Atoms And Nuclei Question 272
Question: In the Bohr’s model of hydrogen-like atom the force between the nucleus and the electron is modified as $ F=\frac{e^{2}}{4\pi {\varepsilon_{0}}}( \frac{1}{r^{2}}+\frac{\beta }{r^{3}} ), $ where $ \beta $ is a constant. For this atom, the radius of the nth orbit in terms of the Bohr radius $ ( a_{0}=\frac{{\varepsilon_{0}}h^{2}}{m\pi e^{2}} ) $ is:
Options:
A) $ r_{n}=a_{0}n-\beta $
B) $ r_{n}=a_{0}n^{2}+\beta $
C) $ r_{n}=a_{0}n^{2}-\beta $
D) $ r_{n}=a_{0}n+\beta $
Show Answer
Answer:
Correct Answer: C
Solution:
- As $ F=\frac{mv^{2}}{r}=\frac{e^{2}}{4\pi {\varepsilon_{0}}}( \frac{1}{r^{2}}+\frac{\beta }{r^{3}} ) $ $ \text{and }mvr=\frac{nh}{2\pi }\Rightarrow v=\frac{nh}{2\pi mr} $
$ \therefore m{{( \frac{nh}{2\pi mr} )}^{2}}\times \frac{1}{r}=\frac{e^{2}}{4\pi {\varepsilon_{0}}}( \frac{1}{r^{2}}+\frac{\beta }{r^{3}} ) $ or, $ \frac{a_{0}n^{2}}{r^{3}}=\frac{1}{r^{2}}+\frac{\beta }{r^{3}}( \therefore a_{0}=\frac{{\varepsilon_{0}}h^{2}}{m\pi e^{2}}Given ) $
$ \therefore r=a_{0}n^{2}-\beta $
