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Expert-verified Found in: Page 228 ### An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279 # 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. What fraction of the nucleons at that time were protons, and what fraction were neutrons?

The fraction of neutron is 0.462 and

The fraction of neutron is 0.538.

See the step by step solution

## Step 1: Given information

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.

## Step 2: Explanation

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:

${P}_{p}=\frac{1}{Z}{e}^{-{E}_{p}/kT}{P}_{n}=\frac{1}{Z}{e}^{-{E}_{n}/kT}\phantom{\rule{0ex}{0ex}}{P}_{p}=\frac{1}{Z}{e}^{-{m}_{p}{c}^{2}/kT}{P}_{n}=\frac{1}{Z}{e}^{-{m}_{n}{c}^{2}/kT}$

The ratio of the probabilities is:

$\frac{{P}_{n}}{{P}_{p}}=\frac{{e}^{-{m}_{n}{c}^{2}/kT}}{{e}^{-{m}_{p}{c}^{2}/kT}}={e}^{-\Delta m{c}^{2}/kT}$

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:

${f}_{n}=\frac{86}{100+86}=0.462\phantom{\rule{0ex}{0ex}}{f}_{n}=0.462$

Fraction of proton is:

${f}_{p}=\frac{100}{100+86}=0.538\phantom{\rule{0ex}{0ex}}{f}_{p}=0.538$ ### Want to see more solutions like these? 