Question

If in Bohr's atomic model, it is assumed that force between electron and proton varies inversely as $r^4$, then the kinetic energy of the electron will be proportional to $n^{\gamma }$, where $\gamma$ is

Answer 1

The force between electron and proton varies inversely as $r^4$

$\Rightarrow F=\frac{K}{r^4}$

This force provides the necessary centripetal force,

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

$\Rightarrow v^2=\frac{K}{m{r}^3}$

According to Bohr's postulate,

$mvr=\frac{nh}{2\pi }$

Squaring both sides, we get,

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

Substituting the values of $v^2$ in above equation, we get,

$\Rightarrow m^2\left(\frac{K}{m{r}^3}\right)r^2=\frac{n^2{h}^2}{4{\pi }^2}$

$\Rightarrow r=\frac{Km4{\pi }^2}{n^2{h}^2}$

So, $K.E=\frac{1}{2}m{v}^2=\frac{1}{2}m\left(\frac{K}{m{r}^3}\right)...\left(1\right)$

Substituting the values of $r$ in equation $\left(1\right)$, we get,

$\Rightarrow K.E=\frac{1}{2}\frac{K\left(n^2{)}^3\right(h^2{)}^3}{K^3{m}^3{4}^3{\pi }^6}$

$\Rightarrow K.E\propto n^6$

Answer : $6$

Why this Question? This question will help in understanding the concept of centripetal force, angular momentum and energy of electron in $n^{th}$ orbit of Bohr model.