 Solution

ID: past-exam-of-the-particle-physics-course-of-the-university-of-oxford/2022/4/b/iii/solution

For a thickness

x

ρ_{n}

of scattering centres, the mean count is

N = ρ_{n} x σ

. Using the total cross section in i and the variance addition in ii,

RMS (θ)^{2} ≃ ρ_{n} x \frac{16 π K}{p ^{2} θ _{m i n}^{2}} 2 θ_{m i n}^{2} ℓ = \frac{32 π K ρ _{n} x ℓ}{p ^{2}} .

Thus

RMS (θ) ≃ 32 π K ρ_{n} ln (θ_{m a x} / θ_{m i n}) \frac{x}{p} .

The stated assumptions make the prefactor independent of

x

and approximately independent of

p

. The lower-cutoff factors cancel except inside the slowly varying logarithm. Treating

N

as a Poisson random variable rather than a fixed count gives the same result after replacing

N

by its mean.

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