Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 53 / 2 / i / b / Solution

At fixed expansion and conversion rates, a smaller neutron half-life reduces ${\tau }_n$ and hence $X_n(t_N)=X_{n,f}{e}^{-(t_N-t_f)/{\tau }_n}$. Fewer neutrons survive, so less helium-4 forms. If the reduced lifetime arises from a stronger weak coupling, its accompanying faster neutron-proton conversion also delays freeze-out and reinforces this decrease.