Atoms And Nuclei Question 284
Question: If the angular momentum of an electron in an orbit is J then the K.E. of the electron in that orbit is
Options:
A) $ \frac{J^{2}}{2mr^{2}} $
B) $ \frac{Jv}{r} $
C) $ \frac{J^{2}}{2m} $
D) $ \frac{J^{2}}{2\pi } $
Show Answer
Answer:
Correct Answer: A
Solution:
Angular momentum $ =mrv=J\text{ }\therefore v=\frac{J^{2}}{2mr^{2}} $ $ K\text{.E}\text{. of electron=}\frac{1}{2}mv^{2}=\frac{1}{2}m{{( \frac{J}{mr} )}^{2}}=\frac{J^{2}}{2mr^{2}} $
