Given:
Z1=2,Z2=79,t=30⋅10−6 m,ρ=19300 km/m3,M=197⋅10−3 kg/mol,θ1=1.5°,θ2=2.5°,θ3=2°,θ4=10°.
Solution
The fraction of particles scattered at angle θ is
f(θ)=πnt(8πϵ0KZ1Z2e2)2cot2(2θ).
In this equation, t - thickness, K - kinetic energy, n is the number of atoms per unit volume of material (gold):
n=MρNA=5.9⋅1028 atoms/m3, where M is the molar mass of Gold and ρ is its density.
i) Between 1.5° and 2.5° the fraction will be
f(θ1)−f(θ2)=πnt(8πϵ0KZ1Z2e2)2[cot2(2θ1)−cot2(2θ2)]= =0.0442.ii) Now it is much easier to find the ratio of particles scattered through angles greater than 2° to the number scattered through angles greater than 10°:
f(θ4)f(θ3)=cot22θ4cot22θ3=25.
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