Atoms And Nuclei Question 440
Question: According to Bohr’s theory, the expressions for the kinetic and potential energy of an electron revolving in an orbit is given respectively by
Options:
A) $ +\frac{e^{2}}{8\pi {\varepsilon_{0}}r} $ and $ -\frac{e^{2}}{4\pi {\varepsilon_{0}}r} $
B) $ +\frac{8\pi {\varepsilon_{0}}e^{2}}{r} $ and $ -\frac{4\pi {\varepsilon_{0}}e^{2}}{r} $
C) $ -\frac{e^{2}}{8\pi {\varepsilon_{0}}r} $ and $ -\frac{e^{2}}{4\pi {\varepsilon_{0}}r} $
D) $ +\frac{e^{2}}{8\pi {\varepsilon_{0}}r} $ and $ +\frac{e^{2}}{4\pi {\varepsilon_{0}}r} $
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Answer:
Correct Answer: A
Solution:
- P.E. $ =-\frac{ke^{2}}{r}=-\frac{e^{2}}{4\pi {\varepsilon_{0}}r} $ ; K.E. $ =-\frac{1}{2}(P\text{.E}\text{.})=\frac{e^{2}}{8\pi {\varepsilon_{0}}r} $
