Question

Suppose the potential energy between electron and proton at a distance $r$ is given by $\frac{k{e}^2}{3r}$ . Then using Bohr's theory radius of $1^{\text{st}}$ bohr orbit is given by

• A. $\frac{3h^2}{2mk{e}^2}$
• B. $\frac{6h^2}{mk{e}^2}$
• C. $\frac{3h^2}{4mk{e}^2}$
• D. $\frac{3h^2}{mk{e}^2}$

Answer 1

The correct option is D $\frac{3h^2}{mk{e}^2}$

Given potential energy between electron and proton is $U=\frac{K{e}^2}{3r}$
Force
$F=-\frac{dU}{dr}$
$\Rightarrow F=-\frac{k{e}^2}{3r^2}$
This force will provide the essential centripetal acceleration.
$\frac{K{e}^2}{3r^2}=\frac{m{v}^2}{r}...\left(1\right)$
$mvr=nh...\left(2\right)$
Putting value of $v$ in equation (1) :-
$\frac{K{e}^2}{3r^2}=\frac{m}{r}\left[\frac{n^2{h}^2}{m^2{r}^2}\right]$
$r=\frac{3n^2{h}^2}{mk{e}^2}$
For $1^{\text{st}}$ orbit- $[n=1]$
$r=\frac{3h^2}{mk{e}^2}$