The figure shows an energy level diagram for the hydrogen atom. Several transitions are marked as I, II, III,_____. The diagram is only indicative and not be scale.

In which transitions is a Balmer photn absorbed?

Options

A. The transition in which a Balmer series photon absorbed is VI
B. The wavelength of the radiation involved in transition II is 486 nm
C. IV transition will occur when a hydrogen atom is irradiated with radiation of wavelength `103 nm`
D. IV transition will emit the longest wavelength line in the visible portion of the hydrogen spectrum

Answer

A, B, D

Explanation

For Balmer series, `underset(("lower"))(n_(1)=2) underset (("higher"))(n_(2)=3,4)……`
`:.` In transition (VI), photon of Balmer series is absorbed.
In transition II
`E_(2)=-3.4 eV, E_(4)=-0.85eV`
`/_\E=2.55eV`
`/_\E=(hc)/(lamda)implieslamda=(hc)/(/_\E)`
`lamda=486nm`
Wavelength of radiation `=103 nm =1030 A^(@)`
` :. /_\ 12400/1030~~12.0eV`
So difference of energy should be `12.0 eV` (approx.)
Hence `n_(1)=1` and `(n_(2)=3`
`(-13.6)eV (-1.51) eV`
`:.` Transition is `V`.
For longest wavelength, energy difference should be minimum.
So, in visible portion of hydrogen atom, minimum energy emitted is in transition IV.