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Q.
If the shortest wavelength of $H$-atom in Lyman series is $x$, then longest wavelength in Balmer series of $He ^{+}$is :-
Solution:
$\frac{1}{\lambda}= R _{ H } \times Z ^{2}\left(\frac{1}{ n _{1}^{2}}-\frac{1}{ n _{2}^{2}}\right)$
$\frac{1}{\lambda_{1}}= R _{ H } \times(1)\left(\frac{1}{(1)^{2}}-\frac{1}{(\infty)^{2}}\right)$
$\lambda_{1}=\frac{1}{ R _{ H }}= X $
$\Rightarrow R _{ H }=\frac{1}{ X }$
$\frac{1}{\lambda_{2}}= R _{ H } \times(2)^{2}\left(\frac{1}{(2)^{2}}-\frac{1}{(3)^{2}}\right)$
$\frac{1}{\lambda_{2}}= R _{ H } \times 4\left(\frac{1}{4}-\frac{1}{9}\right)$
$\frac{1}{\lambda_{2}}= R _{ H } \times 4\left(\frac{9-4}{4 \times 9}\right)$
$\lambda_{2}=\frac{9}{5} R _{ H } $
$\Rightarrow \lambda_{2}=\frac{9}{5} X$