}{4} $
or $\lambda=\frac{4}{R} \ldots .( i )$
For Brackett series,
$\frac{1}{\lambda_{\text {Brackett }}}=R\left(\frac{1}{4^{2}}-\frac{1}{n^{2}}\right)$,
where $n=5,6,7, \ldots$
Again, for shortest wavelength, $n=\infty$
$\Rightarrow \frac{1}{\lambda_{\text {Brackett }}}=R\left(\frac{1}{4^{2}}\right)=\frac{R}{16}$
or $\lambda_{\text {Brackett }}=\frac{16}{R}=4\left(\frac{4}{R}\right)=4 \lambda[$ From Eq.(i)]