Question
The first line of the Balmer series in Hydrogen atom corresponds to the n=3 to n=2 transition, this line is known as H-alpha line. What would be the selection rule for the n=3 to n=2 transition and corresponding lifetimes.
Answers
The n=3 and n=2 levels of H are degenerate, with different angular "L" momentum orbitals (s, p, d, f etc) having the same energy.
n=2 has L = 0 and 1 states (s and p) , and n=3 has L =0, 1 and 2 states (s, p, and d).
From n=3 to 2, only the transitions d->p, p->s and s->p are allowed. The orbital angular momentum quantum number must change by +/- 1. There are also degeneracies associated with the electrons spin direction.
There is a small amount of relativistic splitting of these states, as explained by Willis Lamb, using quantum electrodynamics developed by Palul Dirac.
I suspect this subject goes well beyond the level of the course that you are taking. I am surprised that they would assign such a problem.
The average radiative lifetime of an electron in the n=3 upper level is 1/(4.41*10^7 s^-1) = 2.27*10^-8 seconds .
This must be obtained by averaging over all allowed transitions, with their degeneracies taken into account. There is no way I can lead you though this calculation.
n=2 has L = 0 and 1 states (s and p) , and n=3 has L =0, 1 and 2 states (s, p, and d).
From n=3 to 2, only the transitions d->p, p->s and s->p are allowed. The orbital angular momentum quantum number must change by +/- 1. There are also degeneracies associated with the electrons spin direction.
There is a small amount of relativistic splitting of these states, as explained by Willis Lamb, using quantum electrodynamics developed by Palul Dirac.
I suspect this subject goes well beyond the level of the course that you are taking. I am surprised that they would assign such a problem.
The average radiative lifetime of an electron in the n=3 upper level is 1/(4.41*10^7 s^-1) = 2.27*10^-8 seconds .
This must be obtained by averaging over all allowed transitions, with their degeneracies taken into account. There is no way I can lead you though this calculation.