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)
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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(Consider the speed of light in air $3 \times 10^{8} \,m / s$ )
