Question

Suppose the potential energy between electron and proton at a distance $r$ is given by $\frac{k{e}^2}{3r^3},$ where $k$ is constant. Then using Bohr's theory to hydrogen atom, which of the following is correct

• A. Energy in the $n^{th}$ orbit is propotional to $n^6$
• B. Energy is proportional to $m^{-3}$ ($m=$ mass of electron)
• C. Energy in the $n^{th}$ orbit is proportional to $n^{-2}$
• D. Energy is proportional to $m^3$ ($m=$ mass of electron)

Answer 1

Answer: (A), (B)

Energy is proportional to $m^{-3}$ ($m=$ mass of electron)

Given potential energy between electron and proton is $U=\frac{k{e}^2}{3r^3}$
Force
$F=-\frac{dU}{dr}$
$|F|=\frac{dU}{dr}=\frac{k{e}^2}{r^4}....\left(i\right)$
$\frac{k{e}^2}{r^4}=\frac{m{v}^2}{r}....\left(ii\right)$
and angular momentum is
$mvr=\frac{nh}{2\pi }....\left(iii\right)$
By $\left(ii\right)$ and $\left(iii\right),$
$r=\frac{k{e}^{2}4{\pi }^2}{h^2}\frac{m}{n^2}\propto \frac{K_{1}m}{n^2}....\left(iv\right)$
Total energy $=PE+KE$
$TE=-\frac{k{e}^2}{3r^3}+\frac{1}{2}m{v}^2$
$TE=-\frac{k{e}^2}{3r^3}+\frac{k{e}^2}{2r^3}$
$TE=\frac{k{e}^2}{6r^3}=\frac{-k{e}^2}{6{\left(\frac{K_{1}m}{n^2}\right)}^3}=\frac{-k{e}^2{n}^6}{6K_1^3{m}^3}$
Total energy $\propto n^6$
Total energy $\propto m^{-3}$
$\therefore \left(a\right)$ and (b) are correct.