Question

A particle of mass M is elastically scattered from a stationary proton of mass m. The proton is projected at an angle ϕ = 22.1^◦ while the incident particle is scattered through an angle θ = 5.6^◦ with the incident direction. Calculate M in atomic mass units. (This event was recorded in photographic emulsions in the Wills Lab. Bristol).

Answer 1

We shall find an expression for the ratio of the masses M/m in terms of the angles θ and ϕ. To this end we start with the equation for the transformation of angle from CMS to LS. 

tan θ = sin θ^∗/(cosθ^∗ + M/m) (1) 

θ^∗ = π − ϕ^∗ = π − 2ϕ

(because m and M are oppositely directed in the CMS, and the recoil angle of the proton in the CMS is always twice the angle in the LS) 

Therefore sinθ^∗ = sin(π − 2ϕ) = sin2ϕ 

and cosθ^∗ = cos(π − 2ϕ) = − cos2ϕ 

Equation (1) then becomes 

tanθ = sinθ/ cosθ = sin2ϕ/(M/m − cos 2ϕ) 

Cross multiplying the second equation and re-arranging 

(M/m) sinθ = sinθcos2ϕ + cosθsin2ϕ = sin(θ + 2ϕ) 

M/m = sin(θ + 2ϕ)/ sinθ 

Using θ = 5.6⁰ and ϕ = 22.1⁰, we find M = 7.8 m ≈ 7.8 amu.