Question

A particle of mass $m$ moves around the origin in a potential$\frac{1}{2}m{\omega }^2{r}^2$, where $r$ is the distance from the origin. Applying the Bohr’s model in this case, the radius of the particle in its $n^{th}$ orbit in terms of $a=\sqrt{h/\left(2\pi m\omega \right)}$ is

• A. $a\sqrt{n}$
• B. $an$
• C. $a{n}^2$
• D. an$\sqrt{n}$

Answer 1

Answer: (A)

Energy of particle is
$\frac{1}{2}m{\omega }^2{r}^2=\frac{1}{2}m{v}^2$
where, $v$ = velocity of particle around the path.
$\Rightarrow v=r\omega$
Now, angular momentum of particle will be
$L=mvr-m{r}^2\omega$
By Bohr’s model, we have
$L=\frac{nh}{2\pi }$
$\Rightarrow m{r}^2\omega =\frac{nh}{2\pi }$
$\Rightarrow r^2=\frac{nh}{2\pi m\omega }$
or $r=\sqrt{\frac{h}{2\pi m\omega }}\times \sqrt{n}$
$\Rightarrow r=a\sqrt{n}$
$[\therefore$ given$\sqrt{\frac{h}{2\pi m\omega }=a}]$