University:
Santa Fe College
Course:
PHY 2004 | Applied Physics 1
Academic year:

2024
Views:

126

Pages:

5
Author:

Jabin F.

# Relation b/w Energy and principal quantum no. Bohr-Sommerfeld quantization formula: ∮ pdx = nh where V(x) = kx^n → principal quantum no. n ∝ E^(1/(1+n)) applies for all cases. 1/λ ∝ E^(-1/3 + 1/n) where λ is amplitude Note: Any question of time or distance can be solved using this relation. (related to dependence) Example: Planetary motion (derivation of Kepler's 3rd Law) Solution: V = -GMm/r t ∝ A^(1-n/2) = A^(1+n/2) = A^(3/2) => t² ∝ A³ => t² ∝ a³ P-135 Consider circular orbits in a central force potential V(x) = -k/x^n, where k > 0 and n < 2. If the time period of a circular orbit of radius R is T₁ and that of radius 2R is T₂, then T₂/T₁ is? Solution: V = -kr^(-n) t ∝ A^(1-n/2) when V = kx^m t ∝ A^(1+n/2) T₂/T₁ = (R₂/R₁)^(1+n/2) = (2R/R)^(1+n/2) = 2^(1+n/2) Q: A particle is moving in potential V(x) = kx⁴. If its amplitude is doubled, then its time period of oscillation will become: Solution: V(x) = kx⁴ → n = 4 t ∝ A^(1+n/2) T₂/T₁ = (A₂/A₁)^(1+n/2) = (2A/A)^(1+4/2) = 2³ T₂ = 8T₁ Q: Consider the motion of a classical particle in a one-dimensional double-well potential V(x) = 1/4(x²-2)². If the particle is displaced infinitesimally from the minimum on the positive axis (and friction is neglected), then the particle motion will be: Solution: Let the particle having test mass, m = 1 V(x) = 1/4(x²-2)² → V'(x) = x(x²-2) = 0 x = 0, ±√2 x₀ = 0, ±√2 V''(x) = 3(x²-2) → at x₀ = 0, V''(x) = -9 = -ve (max.) → at x₀ = ±√2, V''(x) = 4 = +ve (min.) ∴ k = V''(x₀=±√2) = 4 ∴ ω = √(k/m) = √(4/1) = 2 ∴ The particle will execute simple harmonic motion in the right well with an angular frequency, ω = 2.

Relation Between Energy and Principle Quantum

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