Question

The potential energy between electron and proton is given by $U=K{e}^2/3r^3$. According to Bohr's theory, the energy in nth orbit of such a hypothetical atom will be proportional to

• A. $n^6$
• B. $n^4$
• C. $n^2$
• D. n

Answer 1

Answer: (A)

$n^6$

The potential energy between the electron and the proton is given as: $U=\frac{K{e}^2}{3r^3}$

Thus the force experienced by the electron: $F=-\frac{dU}{dr}⟹F=\frac{K{e}^2}{r^4}$

This force is balanced by the centripetal force, $\frac{m{v}^2}{r}=\frac{K{e}^2}{r^4}$

$⟹v^2=(\frac{K{e}^2}{m})\frac{1}{r^3}$ ...........(1)

Also $mvr=\frac{nh}{2\pi }$ and equation (1) $⟹r=\frac{4{\pi }^{2}K{e}^{2}m}{n^2{h}^2}$

$⟹r\propto \frac{1}{n^2}$ ..............(2)

Now total energy in $nth$ orbit $E_n=\frac{1}{2}m{v}^2-U=\frac{K{e}^2}{6r^3}$ (using (1))

$\therefore E_n\propto \frac{1}{r^3}⟹E_n\propto n^6$