A small particle of mass m moves in such a way that its potential energy U = 1 2 m ω 2 r 2 where ω…

A small particle of mass m moves in such a way that its potential energy U=12mω2r2 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

  1. n
  2. 1n
  3. n2
  4. n

Solution

The data given is

U=12mω2r2

From Bohr's quantization, 

mvr=nh2πv2=nh2πrm2   ...(i)

In an orbit the value of the kinetic energy is half of the potential energy,

K=mω2r24

Substituting the value of equation (i) in the kinetic energy,

K=12mn2h24π2m2r2=mω2r24n2h22π2m2ω2=r4rn

Asked in: JEE Main 2023 (06 Apr Shift 2)

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