Question

A particle of mass m moves around the origin in a potential $\frac{1}{2}m{w}^2{r}^2$, where r is the distance from the origin. Applying the Bohr 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)

$a\sqrt{n}$

The force at a distance r is $F=-\frac{dV}{dr}=m{e}^{2}r$

If r is the radius of the n-th orbit, centripetal force is equated to F.

$\frac{m{v}^2}{r}=m{e}^{2}r$ .........(1)

Using Bohr's quantization, $mvr=\frac{nh}{2\pi }$ ....(2)

Solving (1) and (2) by eliminating v, we get $r=a\sqrt{n}$