Question

If a hydrogen atom emits a photon of energy $12.1eV$, its orbital angular momentum changes by $△L$. Then, $△L$ equals

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

Answer 1

Answer: (B)

$2.11\times {10}^{-34}Js$

Using the attached formula:
$12.1=13.6\times (\frac{1}{n_1^2}-\frac{1}{n_2^2})$
$(\frac{1}{n_1^2}-\frac{1}{n_2^2})=\frac{8}{9}$

Hence, $n_1=1$ and $n_2=3$
Change in Angular momentum $=(n_2-n_1)\frac{h}{2\pi }$$=2.11\times {10}^{-34}J$