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 = 2 states
B. n = 3 to n = 1 states
C. n = 2 to n = 1 states
D. n = 4 to n = 3 states

Answer

D

Explanation

No. of spectral line = 6
`rArr (n(n-1))/(2)=6 rArr (n=4) and lamda prop (1)/(Delta E)`
Wavelength of emitted radiation will be maximum for transition n = 4 to n = 3.