JEE Main 12 Jan 2019 Morning Question 2
Question: A particle of mass m moves in a circular orbit in a central potential field $ U(r)=\frac{1}{2}kr^{2}. $ If Bohr’s quantization conditions are applied, radii of possible orbitals and energy levels vary with quantum number n as [JEE Main Online Paper Held On 12-Jan-2019 Morning]
Options:
A) $ r _{n}\propto n^{2},E _{n}\propto \frac{1}{n^{2}} $
B) $ r _{n}\propto \sqrt{n},E _{n}\propto n $
C) $ r _{n}\propto \sqrt{n},E _{n}\propto \frac{1}{n} $
D) $ r _{n}\propto n,E _{n}\propto n $
Show Answer
Answer:
Correct Answer: B
Solution:
Force due to this field $ F=-\frac{\partial U}{\partial r} $ $ F=\frac{-\partial }{\partial r}( \frac{1}{2}kr^{2} )=-kr $
For circular orbit, $ \frac{mv^{2}}{r}=-kr\Rightarrow v\propto r $ …(i)
Also, by Bohr’s quantization condition
$ mvr=\frac{nh}{2\pi } $ -(ii)
For eqn. (i) and (ii),
$ r _{n}\propto {n^{1/2}} $ $ U(r)=\frac{1}{2}kr^{2}\Rightarrow E _{n}=-\frac{1}{2}U(r)=-\frac{1}{4}kr^{2} $
$ \Rightarrow E _{n}\propto n $
