Verified Solution • 2025–26 Session
very hard
QJEEA-2024-05: JEE Advanced 2024 JEEA-2024-05 Solution: Q.5 A particle of mass 𝑚 is moving in a circular orbit…
Q.5 A particle of mass 𝑚 is moving in a circular orbit under the influence of the central force
𝐹(𝑟) = −𝑘𝑟, corresponding to the potential energy 𝑉(𝑟) = 𝑘𝑟2/2, where 𝑘 is a positive force constant and 𝑟 is the radial distance from the origin. According to the Bohr’s quantization rule, the angular momentum of the particle is given by 𝐿 = 𝑛ℏ, where ℏ = ℎ/(2𝜋), ℎ is the Planck’s constant, and 𝑛 a positive integer. If 𝑣 and 𝐸 are the speed and total energy of the particle, respectively, then which of the following expression(s) is(are) correct?
Multiple Choice Options
- A (A) 𝑟2 = 𝑛ℏ√ 1 /𝑚𝑘
- B (B) v² = nħ√(k / m³)
- C (C) L = √(k / m)
- D (D) E = nħ√(k / mr²)
Step-by-Step Verified Solution
Solution:
- Force and Centripetal Force:
- The central force provides the centripetal force: F(r) = -kr = -mv²/r.
- Therefore, mv²/r = kr, so mv² = kr².
- Angular Momentum:
- Angular momentum is L = mvr.
- According to Bohr's quantization rule, L = nħ.
- Thus, mvr = nħ.
- Find v and r:
- From mvr = nħ, we get v = nħ/(mr).
- Substitute this into mv² = kr²:
- m(nħ/mr)² = kr²
- m(n²ħ² / m²r²) = kr²
- n²ħ² / mr² = kr²
- n²ħ² = kmr⁴
- r⁴ = n²ħ² / km
- r² = nħ√(1/km) = nħ√(1/mk)
- Find v²:
- From mv² = kr², we have v² = kr²/m.
- Substitute r² = nħ√(1/mk):
- v² = k(nħ√(1/mk)) / m
- v² = nħ√(k²/m²)(1/mk)
- v² = nħ√(k/m³)
- Find Total Energy (E):
- Total energy E = Kinetic Energy (KE) + Potential Energy (PE).
- KE = (1/2)mv² = (1/2)kr²
- PE = (1/2)kr²
- E = (1/2)kr² + (1/2)kr² = kr²
- Substitute r² = nħ√(1/mk):
- E = k(nħ√(1/mk))
- E = nħ√(k²/mk)
- E = nħ√(k/m)
- Analyze the Options:
- (A) r² = nħ√(1 / mk): Correct (derived above).
- (B) v² = nħ√(k / m³): Correct (derived above).
- (C) L = √(k / m): Incorrect. L = nħ, which is independent of k and m.
- (D) E = nħ√(k / mr²): Incorrect. E = nħ√(k/m), which is independent of r.
Correct Options:
(A) and (B)