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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 $a=\sqrt{h /(2 \pi m \omega)}$ is
$a \sqrt{n}$
$a n$
$a n^{2}$
$a n \sqrt{n}$
Solution
$(a)$ 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.
$\Rightarrow \quad v=r \omega$
Now, angular momentum of particle will be
$L=m v r=m r^{2} \omega$
By Bohr's model, we have
$L=\frac{n h}{2 \pi}$
$\Rightarrow \quad m r^{2} \omega=\frac{n h}{2 \pi} \Rightarrow r^{2}=\frac{n h}{2 \pi m \omega}$
or $r=\sqrt{\frac{h}{2 \pi m \omega}} \times \sqrt{n}$
$\Rightarrow \quad r=a \sqrt{n}$
$\left[\because\right.$ given, $\left.\sqrt{\frac{h}{2 \pi m \omega}}=a\right]$
