Question

A particle of mass $m$ moves along a circular orbit in a centro-symmetrical potential field $U\left(r\right)=\frac{K{r}^2}{8}$ (K is a positive constant). According to Bohr's quantization condition, permissible orbital radius is $(n=1,2,3,....)$

• A. $\sqrt{\frac{nh}{\pi mk}}$
• B. $\frac{1}{mK}\sqrt{\frac{nh}{\pi }}$
• C. ${\frac{1}{mK}}^{1/4}\sqrt{\frac{nh}{\pi }}$
• D. $nh\sqrt{\frac{k}{m}}$

Answer 1

The correct option is C ${\frac{1}{mK}}^{1/4}\sqrt{\frac{nh}{\pi }}$

$\text{Given-}\ast \text{Potential Enesgy}U\left(r\right)=\frac{k{r}^2}{8}$

$\text{we know that Force F,}$

$F=-\frac{dU}{dr}=-\frac{d}{dr}\left(\frac{k{r}^2}{8}\right)=\frac{kr}{4}$

$\Rightarrow \text{Force is}F=\frac{kr}{4}\text{in inward direction.}$

$\Rightarrow \text{we get}\frac{kr}{4}=\frac{m{v}^2}{r}$

$\Rightarrow k{r}^2=4m{v}^2-\left(1\right)$

$\text{By Bohr's quantization of angular momentum,}$

$L=mvr=\frac{nh}{2\pi }-\left(2\right)$

$\text{From eqn}\left(1\right)\&\left(2\right)\text{we get -}$

$r={\left(\frac{1}{mk}\right)}^{1/4}\cdot \sqrt{\frac{nh}{\pi }}$

$\text{Hence, option (c) is correct.}$