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)

• A. $\frac{3 h}{\pi}$
• B. $\frac{2 h}{\pi}$
• C. $\frac{h}{\pi}$
• D. $\frac{4 h}{\pi}$

Answer 1

Answer: (C)

This is the case of electron jumping from $3^{\text {rd }}$ orbit to $1^{15 t}$ orbit
$\Delta L =\frac{n_{1} h}{2 \pi}-\frac{n h}{2 \pi} $
$=\frac{3 h}{2 \pi}-\frac{h}{2 \pi}=\frac{h}{\pi}$