The ionization energy of the electron in the hydrogen atom in its ground state is 13.6 eV. The atoms are excited to higher energy levels to emit radiations of 6 wavelengths. Maximum wavelength of emitted radiation corresponds to the transition between

Number of spectral lines obtained due to transition of electron from nth orbit to ground state is

$\mathrm{N}=\frac{\mathrm{n}\left(\mathrm{n}-1\right)}{2}$ and for maximum wavelength the difference between the orbits of the series should be minimum.

Number of spectral lines $\mathrm{N}=\frac{\mathrm{n}\left(\mathrm{n}-1\right)}{2}$

$\begin{array}{l}\Rightarrow \frac{\mathrm{n}\left(\mathrm{n}-1\right)}{2}=6\\ \mathrm{or} {\mathrm{n}}^2-\mathrm{n}-12=0\\ \mathrm{or} \left(\mathrm{n}-4\right)\left(\mathrm{n}+3\right)=0\\ \mathrm{or}\mathrm{n}=4\end{array}$

Now as the difference in energy between 4 th orbit and 3 rd orbit is minimum , therefore electron jumps from the 4 th orbit to the 3 rd orbit .