Question

In hydrogen spectrum, a hydrogen atom emits a photon of wavelength $1027\,\mathring{A}$ . Its angular momentum changes by

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

Answer 1

Answer: (A)

The energy of the photon emitted $E=\frac{12430}{1027}=12.1\,eV$
This corresponds to transition from
$n=3$ with $E_{3}=-1.5\,eV$ to $n_{f}=1$
with $E_{1}=-13.6\,eV$
Hence $\Delta L=3\left(\frac{h}{2 \pi }\right)-1\left(\frac{h}{2 \pi }\right)=\frac{h}{\pi }$