JEE Main 17 March 2021 Shift 2 Solved Paper

As we know, the force on the particle is given as
F=−‌

dU(r)

dr

Given, U(r)=U0r4
⇒‌‌F=−‌

d(U0r4)

dr

⇒F=−4U0r3
As we know, the force on the particle moving in a circular orbit of radius r will be centripetal force
|‌

mv2

r

|=4U0r3
⇒v2∝r4
v∝r2 .....(i)
Hence, the velocity of the particle is square of the radius of the orbit. Using the Bohr's quantisation condition, which states that the electron revolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of ‌

h

2π

, here, h is
Planck's constant.
i.e,
n‌

h

2π

=mvr  [From Eq. (i)]
⇒‌‌r3∝n⇒r∝n1∕3
Comparing the above relation with n1/α, we get
α=3