Question

In the Bohr’s model of hydrogen-like atom the force between the nucleus and the electron is modified as $F=\frac{e^{2}}{4\pi\varepsilon_{0}}\left(\frac{1}{r^{2}}+\frac{\beta}{r^{3}}\right),$ where $\beta$ is a constant. For this atom, the radius of the $n^{th}$ orbit in terms of the Bohr radius $\left(a_{0}=\frac{\varepsilon_{0}h^{2}}{m\pi\,e^{2}}\right)$ 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 $F=\frac{mv^{2}}{r}=\frac{e^{2}}{4\pi\,\in_{0}}\left(\frac{1}{r^{2}}+\frac{B}{r^{3}}\right)$
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, $\frac{1}{r^{2}}+\frac{B}{r^{3}}=\frac{mn^{2}h^{2}4\pi \,\in_{0}}{4\pi^{2}m^{2}e^{2}r^{3}}$
or, $\frac{a_{0}n^{2}}{r^{3}}=\frac{1}{r^{2}}+\frac{B}{r^{3}}$
$\left(\because a_{0}=\frac{\in_{0}h^{2}}{m\pi e^{2}}Given\right)$
$\therefore r=a_{0}n^{2}-B$