Physics - Modern Physics Question with Solution | TestHub

$F_r=\frac{-dU}{dr}=-kr$For circular motion

$|F_r|=kr=\frac{m{v}^2}{r}\Rightarrow k{r}^2=m{v}^2$ ........(i)

Bohr's quantization $\Rightarrow mvr=\frac{nh}{2\pi }$ ....(ii) From (i) and (ii)

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

$\Rightarrow \frac{1}{m}{\left(\frac{nh}{2\pi r}\right)}^2=k{r}^2\Rightarrow \frac{n^2{h}^2}{4{\pi }^{2}mk}=r^4$

$\Rightarrow r={\left(\frac{h^2}{4{\pi }^{2}mk}\right)}^{1/4}{n}^{1/2}$

$r\propto \sqrt{n}$

From equation (i) $U\propto \sqrt{n}$

$KE=\frac{1}{2}m{v}^2$ $PE=\frac{1}{2}k{r}^2$

$E=K+U=\frac{1}{2}m{v}^2+\frac{1}{2}k{r}^2=k{r}^2\propto n$