 Solution

ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2013-resit/9/b/solution

Let

p

be the incident momentum and

q_{1}, q_{2}

the outgoing magnitudes in the laboratory frame. Transverse conservation of momentum gives

(q_{1} - q_{2}) sin θ = 0

. For the nondegenerate symmetric scattering,

q_{1} = q_{2} = q

, and the two outgoing energies are equal,

ϵ = (E + E_{0}) /2

. Then

p = 2 q cos θ, p^{2} c^{2} = E^{2} - E_{0}^{2}, q^{2} c^{2} = \frac{( E + E _{0} ) ^{2}}{4} - E_{0}^{2} .

Combining these mass shell relations yields

cos^{2} θ = \frac{E + E _{0}}{E + 3 E _{0}}, tan^{2} θ = \frac{2 E _{0}}{E + E _{0}} .

Here

E

includes the incident electron's rest energy. As

E ↓ E_{0}

the limiting angle is

4 5^{\circ}

, consistent with the nonrelativistic equal-mass elastic collision; at large

E / E_{0}

the outgoing tracks become forward directed.

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