Q. The product of linear momentum and angular momentum of an electron of the hydrogen atom is proportional to$\ n^{x}$, where $x$ is
A) 0
B) 1
C) $- 2$
D) 2
Explanation
(a) Linear momentum = $mv\ = \ \frac{mcZ}{137\ n}$
Angular momentum = $\frac{nh}{2\ \pi}$
Given,
Linear momentum $\times$ angular momentum $\propto \ n^{x}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{mcZ}{137\ n}\ \times \ \frac{nh}{2\pi} \propto \ n^{x}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n^{0} \propto \ n^{x}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.