JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Modern Physics

A particle of mass

m

moves in a circular orbit in a central potential field

U (r) = \frac{1}{2} k r^{2}

. If Bohr's quantization conditions are applied, radii of possible orbits and energy levels vary with quantum number n as:

Answer: (A)

Solution

In this potential the restoring force is

F = - \frac{d U}{d r} = - k r

, so for a circular orbit the central-force balance is

\frac{m v ^{2}}{r} = k r

, i.e.

m v^{2} = k r^{2}

. Bohr's rule quantises the angular momentum,

m v r = \frac{nh}{2 π}

. Eliminating

v

between these two relations gives

r^{2} \propto n

, so

r_{n} \propto n

. The total energy is kinetic plus potential,

\frac{1}{2} m v^{2} + \frac{1}{2} k r^{2} = \frac{1}{2} k r^{2} + \frac{1}{2} k r^{2} = k r^{2}

, which is proportional to

r^{2}

and hence to

n

, so

E_{n} \propto n

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Single correct option