Question

When a hydrogen atom emits a photon of energy 12.1 ev, the oribital angular monetum changes by?

Answer 1

Emission of photon of 12.1eV corresponds to the transition from n=3 to n=1

∴∴ Change in angular momentum

=(n2−n1)h/2π

=(3−1)h/2π

=h/π

=6.626×10^−343.14

=2.11×10^−34JSec

Answer 2

We must know that the emission of photon of 12.1 eV is related to the transition from n=3 to n=1.

The change in angular momentum can be calculated with the help of below formula:

Change in angular momentum $\bold{=\left(\mathrm{n}_{2}-\mathrm{n}_{1}\right) \mathrm{h} / 2 \pi}$

The given value of $n_{1} \text { and } n_{2}$ along with value of $h \text { and } \pi$ which are constant.  

Change in angular momentum $=\left(\mathrm{n}_{2}-\mathrm{n}_{1}\right) \mathrm{h} / 2 \pi$

Change in angular momentum $=(3-1) \times 6.126 \times 10^{-34} \mathrm{J Sec} / 2 \times 3.14$

Change in angular momentum $\bold{=2.11 \times 10^{-34} \mathrm{J Sec}}$