Question

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

The correct option is C $r_n=a_0{n}^2-\beta$

$\frac{e^2}{4\pi {\epsilon }_0}(\frac{1}{r^3}-\frac{\beta }{r^3})$
$\therefore v^2=\frac{e^2}{4\pi {\epsilon }_{0}mr}(\frac{1}{r^{}}+\frac{\beta }{r^2})$
From Bohr's hypothesis $v=\frac{nh}{2\pi mr}\therefore \frac{n^2{hn}^2}{4{\pi }^2{m}^2{r}^2}=\frac{e^2}{4\pi {\epsilon }_{0}mr}(\frac{1}{r^{}}+\frac{\beta }{r^2})$
$\therefore r+\beta =a_0{n}^2$