Particle Physics: Question 2

Worked solution

Step 1: Read off what the equation claims

The proposed equation is $u \rightarrow d + e^{-} + \bar\nu_e$ This states that a single up quark converts into a down quark, releasing an electron and an electron antineutrino.

Step 2: The lepton pairing is not the problem

The electron and antineutrino pairing given is genuinely the correct pair of leptons for $\beta^-$ decay (a neutron, $udd$, decaying to a proton, $uud$, releases $e^{-}$ and $\bar\nu_e$, not $e^{+}$ and $\nu_e$). So whatever is wrong with this equation, it is not the choice of leptons.

Step 3: Check charge conservation directly

Using quark and lepton charges in units of the elementary charge $e$: up quark $=+\frac23$, down quark $=-\frac13$, electron $=-1$, antineutrino $=0$.

Left-hand side: $+\frac23$

Right-hand side: $-\frac13 + (-1) + 0 = -\frac43$

$+\frac23 \neq -\frac43$

so charge is not conserved by this equation.

Step 4: Why this happens

The quark change $u \rightarrow d$ is the correct change for $\beta^+$ decay (a proton, $uud$, becoming a neutron, $udd$), not for $\beta^-$ decay (a neutron, $udd$, becoming a proton, $uud$), which instead requires $d \rightarrow u$. Pairing the $\beta^+$ quark change with the $\beta^-$ lepton pair reverses the sign of the quark-side charge without changing the lepton-side charge to match, so the two sides no longer balance.

Step 5: Why the other options are not what fails here

• Nucleon number: exactly one quark converts into one other quark, so a single nucleon (three quarks) remains a single nucleon throughout; nucleon number is unaffected by which flavour change is written.
• Momentum and mass-energy: these govern the actual kinetic energies and directions of motion of the particles produced in a real decay, which cannot be checked from the particle identities written in a symbolic equation, they are not what this wrong quark/lepton pairing violates.

Final answer

• The physical quantity that is not conserved is $\boxed{\text{charge}}$, option C.