Question 10.7: Using the elastic scattering formulae In an experiment, part...
Using the elastic scattering formulae
In an experiment, particles of mass m and energy E are used to bombard stationary target particles of mass 2m.
Q. The experimenters wish to select particles that, after scattering, have energy E/3. At what scattering angle will they find such particles?
A. If E_{1} / E_{0}=1 / 3, then by energy conservation E_{2} / E_{0}=2 / 3 . First use formula D to find ψ. Since the mass ratio γ = 1/2, this gives
\frac{2}{3}=\frac{8}{9} \sin ^{2}\left(\frac{1}{2} \psi\right),
so that ψ = 120°. Now use formula A to find the scattering angle \theta_{1} . This gives tan \theta_{1}=\infty \text { so that } \theta_{1}=90^{\circ} . Particles scattered with energy E/3 will therefore be found emerging at right angles to the incident beam.
Q. In one collision, the opening angle was measured to be 45°. What were the individual scattering and recoil angles?
A. First use formula C to find ψ. This gives
\cot \left(\frac{1}{2} \psi\right)=\frac{1}{3} ,
so that \frac{1}{2} \psi=72^{\circ} , to the nearest degree. Now use formula B to find the recoil angle \theta_{2} \text {. This gives } \theta_{2}=90^{\circ}-72^{\circ}=18^{\circ} . The scattering angle \theta_{1} must therefore be \theta_{1}=\theta-\theta_{2}=45^{\circ}-18^{\circ}=27^{\circ} .
Q. In another collision, the scattering angle was measured to be 45°. What was the recoil angle?
A. First use formula A to find ψ. This shows that ψ satisfies the equation
2 \cos \psi-2 \sin \psi=1,
which can be written in the form*
\sqrt{8} \cos \left(\psi+45^{\circ}\right)=1.
This gives ψ = 24°, to the nearest degree. Formula B now gives the recoil angle \theta_{2} to be \theta_{2}=78^{\circ} , to the nearest degree.