So \(\quad \lambda=\frac{ hc }{\Delta E } \quad\) for \(\lambda\) minimum i.e.

shortest; \(\Delta E=\) maximum

for Lyman series \(n=1 \&\) for \(\Delta E_{\text {max }}\)

Transition must be form \(n=\infty\) to \(n=1\)

So \(\quad \frac{1}{\lambda}= R _{ H } Z^{2}\left(\frac{1}{ n _{1}^{2}}-\frac{1}{ n _{2}^{2}}\right)\)3

\(\frac{1}{\lambda}= R _{ H } Z^{2}(1-0)\)

\(\frac{1}{\lambda}= R \times(1)^{2} \Rightarrow \lambda_{1}=\frac{1}{ R }\)

For longest wavelength \(\Delta E =\) minimum for Balmer series \(n =3\) to \(n =2\) will have \(\Delta E\) minimum

for \(He ^{+} Z =2\)

So \(\frac{1}{\lambda_{2}}= R _{ H } \times Z^{2}\left(\frac{1}{ n _{1}^{2}}-\frac{1}{ n _{2}^{2}}\right)\)

\(\frac{1}{\lambda_{2}}= R _{ H } \times 4\left(\frac{1}{4}-\frac{1}{9}\right)\)

\(\frac{1}{\lambda_{2}}= R _{ H } \times \frac{5}{9}\)

\(\lambda_{2}=\lambda_{1} \times \frac{9}{5}\)