Question
When a hydrogen atom emits a photon of energy $12.09 \,eV$, its orbital angular momentum changes by (where $h$ is Planck's constant)

Answer

(c)

This is the case of electron jumping from $3^{\text {rd }}$ orbit to $1^{\text {st }}$ orbit

$\Delta L=\frac{n_1 h}{2 \pi}-\frac{n h}{2 \pi}$

$=\frac{3 h}{2 \pi}-\frac{h}{2 \pi}=\frac{h}{\pi}$

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