1.

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

Answer»

The TRANSITION in which a Balmer series PHOTON absorbed is VI
The wavelength of the radiation involved in transition II is 486 nm
IV transition will occur when a hydrogen atom is IRRADIATED with radiation of wavelength `103 nm`
IV transition will emit the longest wavelength line in the visible PORTION of the hydrogen spectrum

Solution :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.


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