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}} $



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