Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 2 / 9E / c / Solution

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} .

Neglecting the neutron--proton mass difference,

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} .

 Ancestors (11)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version