Question

In the Bohr’s model of hydrogen-like atom the force between the nucleus and the electron is modified as $F=\frac{e^2}{4\pi {\epsilon }_0}\left(\frac{1}{r^2}+\frac{\beta }{r^3}\right),$ where $\beta$ is a constant. For this atom, the radius of the $n^{th}$ orbit in terms of the Bohr radius $\left(a_0=\frac{{\epsilon }_0{h}^2}{m\pi e^2}\right)$ is :

• A. $r_n=a_{0}n-\beta$
• B. $r_n=a_0{n}^2+\beta$
• C. $r_n=a_0{n}^2-\beta$
• D. $r_n=a_{0}n+\beta$

Answer 1

Answer: (C)

As $F=\frac{m{v}^2}{r}=\frac{e^2}{4\pi {\in }_0}\left(\frac{1}{r^2}+\frac{B}{r^3}\right)$
and $mvr=\frac{nh}{2\pi }\Rightarrow v=\frac{nh}{2\pi mr}$
$\therefore {\left(\frac{nh}{2\pi mr}\right)}^2\times \frac{1}{r}=\frac{e^2}{4\pi {\in }_0}\left(\frac{1}{r^2}+\frac{B}{r^3}\right)$
or, $\frac{1}{r^2}+\frac{B}{r^3}=\frac{m{n}^2{h}^{2}4\pi {\in }_0}{4{\pi }^2{m}^2{e}^2{r}^3}$
or, $\frac{a_0{n}^2}{r^3}=\frac{1}{r^2}+\frac{B}{r^3}$
$\left(\because a_0=\frac{{\in }_0{h}^2}{m\pi e^2}Given\right)$
$\therefore r=a_0{n}^2-B$