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} $
BETA
