Question

If potential energy between a proton and an electron is given by $|U|=k{e}^2/2R^3$, 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. $\frac{6{\pi }^2}{n^2}\frac{k{e}^{2}m}{h^2}$
• C. $\frac{2\pi }{n}\frac{k{e}^{2}m}{h^2}$
• D. $\frac{4{\pi }^{2}k{e}^{2}m}{n^2{h}^2}$

Answer 1

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

Potential energy at a distance $R$ is given as, $U=\frac{k{e}^2}{2R^3}$

Thus force between proton and electron $=-\frac{dU}{dR}$ $=\frac{3k{e}^2}{2R^4}$

This force causes the centrifugal acceleration of the system.

Thus, $\frac{3k{e}^2}{2R^4}=\frac{m{v}^2}{R}$

Bohr's postulate is: $mvR=\frac{nh}{2\pi }$

Substituting the expression for $v$ from above equation gives,

$\frac{3k{e}^2}{2R^4}=\frac{m}{R}(\frac{nh}{2\pi mR}{)}^2$

$⟹R=\frac{6k{e}^2{\pi }^{2}m}{n^2{h}^2}$