Hydrogen atom in its ground state is excited by a radiation of wavelength 97.26 nm. The longest wavelength it emits is

$\mathrm{\Delta }E=\frac{hc}{\lambda }=\frac{\left(6.626\times {10}^{-34}\mathrm{Js}\right)\left(3\times {10}^8{\mathrm{ms}}^{-1}\right)}{\left(97.26\times {10}^{-9}\mathrm{m}\right)}=2.0438\times {10}^{-18}\mathrm{J}$

Also $\mathrm{\Delta }E=\left(2.18\times {10}^{-18}\mathrm{J}\right)\left(\frac{1}{1^2}-\frac{1}{n_2^2}\right)$

Hence $\begin{array}{l}\frac{1}{n_2^2}=1-\left(\frac{\mathrm{\Delta }E}{2.18\times {10}^{-18}\mathrm{J}}\right)=1-\frac{2.0436\times {10}^{-18}}{2.18\times {10}^{-18}}=1-0.9374\\ n_2=\sqrt{1/(1-0.9374)}\simeq 4\end{array}$

The transition $n_2=4\to n_1=3$ will emit the longest wavelength. Hence

$\begin{array}{l}\mathrm{\Delta }E=\left(2.18\times {10}^{-18}\mathrm{J}\right)\left(\frac{1}{3^2}-\frac{1}{4^2}\right)=1.06\times {10}^{-19}\mathrm{J}\\ \lambda =\frac{hc}{\mathrm{\Delta }E}=\frac{\left(6.626\times {10}^{-34}\mathrm{Js}\right)\left(3\times {10}^8{\mathrm{ms}}^{-1}\right)}{\left(1.06\times {10}^{-19}\mathrm{J}\right)}=1.875\times {10}^{-6}\mathrm{m}=1875\mathrm{nm}\end{array}$