2
$\begingroup$

Since a neutron and a proton are made up of quarks and an electron is a lepton, how can a neutron yield an electron?

$\endgroup$

2 Answers 2

5
$\begingroup$

A neutron is made of 3 quarks, two down quarks and one up. The process you are talking about is called beta decay. It is a weak nuclear interaction the can be summarized like this:

$$(u+d+d) \rightarrow (u+d+u) + e^- + \bar{\nu}$$

One of the down quarks ($d$) decays producing an up quark ($u$), an electron ($e^-$), and an anti-neutrino ($\bar\nu$). The final baryon state has two up quarks and one down, that's a proton. This process follows a few conservation laws.

  • conservation of baryon number: $1 \rightarrow 1$, Each quark counts for $1/3$ of a baryon number. The three-quark proton and neutron are each $1$ baryon. This maintains the same number of quarks on each side of the arrow.
  • conservation of lepton number $0 \rightarrow +1 -1$. A neutrino is also a lepton, and an anti-particle counts as $-1$ of it's type. So the electron and anti-neutrino add up to zero total leptons on the right-hand-side.
  • conservation of charge $0 \rightarrow +1 -1 +0$. The $(udd)$ neutron and the anti-neutrino have zero charge. The $(uud)$ proton has $+1$ charge and the electron has $-1$, so the charge is zero on both sides of the arrow.
  • conservation of energy $E_n \rightarrow E_p + E_e + E_\nu$. The total energy of the particles on each side is the same.

Energy conservation is a bit more that just that. All of the particles have mass, so they have energy following following $E=mc^2$. Additionally, they could have kinetic energy if they are moving.

If the initial neutron is at rest, then its energy is $E_n=(1.008664\,\mathrm{u})c^2 = 939.6$ MeV, where $\mathrm{u}$ is the atomic mass unit and MeV is a mega-electron-volt, a unit of energy. A proton at rest has an energy of $E_p = 938.3$ MeV, and an electron at rest has $E_e = 0.5$ MeV. Nobody knows the rest mass of a neutrino, but it's at least a million times less than an electron. We'll call it zero, even though it's not...

When we put this together assuming everything is at rest, we get: $$ 939.6\,\mathrm{MeV} \rightarrow 938.3\,\mathrm{MeV} + 0.5\,\mathrm{MeV} + 0\,\mathrm{MeV} $$ But wait, the two sides aren't equal. The right-hand-side is missing $0.8$ MeV. To conserve energy the new particles must be moving, so they have a little bit of kinetic energy to make up the mass-Energy difference.

$\endgroup$
2
  • $\begingroup$ good point. I'll edit to address conservation of mass/energy too $\endgroup$
    – Paul T.
    Commented Oct 12, 2019 at 12:19
  • $\begingroup$ First answer has a typo in that the structure of the neutron and proton are the other way around A neutron is made of 3 quarks, but it is two down quarks and one up A proton is made of 3 quarks, two up quarks and one down Then you need to change the equation appropriately (u+d+d)→(u+d+u)+e−+ν¯ here is a Q and A from Fermilab that partly covers the question fnal.gov/pub/science/inquiring/questions/antineutron.html $\endgroup$
    – From Geoff
    Commented Mar 26, 2023 at 18:43
2
$\begingroup$

A nuclear decay is not something that already exists being pulled out of a particle. It is the creation of new particles. As long as the transition respects all of the conservation laws there is a probability of it happening.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.