Question

A particle of mass $m$ moves around in a circular orbit in a centro symmetric potential field u(r)$=\frac{k{r}^2}{2}$. Using Bohr’s quantization rule, the permissible energy levels are

• A. $\sqrt{\frac{k}{m}}$
• B. $n\sqrt{\frac{k}{m}}$
• C. $n$
• D. $\sqrt{\frac{nk}{m}}$

Answer 1

The correct option is B $n\sqrt{\frac{k}{m}}$

$U=\frac{k{r}^2}{2}$

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

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

Total energy $=\frac{1}{2}m{v}^2+U=k{r}^2$

From bohr's quantization, $mvr=\frac{nh}{2\pi }$

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

$⟹mk{r}^4=\frac{n^2{h}^2}{4{\pi }^2}$

$⟹r^2=\frac{nh}{2\pi \sqrt{mk}}$

Energy $=k{r}^2=k\frac{nh}{2\pi \sqrt{mk}}\propto n\sqrt{\frac{k}{m}}$