bandikizaui

2022-07-14

Deexcitation of electron gives us emission spectrum but in my book states it's only possible if
$\mathrm{\Delta }n\ne 0$
$\mathrm{\Delta }\ell =±1$
$\mathrm{\Delta }m=0,±1$

Karissa Macdonald

The first condition on $\mathrm{\Delta }n$ guarantees the initial and final states do not have the same energy, so the photon can carry the difference.
The second condition arises because in atoms, the emitted radiation is usually of the dipole type. The dipole is a vector, with angular momentum $\ell =1$, so the change in $\ell$ is due to momentum carried by radiation. Note it is also technically possible to have $\mathrm{\Delta }\ell =0$ transition that does not change $\ell$ is prevented by Laporte’s rule (i.e. by parity argument). (Note also that, in deformed nuclei, quadrupole transitions can occur with $\mathrm{\Delta }\ell =±2$.)
The last condition is related to the polarization of the radiation. For $\mathrm{\Delta }m=±1$, the emitted radiation will have circular polarization, whereas for $\mathrm{\Delta }m=0$ the radiation will be linearly polarized.

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