Atoms And Nuclei Question 91
Question: The time of revolution of an electron around a nucleus of charge Ze in nth Bohr orbit is directly proportional to [MP PET 2003]
Options:
A) n
B) $ \frac{n^{3}}{Z^{2}} $
C) $ \frac{n^{2}}{Z} $
D) $ \frac{Z}{n} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ T=\frac{2\pi r}{v} $ ; r = radius of nth orbit $ =\frac{n^{2}h^{2}}{\pi mZe^{2}} $
v = speed of $ {{e}^{-}} $ in nth orbit $ =\frac{ze^{2}}{2{\varepsilon_{0}}nh} $ \ $ T=\frac{4\varepsilon _{0}^{2}n^{3}h^{3}}{mZ^{2}e^{4}} $
Þ $ T\propto \frac{n^{3}}{Z^{2}} $
