Question

When a hydrogen atom emits a photon of energy $12.1eV$, the orbital angular momentum changes by:

• A. $1.05\times {10}^{-34}$ $J$ $sec$
• B. $2.11\times {10}^{-34}$ $J$ $sec$
• C. $3.16\times {10}^{-34}$ $J$ $sec$
• D. $4.22\times {10}^{-34}$ $J$ $sec$

Answer 1

The correct option is B $2.11\times {10}^{-34}$ $J$ $sec$

Emission of photon of 12.1 eV , this corresponds to $n_1=3$ and $n_2=1$

Orbital angular momentum $=\frac{nh}{2\pi }$

$\therefore$ Change in orbital angular momentum $=\frac{(3-1)h}{2\pi }=\frac{h}{\pi }=\frac{6.63\times {10}^{-34}}{3.14}=2.11\times {10}^{-34}Jsec$