A small particle of mass m moves in such a way that its potential energy $U = \frac{1}{2} {mω}^{2} r^{2}$ where $ω$ is constant and r is the distance of the particle from origin. Assuming Bohr’s quantization of momentum and circular orbit, the radius of nth orbit will be proportional to

see full answer

Take the SCORE scholarship exam from home and compete for scholarships worth ₹1 crore!*

An Intiative by Sri Chaitanya

n²

1/n

$\sqrt{n}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$\begin{array}{r} U = \frac{1}{2} {mω}^{2} r^{2} \\ F = - \frac{dU}{dr} = - {mω}^{2} r \end{array}$

$m ω^{2} r = \frac{m v^{2}}{r}$

$\begin{aligned} \Rightarrow v = ω r \\ & m v r = \frac{n h}{2 π} \end{aligned}$

$m ω r^{2} = \frac{n h}{2 π}$

$\Rightarrow r \propto \sqrt{n}$

Watch 3-min video & get full concept clarity