# Why can we make those approximations regarding the Big Bang Nucleosynthesis (BBN)?

(About neutron decoupling) Consider $$p+e^{-}\rightarrow n+\nu_{e}.$$ In particular if we consider temperatures about $$\simeq1\mbox{MeV}$$ both neutrons and protons are non relativistic and we can find the neutron to proton ratio as $$\frac{n_{n}}{n_{p}}=\frac{g_{n}}{g_{p}}\left(\frac{m_{n}}{m_{p}}\right)^{3/2}\exp\left(\frac{\mu_{e}-\mu_{\nu}}{T}\right)\exp\left[-\frac{Δ m}{T}\right]$$ Then one usually neglects the exponential with the chemical potentials since $$\frac{\mu_{e}}{T}\simeq\frac{n_{e}}{n_{\gamma}} \simeq \frac{n_{p}}{n_{\gamma}}=\eta_{B}\ll1$$ My first question is about the last equation: Why can we say that? In particular where does the first passage come from?

(About nuclei formation) To study the deuteron formation $$p+n\rightarrow d$$ one usually consider the Saha equation $$\frac{n_{d}}{n_{p}n_{n}}\simeq\frac{3}{4}\left(\frac{4\pi}{m_{p}T}\right)^{3/2}\exp\left(\frac{B_{d}}{T}\right)$$ where we used $$g_{p}=g_{n}=2,g_{d}=3$$ and $$m_{d}\simeq2m_{p}$$. At this point one usually say that, if $$T_{nuc}$$ is such that $$n_{d}\simeq n_{p}$$, it reduces to (neglecting numerical factors) $$1\simeq\eta_{B}n_{\gamma}\left(\frac{1}{m_{p}T_{nuc}}\right)^{3/2}\exp\left[\frac{B_{d}}{T_{nuc}}\right]\simeq\eta_{B}\left(\frac{T_{nuc}}{m_{p}}\right)^{3/2}\exp\left(\frac{B_{d}}{T_{nuc}}\right)$$ Well I don't understand why we can approximate the baryon density with the neutron density, can someone help me?