Question
If in Bohr's atomic model, it is assumed that force between electron and proton varies inversely as $r^4$, energy of the system will be proportional to

Answer

$\frac{m v^2}{r}=\frac{k}{r^4}$

$V^2=\frac{k}{m r^3}$

we know

$m v r=\frac{n h}{2 x}$

$m^2 v^2 r^2=\frac{n^2 h^2}{4 \pi^2}$

$\frac{m^2 k r}{m r}=\frac{n^2 h^2}{4 x^2}$

$r=\frac{m k^3 \pi^2}{n^2 h^2}$

$k \cdot f=\frac{1}{2} m v^2=\frac{1}{2} m / \frac{k}{m r^3}$

$=\frac{1}{2} m^m \frac{K}{m^1 m^3}$

$E \alpha n^6$

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