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PhyMedAdv 2020 • Paper 1

Bohr Quantization for Linear Potential Energy System

Suppose a particle of mass mm travels in circular paths under a potential energy given by V(r)=FrV(r) = Fr, where FF is a positive constant and rr represents the radial distance from the origin. Quantization rules of the Bohr model are applied to calculate its energy levels. If RR denotes the orbital radius, vv its orbital speed, and EE the total energy in the nthn^{\text{th}} stationary orbit (hh being Planck's constant), then:

Options

A

R∝n1/3R \propto n^{1/3} and v∝n2/3v \propto n^{2/3}

B

R∝n2/3R \propto n^{2/3} and v∝n1/3v \propto n^{1/3}

Correct
C

E=32(n2h2F24π2m)1/3E = \frac{3}{2} \left( \frac{n^2 h^2 F^2}{4 \pi^2 m} \right)^{1/3}

Correct
D

E=2(n2h2F24π2m)1/3E = 2 \left( \frac{n^2 h^2 F^2}{4 \pi^2 m} \right)^{1/3}

Topics & Concepts

AtomsBohr Model

Step-by-Step Solution

To determine the correct options, we apply the principles of classical mechanics along with the Bohr quantization hypothesis for circular orbits.

1. Dynamics of the Circular Motion

The potential energy of the particle at a radial distance rr is given by: V(r)=FrV(r) = Fr

The radial central force f(r)f(r) acting on the particle is: f(r)=−dVdr=−Ff(r) = -\frac{dV}{dr} = -F

The magnitude of this inward central force provides the necessary centripetal force for circular motion of radius RR: mv2R=F  ⟹  mv2=FR\frac{m v^2}{R} = F \implies m v^2 = F R   ⟹  v=FRm\implies v = \sqrt{\frac{F R}{m}}

2. Bohr Quantization Condition

According to Bohr's quantization rule, the angular momentum LL of the particle in the nthn^{\text{th}} orbit is quantized: L=mvR=nh2πL = m v R = \frac{n h}{2\pi}

Substituting v=FRmv = \sqrt{\frac{F R}{m}} into the quantization equation: mFRmR=nh2πm \sqrt{\frac{F R}{m}} R = \frac{n h}{2\pi} mFR3=nh2π\sqrt{m F R^3} = \frac{n h}{2\pi}

Squaring both sides: mFR3=n2h24π2m F R^3 = \frac{n^2 h^2}{4 \pi^2} R3=n2h24π2mFR^3 = \frac{n^2 h^2}{4 \pi^2 m F} R=(n2h24π2mF)1/3R = \left( \frac{n^2 h^2}{4 \pi^2 m F} \right)^{1/3}

From this expression, we observe that: R∝n2/3R \propto n^{2/3}

Now, substituting R∝n2/3R \propto n^{2/3} back into the speed relation v=FRmv = \sqrt{\frac{F R}{m}}: v∝R1/2∝(n2/3)1/2=n1/3v \propto R^{1/2} \propto \left(n^{2/3}\right)^{1/2} = n^{1/3} v∝n1/3v \propto n^{1/3}

Therefore, Option B is correct and Option A is incorrect.

3. Total Energy of the System

The total energy EE in the nthn^{\text{th}} orbit is the sum of its kinetic energy KK and potential energy VV:

  • Kinetic Energy (KK): K=12mv2=12FRK = \frac{1}{2} m v^2 = \frac{1}{2} F R

  • Potential Energy (VV): V=FRV = F R

  • Total Energy (EE): E=K+V=12FR+FR=32FRE = K + V = \frac{1}{2} F R + F R = \frac{3}{2} F R

Substituting the value of RR: E=32F(n2h24π2mF)1/3E = \frac{3}{2} F \left( \frac{n^2 h^2}{4 \pi^2 m F} \right)^{1/3} E=32(F3⋅n2h24π2mF)1/3E = \frac{3}{2} \left( F^3 \cdot \frac{n^2 h^2}{4 \pi^2 m F} \right)^{1/3} E=32(n2h2F24π2m)1/3E = \frac{3}{2} \left( \frac{n^2 h^2 F^2}{4 \pi^2 m} \right)^{1/3}

Therefore, Option C is correct and Option D is incorrect.

Conclusion

The correct options are B and C.

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