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GR0177 #84
Problem
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Quantum Mechanics$\Rightarrow$}Selection Rules

The selection rules are $\Delta l = \pm 1$ and $\Delta m =0,\pm 1$.

The transitions have the following quantum numbers:

A: $\Delta l = 0$ and $\Delta j = 0$

B: $\Delta l = -1$ and $\Delta j = -1$

C: $\Delta l = -1$ and $\Delta j = 0$

By the selection rules above, transition A is forbidden in $\Delta l$. Transitions B and C both work out, since they do not violate the $\Delta l$ rule.

Take choice (D).

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grephy
2012-11-06 09:39:13
why s is not taken into account?

In transition C, there IS a change in s, right? Then why this transition can happen?!
 jonrichens2012-11-08 02:39:37 I believe that the spin selection rule only comes into play for multi-electron atoms (where the directional spin breaks the symmetry, or something). This question is about hydrogen and the orientation of spin is not conserved in the same way. Hydrogen selection rules (delta)j = 0, +- 1 not 0 -> 0 (delta)l = +- 1 (delta)mj = 0, +- 1 Helium selection rule (spin orientation important) (delta)j = 0, +- 1 not 0 -> 0 (delta)l = 0, +- 1 (delta)mj = 0, +- 1 (delta)s = 0 not 0 -> 0 hope that helps
antithesis
2007-10-01 17:36:51
Can you clarify?
You state the selection rules for l and m, but when you go through the ABC options, you only mentions the l and j rules (I know they are related).

Thanks
 Richard2007-10-30 11:12:31 The condition on the total angular momentum quantum number comes from the fact that $j=l\pm\frac{1}{2}$ I think it may be best to remember that for an electric-dipole transition the following selection rules apply: $\Delta S=0$ $\Delta l =\pm 1$ $\Delta m = 0,\pm 1$ $\Delta j = 0,\pm 1$ with the exception of a $j=0$ to $j=0$ transition. Of course, these rules are different when you have a quadrupole or octupole transition. I believe the only difference is an additional $\pm 2$ or $\pm 3$ possibility on $l$ $m$ and $j$ for quadrupole and octupole respectively.

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