A particle of mass m moves in a circular orbit in a
central potential field U(r) = U₀r⁴. If Bohr's quantization conditions
are applied, radii of possible
orbitals rₙ vary with n^(1/α), where α is ____________.

Solution

Physics sir · 4 steps

Turn the potential into a force first, then let mvr = nh/2π fix the speed in each orbit.

1. Force from potential. The force is F = -dU/dr = -4U₀ r³. The minus sign means it points towards the centre, so it supplies the centripetal force: mv²/r = 4U₀ r³, which gives mv² = 4U₀ r⁴.
2. Bohr condition. Quantised angular momentum gives mvr = nh/2π, so v = nh/(2π m r).
3. Eliminate v. Put this v into mv² = 4U₀ r⁴. Everything except n and r is a constant.

   (n² h²)/(4π² m r²) = 4U₀ r⁴ ⇒ r⁶ = (n² h²)/(16π² m U₀) ⇒ r ∝ n^(2/6) = n^(1/3)

4. So. rₙ ∝ n^(1/3), so 1/α = 1/3 and α = 3.

The common slip: 3.5. Using U₀ r⁴ itself as the force, without differentiating, gives r⁷ ∝ n², so r ∝ n^(2/7) and α = 3.5.

The idea to keep. For any power-law potential, find the force, match it to mv²/r, then substitute v from mvr = nh/2π. The power of r that collects on one side tells you how rₙ scales with n.

Answer from the NTA final answer key. Solution by PrepWiser.