Calculate the change in magnetic dipole moment when the electron in a `H-`atom makes a transition from the `n=4` orbit to the `n=3` orbit. Suppose that the atoms is in a magnetic field of induction. Calculate the change in energy when it makes the same transition and the corresponding change in frequency of the emitted photon due to the magnetic field.

Explanation

We assume that an electron moves in a circular orbit of radius `r` with speed `v`. The current due to the electron is (`e` being the charge on the electron),
`i=(e)/(T)=(ev)/(2pir)`, where `T` is the time period
The dipole moment due to this current is
`mu=i.pir^(2)=(ev)/(2pir)xxpir^(2)=(evr)/(2)`
`=(e)/(2m)xxmvr=(e)/(2m)xxnh`(`h=h//2pi`)
when the electron is in the `n^(th)` orbit.
When it makes a transition from the `n=4` to the `n=3` level, the change in the dipole moment equals
`Deltamu=mu(n=4)-mu(n=3)=(eh)/(2m)`(also known as the Bohr magneton)
If the atom is in a magnetic field of induction `B`, then the energy of the electron is given by
`E'_(n)=E_(n)-vecmu.vecB=-(R_(E))/(n^(2))-((eBh)/(2m))n`
(where `R_(E)` represents the ionisation energy in the absence of a magnetic field)
When it makes a transition,
`E'_(n_(1))=E'_(n_(2))=DeltaE'_(n_(1)n_(2))=(E_(n_(1))-E_(n_(2)))-(eBh)/(2m)(n_(1)-n_(2))`
`=R_(E)((1)/(n_(2)^(2))-(1)/(n_(1)^(2)))-(eh)/(2m)B(n_(1)-n_(2))`
Due to the magnetic field, the change in energy equals `(eh)/(2m)B(n_(1)-n_(2))`, and the corresponding change in the frequency of the emitted photon equals
`(ehB)/(hxx2m)(n_(1)-n_(2))=(eB)/(4pim)(n_(1)-n_(2))`