MCQ
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
  • $a \sqrt{n}$
  • B
    $a n$
  • C
    $a n^{2}$
  • D
    $a n \sqrt{n}$

Answer

Correct option: A.
$a \sqrt{n}$
a
$(a)$ 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, $\left.\sqrt{\frac{h}{2 \pi m \omega}}=a\right]$

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