The Helium mass fraction from the Big Bang Nucleosynthesis

Question

In Perkin's book Particle Astrophysics (page 144): I do not understand how one comes to the following expression (the second equality with $r$) for the Helium mass fraction due to the Big Bang Nucleosynthesis:

$$Y= \frac{4N_\text{He}}{4N_\text{He}+N_\text{H}}= \frac{2r}{1+r} $$ where $r=N_\text{n}/N_\text{p}$.

The first equality follows from the fact that He is (approximately) 4 times heavier than H: $$Y = \frac{m_\text{He}}{m_\text{He}+m_\text{H}}= \frac{4N_\text{He}}{4N_\text{He}+N_\text{H}}.$$

However I can't derive the second equality relating $Y$ to $r$:

$N_\text{He}= 2N_\text{p} + 2N_\text{n}$ and $N_\text{H}= N_\text{p} + N_\text{n}$

$$Y= \frac{4N_\text{He}}{4N_\text{He}+N_\text{H}}= \frac{8(N_\text{n}+N_\text{p})}{9(N_\text{n}+N_\text{p})}\quad???$$

Answer 1

The problem is you have the wrong relations between $\{N_\mathrm{H}, N_\mathrm{He}\}$ and $\{N_\mathrm{p}, N_\mathrm{n}\}$. Every hydrogen contains 1 proton, and every helium contains 2, so $N_\mathrm{p} = N_\mathrm{H} + 2 N_\mathrm{He}$. The neutrons are only contributed to by helium in the accounting: $N_\mathrm{n} = 2 N_\mathrm{He}$. Inverting these relations yields \begin{align} N_\mathrm{H} & = N_\mathrm{p} - N_\mathrm{n} \\ N_\mathrm{He} & = \frac{1}{2} N_\mathrm{n}. \end{align}

It looks like you got the direction wrong, in the sense that there should be fewer helium nuclei than protons or neutrons (the 2's are on the wrong side). Also, "hydrogen" means ${}^1\mathrm{H}$ not ${}^2\mathrm{D}$ unless otherwise stated.