Question

When a hydrogen atom emits a photon of energy $12.09eV$,it's orbit's angular momentum changes by (where $h$ is Planck's constant) ?

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

Answer 1

The correct option is C $\frac{h}{\pi }$

Given that,

Emission of photon of 12.1eV corresponds to the transition from

$n=3,n=1$

Now, change in angular momentum

$=(n_2-n_1)\times \frac{h}{2\pi }$

$=(3-1)\times \frac{h}{2\pi }$

$=\frac{h}{\pi }$

Hence, this is the required solution