Atoms And Nuclei Question 28

Question: If m is mass of electron, v its velocity, r the radius of stationary circular orbit around a nucleus with charge Ze, then from Bohr’s first postulate, the kinetic energy $ K=\frac{1}{2}mv^{2} $ of the electron in C.G.S. system is equal to [NCERT 1977]

Options:

A) $ \frac{1}{2}\frac{Ze^{2}}{r} $

B) $ \frac{1}{2}\frac{Ze^{2}}{r^{2}} $

C) $ \frac{Ze^{2}}{r} $

D) $ \frac{Ze}{r^{2}} $

Show Answer

Answer:

Correct Answer: A

Solution:

In the revolution of electron, coulomb force provides the necessary centripetal force Þ $ \frac{ze^{2}}{r^{2}}=\frac{mv^{2}}{r} $
Þ $ mv^{2}=\frac{ze^{2}}{r} $ \ K.E. $ =\frac{1}{2}mv^{2}=\frac{ze^{2}}{2r} $



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