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Can you say that the lifetime of the 3d state in the hydrogen atom is shorter than the one of the 3s state because the centrifugal energy associated with 3d is higher than the one associated with 3s? By centrifugal energy I mean the contribution given by

$E_{rot}=\dfrac{l(l+1)}{2r^2}$

to the total energy. I am trying to explain atomic transitions intuitively (without having to calculate the transition strengths) to somebody, and I would like to know if I can use that argument or where it fails. Cheers

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There are more quantum numbers governing transitions anyway. this might help tapir.caltech.edu/~chirata/ay102/Atomic.pdf –  anna v Jun 11 at 4:16

1 Answer 1

No, the difference in lifetimes shouldn't be due to energy difference. For one thing, these energy differences are nearly zero. The states of a given $n$ in a hydrogen atom should be nearly degenerate regardless of $l$-value. (I say "nearly" since there is some energy splitting due to the Lamb shift, spin-orbit coupling, etc., but these are expected to be small.) Instead, the difference is due to the different matrix elements for the dipole operator between the different orbitals. Roughly speaking, d- and s-orbitals look much different from each other and occupy different regions in space, so it makes sense that integrals involving each of them will end up with much different values, resulting in different lifetimes.

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