 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-2/9e/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Once deuterium survives photodissociation, nuclear reactions rapidly assemble nearly every available neutron into tightly bound helium-4. Protons are more abundant, so neutrons are the limiting constituent. If

r = n_{n} / n_{p} = 1/7

, then

n_{He} ≃ \frac{n _{n}}{2} .

Y_{p} = \frac{ρ _{He}}{ρ _{B}} ≃ \frac{4 m _{p} ( n _{n} /2 )}{m _{p} ( n _{n} + n _{p} )} = \frac{2 r}{1 + r} .

Thus

Y_{p} ≃ \frac{1}{4} .

 New to topics? Read the docs here!