Question

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

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

Answer 1

The correct option is A $n^6$

$U=\frac{K{e}^2}{3r^3}$

$⟹F=\frac{m{v}^2}{r}=\frac{-dU}{dr}\propto \frac{1}{r^4}$

$⟹v\propto r^{-3/2}$

From Bohr' quantization, $mvr=nh/2\pi$

$⟹n\propto r^{-3/2}r=r^{-1/2}$.................... 1

Energy in nth orbit=$\frac{1}{2}m{v}^2+U=\frac{K{e}^2}{2r^3}+\frac{K{e}^2}{3r^3}=\frac{5K{e}^2}{6r^3}\propto r^{-3}\propto n^6$(Using equation 1)