### Video Transcript

Which of the following equations
shows a particle interaction that would violate the conservation of lepton
number? (A) A gamma-ray photon produces a
muon and an antimuon. (B) A muon decays into an electron,
an electron antineutrino, and a mu neutrino. (C) Carbon-14 decays into
nitrogen-14 and an electron. (D) A tauon and an antitauon
combine to form a gamma-ray photon. And (E) oxygen-15 decays into
nitrogen-15 plus a positron plus an electron neutrino.

Considering these five
interactions, we want to identify which one would violate the conservation of lepton
number. Lepton number, we can recall, is a
property of elementary particles. If a given particle is classified
as a lepton, then that means its lepton number is positive one. The particles that make up this
class are the electron, the muon, the tauon, and three different types of
neutrino. Now, on the other hand, any
antiparticles of leptons, that is, antileptons, have a lepton number of negative
one. These particles include the
positron, the antimuon, the antitauon, and then the electron antineutrino, the mu
antineutrino, and the tau antineutrino.

Of course, it is possible for a
particle to be neither a lepton nor an antilepton. In that case, its lepton number is
simply zero. This list includes particles such
as photons, neutrons, and protons. All of these have a lepton number
of zero. Any particle that we could come up
with then has a lepton number of plus or minus one or zero. And this fits in to what’s called
the conservation of lepton number. This conservation is a physical law
that says that in any nuclear reaction, the total lepton number before the
interaction must equal the total lepton number after it.

So to figure out which of our five
answer options shows a particle interaction that would violate this conservation,
what we’ll need to do is, for each interaction, figure out the total lepton number
on the left- and on the right-hand sides. Let’s start with this first
interaction, a gamma-ray photon producing a muon and an antimuon. Since a photon is neither a lepton
nor an antilepton, it has a lepton number of zero, while on the product side, a
muon, being a lepton, has a lepton number of positive one and an anti muon has a
lepton number of negative one. One minus one is zero. So we can say the total lepton
number on either side of this interaction is the same, and therefore it’s
conserved.

This first interaction then would
not violate the conservation of lepton number. Considering the next interaction,
here we have a muon, which we see is a lepton and so it has a lepton number of
positive one, decaying into an electron, which also has a lepton number of positive
one, then an electron antineutrino, which as an antilepton has a lepton number of
negative one, plus a mu neutrino, which we can see from our list is a lepton and
therefore has a lepton number of positive one. Looking at all the numbers on the
product side, we have one minus one plus one. That adds up to one. And so, once again, we have a
conservation of lepton number in this interaction.

Now, let’s look at option (C) with
carbon-14 decaying into nitrogen-14 and an electron. This interaction involves two
atomic nuclei, carbon and nitrogen, without their electrons. And so, other than this electron
here, all we’re considering are the protons and neutrons that make up carbon-14 and
nitrogen-14. We’ve seen that both protons and
neutrons are neither leptons nor antileptons and therefore have a lepton number of
zero. So that means our entire carbon
nucleus has a lepton number of zero, as does our nitrogen nucleus.

This is because the particles that
make up these nuclei all have lepton numbers of zero. An electron, on the other hand, is
a lepton and therefore has a lepton number of positive one. And this shows us that, in this
potential interaction, we do not have a conservation of lepton number. Therefore, this interaction
violates that law. Let’s continue on and see if any of
the remaining interactions also violate this conservation law.

In our next interaction, we have a
tauon combining with an antitauon to form a gamma-ray photon. And from our table, we see that a
tauon, being a lepton, has a lepton number of positive one, while an antitauon has a
lepton number of negative one. And then a photon has a lepton
number of zero. One minus one equals zero. So this interaction does not
violate the conservation of lepton number.

Lastly, considering interaction
(E), here we have oxygen-15 decaying into nitrogen-15 plus a positron plus an
electron neutrino. Once again, the atomic symbols we
see, oxygen and nitrogen, refer to nuclei of those atoms; no electrons included. Therefore, they’re comprised
entirely of protons and neutrons. And so each have a total lepton
number of zero.

Our positron though, being an
antilepton, has a lepton number of negative one, while the electron neutrino, being
a lepton, has a lepton number of positive one. On the product side of this
interaction then, we have zero minus one plus one, which is simply zero. So this interaction does not
violate the conservation of lepton number. So the only interaction that does
violate this conservation is carbon-14 decaying into nitrogen-14 plus an
electron. And the fact that this interaction
would violate the conservation of lepton number means that it’s not a possible
interaction. That is, carbon-14 can’t decay into
just a nitrogen-14 nucleus and an electron. As we’ve seen, that’s because it
violates lepton number conservation.