University:
Santa Fe College
Course:
PHY 2048 | General Physics 1 with Calculus
Academic year:

2024
Views:

48

Pages:

6
Author:

Jabin F.

Born Approximation --> (First Born approximation): - The first Born approximation is used to calculate scattering amplitude f(019) in the scattering if scattering poth is weak and incident particle has large energy, By neglecting re scattering. Ψsc << Ψin or Ψsc << Ψtotal Scattered Angle f(θ,φ) = (-m / 2πħ²) ∫ e^(i(k-k')γ) V(γ) dτ k = k' + kk = k - k' f(θ,φ) = (-m / 2πħ²) ∫∫∫ e^(ikγ cosθ') V(γ) γ² dγ sinθ' dθ' dφ For Spherically symmetric or central pot. V(γ) = V(γ) (independent of θ and φ) f(θ,φ) = (-m / ħ²) ∫ (∫ e^(ikγ cosθ') sinθ' dθ') V(γ) γ² dγ = (-m / ħ²) ∫ (∫ e^(ikγt) dt) V(γ) γ² dγ = (-m / ħ²) ∫ (e^(ikγt) / (ikγ) + 1) V(γ) γ² dγ -m² ∫(e^(ikγ) - e^(-ikγ)) V(γ) γ² dγ K²/4π² 0 ħ²/(2m) Foγ lasting scatteγing: ħ²/2m² -> ħ²/2m²K² K² = K² + k² - 2Kk cosθ |K'|² = |K|² + |k|² - 2|K||k|cosθ |K'|² = 2k²(1-cosθ) K² = 2k²x²sin²(θ/2) K = 2k sin(θ/2) - scatteγing angle f(θ) = -2m/(ħ²K) ∫sin(Kγ) V(γ) γ dγ ↳ foγ centγal pot. V(γ) = V(γ) Pγob. find the scatteγing amplitude and diffeγential cγoss-section by using Boγn appγoximation in the scatteγing of a paγticle of mass m' scatteγed by a scatteγing pot. - V(γ) (i) V(γ) = Voe^(-uγ) | Vo and u > 0 and γeal const. (ii) V(γ) = Voe^(-uγ²) (iii) V(γ) = Vo delta(γ - a), Va and |a| γead as Sol (i) V(γ) = .Voe^(-Mγ)/γ f(θ) = -2m/(ħ²K) ∫sin(Kγ)(Voe^(-Mγ)/γ) dγ = -2mVo/(ħ²K) ∫sin(Kγ)e^(-Mγ)/γ dγ = -2mVo/(ħ²K) ∫(e^(ikγ) - e^(-ikγ))/2i * e^(-Mγ)/γ dγ = -mVo/(iħ²K) ∫(e^(-(M-ik)γ) - e^(-(M+ik)γ)) dγ (γ^0) = -mVo/(ħ²K) (1/(M-ik) - 1/(M+ik)) = -mVo/(ħ²K) ((M+ik - M+ik)/(M²-k²)) f(θ) = -mVo/(ħ²K) (2ik/(M²-k²)) with K = 2ksin(θ/2) dσ/dΩ = |f(θ)|² = 4m²V₀²/ħ⁴(μ² + K²)² σ = ∫∫(dσ/dΩ) sinθ dθ dφ π 2π 0 0 = 2π ∫(4m²V₀²/ħ⁴(μ² + 4k²sin²(θ/2)²)) sinθ dθ 0 (ii) V(γ) = V₀e^(-μγ) central pot f(θ) = -2mV₀/(ħ²K) ∫sin(Kγ)e^(-μγ) dγ ∞ 0 = mV₀/(iħ²K) ∫(e^(-(μ-ik)γ) - e^(-(μ+ik)γ))/γ dγ ∞ 0 = -mV₀/(ħ²K) [1/(μ-ik) - 1/(μ+ik)] = -mV₀/(ħ²K) [(μ+ik - μ+ik)/(μ²-k²)] = -mV₀/(ħ²K) (2ik/(μ²-k²)) f(θ) = -4mV₀μ/(ħ²(μ²+k²)) dσ/dΩ = |f(θ)|² = 16m²V₀²μ²/ħ⁴(μ²+K²)⁴ (iii) V(γ) = V₀δ(γ-a) f(θ,φ) = -m/2πħ² ∫e^(ikγ)V(γ)dγ = -mV₀/2πħ² ∫e^(ikγ)δ(γ-a)dγ f(θ,φ) = -mV₀/(2πħ²)e^(ika) dσ/dΩ = -mV₀²/4π²ħ⁴ π2π -> σ_T = ∫∫(dσ/dΩ)sinθdθdφ 00 σ_T = m²V₀²/πħ⁴