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. $an^2$
• D. an$\sqrt{n}$

Answer 1

Answer: (A)

Energy of particle is
$\frac{1}{2}m\omega^{2}\,r^{2}=\frac{1}{2}mv^{2}$
where, $v$ = velocity of particle around the path.
$\Rightarrow v=r\omega$
Now, angular momentum of particle will be
$L=mvr-mr^{2}\omega$
By Bohr’s model, we have
$L=\frac{nh}{2\pi}$
$\Rightarrow mr^{2}\omega=\frac{nh}{2\pi}$
$\Rightarrow r^{2}=\frac{nh}{2\pi m\omega}$
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}]$