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

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

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

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

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 .

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