The shortest wavelength of Hatom in the Lyman series is $\lambda_{1}$. The longest wavelength in the Balmer series is $\mathrm{He}^{+}$is :
Correct Option:
Shortest wavelength $\rightarrow$ Max. energy $(\infty \rightarrow 1)$
For Lyman series of H atom,
$\frac{1}{\lambda_{1}}=R_{\mathrm{H}}(1)^{2}\left[\frac{1}{1}-0\right]$
$\Rightarrow \frac{1}{\lambda_{1}}=R_{\mathrm{H}} \Rightarrow R_{\mathrm{H}}=\frac{1}{\lambda_{1}}$
For Balmer series of $\mathrm{He}^{+}$,
$\frac{1}{\lambda}=R_{\mathrm{H}}(2)^{2}\left[\frac{1}{2^{2}}-\frac{1}{3^{2}}\right] \Rightarrow \frac{1}{\lambda}=R_{\mathrm{H}}(4)\left(\frac{9-4}{36}\right)$
$\Rightarrow \frac{1}{\lambda}=\frac{5 R_{\mathrm{H}}}{9} \Rightarrow \lambda=\frac{9}{5 R_{\mathrm{H}}}=\frac{9 \lambda_{1}}{5}$