The ionization energy of the electron in the hydrogen atom in its...

Key Idea Number of spectral lines abtained due to transition of electron from $\boldsymbol{n}^{\text{th }}$ orbit to lower orbit is $\mathrm{N}=\frac{\mathrm{n}(\mathrm{n}-1)}{2}$ and for maximum wavelength the difference between the orbits of the series should be minimum. Number of spectral lines $\mathrm{N}=\frac{\mathrm{n}(\mathrm{n}-1)}{2}$$\begin{aligned} \Rightarrow & \frac{n(n-1)}{2} & =6 \\ \text{ or } & n^2-n-12 & =0 \\ \text{ or } & (n-4)(n+3) & =0 \\ \text{ or } & n & =4 \end{aligned}$$Now as the first line of the series has the maximum wavelength, therefore electron jumps from the$4^{\text{th } } orbit to the third orbit.