Question

The ionization enegry 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
A. `n = 3` to `n = 2` states
B. `n = 3` to `n = 1` states
C. `n = 2` to `n = 1` states
D. `n = 4` to `n = 3` states

Answer 1

Correct Answer - D
Number of spectral lines obtained due to transition of electron from `nth` orbit to lower orbit is `N = (n(n - 1))/(2)` and for maximum wavelength the difference between the orbits of the series should be minimum.
`rArr (n(n - 1))/(2) = 6`
or `n^(2) - n - 12 = 0`
or `(n - 4) (n + 3) = 0`
or `n = 4`
Now as the first line of the series has the amximum wavelength, therefore electron jumps from the `4th` orbit to the third orbit.