Atomic Structure Question 367
Question: The angular momentum of an electron in a given orbit is J. Its kinetic energy will be:
Options:
A) $ \frac{1}{2},\frac{J^{2}}{mr^{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 $ J=mvr $ $ J^{2}=m^{2}v^{2}r^{2} $ or $ \frac{J^{2}}{2}=( \frac{1}{2}mv^{2} )mr^{2} $ or $ KE=\frac{J^{2}}{2mr^{2}} $
BETA
