91Ó°ÊÓ

At very high temperatures (as in the very early universe), the proton and the neutron can be thought of as two different states of the same particle, called the "nucleon." (The reactions that convert a proton to a neutron or vice versa require the absorption of an electron or a positron or a neutrino, but all of these particles tend to be very abundant at sufficiently high temperatures.) Since the neutron's mass is higher than the proton's by 2.3 x 10^-30 kg, its energy is higher by this amount times c². Suppose, then, that at some very early time, the nucleons were in thermal equilibrium with the rest of the universe at 10¹¹ K.

The neutron and the proton are two states of the nucleon in the early cosmos, and the energies of the proton and neutron are:

$E_p=m_p{c}^2{E}_n=m_n{c}^2$

The probabilities of two states are:

$$\begin{gathered} P_p=\frac{1}{Z}{e}^{-E_p/kT}{P}_n=\frac{1}{Z}{e}^{-E_n/kT} \\ {P}_p=\frac{1}{Z}{e}^{-m_p{c}^2/kT}{P}_n=\frac{1}{Z}{e}^{-m_n{c}^2/kT} \end{gathered}$$

The ratio of the probabilities is:

$\frac{P_n}{P_p}=\frac{e^{-m_n{c}^2/kT}}{e^{-m_p{c}^2/kT}}=e^{-\mathrm{Δ}m{c}^2/kT}$

Where $∆m$is the mass difference between neutron and proton.

$\frac{P_n}{P_p}=e^{-\left(2.3×{10}^{-30}\mathrm{kg}\right){\left(3.0×{10}^8\mathrm{m}/\mathrm{s}\right)}^2/\left(1.38×{10}^{-23}\mathrm{J}/\mathrm{K}\right)\left({10}^{11}\mathrm{K}\right)}$

$\frac{P_n}{P_p}=0.86$

This means that for every 100 protons, there are 86 neutrons; the total number is 100+86 =186; the neutron fraction is:

$$\begin{gathered} f_n=\frac{86}{100+86}=0.462 \\ {f}_n=0.462 \end{gathered}$$

Fraction of proton is:

$$\begin{gathered} f_p=\frac{100}{100+86}=0.538 \\ {f}_p=0.538 \end{gathered}$$