Question

If potential energy between a proton and an electron is given by $U=\frac{-k{e}^2}{2R^3}$, where e is the charge of electron and R is the radius of atom, then radius of Bohr's orbit is given by ($h$ = Planck's constant, $k$ = constant)

• A. $\frac{k{e}^{2}m}{h^2}$
• B. $\frac{6{\pi }^{2}k{e}^{2}m}{n^2{h}^2}$
• C. $\frac{2\pi k{e}^{2}m}{n{h}^2}$
• D. $\frac{4{\pi }^{2}k{e}^{2}m}{n^2{h}^2}$

Answer 1

The correct option is B $\frac{6{\pi }^{2}k{e}^{2}m}{n^2{h}^2}$

$U=-\frac{k{e}^2}{2R^3}$, $F$ = -$\frac{dU}{dR}$ =$\frac{3k{e}^2}{2R^4}$
But $F$ = $\frac{m{v}^2}{R}$$\Rightarrow \frac{m{v}^2}{R}=\frac{3k{e}^2}{2R^4}$
also, $mvR=\frac{nh}{2\pi }$
After solving both equations, we get $R$ = $\frac{6{\pi }^{2}k{e}^{2}m}{n^2{h}^2}$