Math Exploration 16.1

1. Proton—Neutron Equilibrium

 

We have learned that protons and neutrons were able to easily interconvert in the hot, early Universe. Our goal is to determine the energy and corresponding temperature at which random collisions in a gas will be able to convert a proton to a neutron. The reaction can be written as:

 

\begin{equation}\textsf{p} + e^{-} +\Delta E \rightarrow  \textsf{n} + \nu_e\end{equation}

 

The particle symbols are: p (proton), e^- (electron), n (neutron), and ν_e (electron neutrino). The ΔE in the equation represents the energy required to convert one of the up quarks in a proton into a down quark, which converts the proton to a neutron. Essentially, what the collision must do is provide the mass difference between the proton and neutron, plus whatever additional kinetic energies are in the particles on the right. This energy can be provided by collisions within the hot gas.

 

a. The mass of a proton is 1.672621 × 10^-27 kg. The mass of the electron is 9.109382 × 10^-31 kg. The neutron mass, 1.6749273 × 10^-27 kg, is slightly larger than the sum of the proton and electron masses. Calculate the difference in mass, Δm.

• Given: m_p = 1.672621e-27 kg, 9.109382e-31 kg, m_n = 1.6749273e-27 kg
• Find: Δm
• Concept: Δm = m_n – (m_p + m_e)
• Solution: Δm = 1.674927351e-27 kg – (1.67262177e-27 kg + 9.109382e-31 kg) = 1.395e-30 kg

 

b. What is the (minimum) temperature at which it was possible to turn protons into neutrons?

• Given: Δm = 1.395e-30
• Find: T
• Concept: T ~ Δmc²/k
• Solution: T ~ (1.395e-30 kg)(3e8 m/s)²/(1.38e-23 J/K) = 9.1e9 K

 

So when the Universe had a temperature of about 9 billion kelvin or higher, it was possible to constantly turn protons into neutrons, and there were approximately equal numbers of the two types of particles.

 

We do not have to worry about converting a neutron to a proton; that happens spontaneously because the proton has less mass; energy is released when the neutron decays to a proton.