Ionisation potential of hydrogen atom is `13.6 eV`. Hydrogen atom in ground state is excited by monochromatic light of energy `12.1 eV`. The spectral lines emitted by hydrogen according to Bohr's theory will be

Options

A. One
B. two
C. three
D. four

Answer

C

Explanation

`I.E. = -E_(1)`
`implies E_(1)=-I.E.=-13.6eV`
`E_(f)-E_(1)=12.1eV`
`-(13.6)/(n^2)-(-13.6)=12.1`.
`-(13.6)/(n^2)=12.1-13.6=-1.5`
`n^(2)=(13.6)/(1.5)=9.06 ~= 9`
`n=3`
`implies e^(-)` is excited to `n=3`
Max. Lines emitted `= ((n_(2)-n_(1))(n_(2)-n_(1)+1))/(2)`
`=((3-1)(3-1+1))/(2)=(2xx3)/(2)=3`.