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PhyMedAdv 2024 • Paper 1

Bohr Quantized Circular Motion under Harmonic Central Force

A particle of mass mm is moving in a circular orbit under the influence of the central force F(r)=−krF(r) = -kr, corresponding to the potential energy V(r)=kr2/2V(r) = kr^2/2, where kk is a positive force constant and rr is the radial distance from the origin. According to the Bohr's quantization rule, the angular momentum of the particle is given by L=nℏL = n\hbar, where ℏ=h/(2π)\hbar = h/(2\pi), hh is the Planck's constant, and nn a positive integer. If vv and EE are the speed and total energy of the particle, respectively, then which of the following expression(s) is(are) correct?

Options

A

r2=nℏ1mkr^2 = n\hbar \sqrt{\frac{1}{mk}}

Correct
B

v2=nℏkm3v^2 = n\hbar \sqrt{\frac{k}{m^3}}

Correct
C

Lmr2=km\frac{L}{mr^2} = \sqrt{\frac{k}{m}}

Correct
D

E=nℏ2kmE = \frac{n\hbar}{2} \sqrt{\frac{k}{m}}

Correct

Topics & Concepts

AtomsBohr Model

Step-by-Step Solution

To determine which expressions are correct, we analyze the circular motion of the particle of mass mm under the central force F(r)=−krF(r) = -kr and potential energy V(r)=12kr2V(r) = \frac{1}{2}kr^2.

1. Condition for Circular Orbit

For a circular orbit of radius rr, the centripetal force is provided by the magnitude of the central force: mv2r=kr  ⟹  mv2=kr2— (1)\frac{mv^2}{r} = kr \implies mv^2 = kr^2 \quad \text{--- (1)}

2. Bohr's Quantization Rule

According to the quantization rule for angular momentum: L=mvr=nℏ  ⟹  v=nℏmr— (2)L = mvr = n\hbar \implies v = \frac{n\hbar}{mr} \quad \text{--- (2)}

3. Verification of Option (A): Expression for r2r^2

Substituting equation (2) into equation (1): m(nℏmr)2=kr2m \left(\frac{n\hbar}{mr}\right)^2 = kr^2 n2ℏ2mr2=kr2  ⟹  r4=n2ℏ2mk\frac{n^2\hbar^2}{mr^2} = kr^2 \implies r^4 = \frac{n^2\hbar^2}{mk} Taking the square root on both sides: r2=nℏ1mkr^2 = n\hbar \sqrt{\frac{1}{mk}}

Thus, Option (A) is correct.

4. Verification of Option (B): Expression for v2v^2

From equation (2), we have v2=n2ℏ2m2r2v^2 = \frac{n^2\hbar^2}{m^2r^2}. Substituting the expression for r2r^2 from Option (A): v2=n2ℏ2m2(nℏ1mk)=nℏm2mk=nℏmkm4=nℏkm3v^2 = \frac{n^2\hbar^2}{m^2 \left(n\hbar \sqrt{\frac{1}{mk}}\right)} = \frac{n\hbar}{m^2} \sqrt{mk} = n\hbar \sqrt{\frac{mk}{m^4}} = n\hbar \sqrt{\frac{k}{m^3}}

Thus, Option (B) is correct.

5. Verification of Option (C): Expression for Lmr2\frac{L}{mr^2}

Using L=mvrL = mvr: Lmr2=mvrmr2=vr\frac{L}{mr^2} = \frac{mvr}{mr^2} = \frac{v}{r} From equation (1), we know that v2r2=km\frac{v^2}{r^2} = \frac{k}{m}, which gives vr=km\frac{v}{r} = \sqrt{\frac{k}{m}}. Therefore: Lmr2=km\frac{L}{mr^2} = \sqrt{\frac{k}{m}}

Thus, Option (C) is correct.

6. Verification of Option (D): Expression for Total Energy EE

The total mechanical energy EE is the sum of kinetic energy (TT) and potential energy (VV): T=12mv2=12kr2T = \frac{1}{2}mv^2 = \frac{1}{2}kr^2 V(r)=12kr2V(r) = \frac{1}{2}kr^2 E=T+V=12kr2+12kr2=kr2E = T + V = \frac{1}{2}kr^2 + \frac{1}{2}kr^2 = kr^2

Substituting r2=nℏ1mkr^2 = n\hbar \sqrt{\frac{1}{mk}}: E=k(nℏ1mk)=nℏk2mk=nℏkmE = k \left(n\hbar \sqrt{\frac{1}{mk}}\right) = n\hbar \sqrt{\frac{k^2}{mk}} = n\hbar \sqrt{\frac{k}{m}}

Thus, Option (D) is correct.

Conclusion

The correct options are (A), (B), (C), and (D).

Bohr Quantized Circular Motion under Harmonic Central Force | Physics PYQ Solution - JEE Challenger