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

Options

A. n=3 to n= 2state
B. n=3 to n=1 state
C. n=2 to n=1 state
D. n=4 to n=3 state

Answer

D

Explanation

The number of spectral line emitted when the electron in the high energy orbit jumps to the ground state orbit is 6.
Thus N=6= `(n(n-1))/(2)`
`therefore 12=n^2-n therefore n^2 - n -12 = 0 therefore (n-4) (n+3) = 0`
`therefore n=4 " as " n=-3` is not possible .
`therefore` The electron is in the 4th orbit .
Out of the six transition (n=4 to 3, 4 to 2, 4 to 1, 3 to 2, 3 to 1 and 2 to 1) the maximum wavelength will correspond to the first line of the series when the electron jumps from 4th orbit to 3rd orbit .