Conservation laws determine whether an interaction is allowed

Quarks, Leptons & Conservation Laws — AQA A-Level Physics

All particle interactions must conserve: chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). ($Q$), baryon numberA quantum number conserved in all interactions. Baryons (protons, neutrons) have baryon number +1, antibaryons have -1, and mesons/leptons have 0. ($B$), lepton numberA quantum number conserved in all interactions. Leptons (electrons, neutrinos) have lepton number +1, antileptons have -1, and hadrons have 0. ($L$), energyThe capacity to do work. Measured in joules (J). and momentum.
Strangeness ($S$) must be conserved in strong and electromagnetic interactions.
Strangeness does not need to be conserved in weak interactions — it can change by $0$, $+1$ or $-1$.
If even one of $Q$, $B$ or $L$ is not conserved, the interaction is forbidden.

How to check conservation

Write out $Q$, $B$, $L$ and $S$ for every particle on both sides of the equation.
Sum each quantum number on the left. Sum each on the right. They must be equal.
If $S$ is not conserved, the interaction must be a weak interaction (or it is forbidden).
If $B$ or $L$ is not conserved, the interaction is forbidden — there is no exception.

Worked Example

The $\Lambda^{0}$ particle has quark composition $uds$. Determine whether $\Lambda^{0} \rightarrow \pi^{-} + e^{+} + \nu_e$ is permitted.

Show Solution

Check charge

LHS: $\Lambda^{0}$ has $Q = 0$.

RHS: $\pi^{-}$ has $Q = -1$, $e^{+}$ has $Q = +1$, $\nu_e$ has $Q = 0$.

$-1 + 1 + 0 = 0$. ChargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). is conserved.

Check strangeness

$\Lambda^{0}$ contains an $s$ quark, so $S = -1$.

$\pi^{-}$, $e^{+}$, $\nu_e$ all have $S = 0$.

$-1 \neq 0$. Strangeness is not conserved — this could be a weak interaction.

Check baryon number

$\Lambda^{0}$ is a baryon: $B = +1$.

$\pi^{-}$ is a meson: $B = 0$. $e^{+}$: $B = 0$. $\nu_e$: $B = 0$.

$+1 \neq 0$. Baryon numberA quantum number conserved in all interactions. Baryons (protons, neutrons) have baryon number +1, antibaryons have -1, and mesons/leptons have 0. is not conserved.

Conclusion

Baryon number is not conserved. This interaction is not permitted.

Answer

The interaction is forbidden because baryon number is not conserved.

Worked Example

Using the quark model, prove that charge is conserved in $\beta^{-}$ decay: $n \rightarrow p + e^{-} + \bar{\nu}_e$.

Show Solution

Write quark compositions

Neutron = $udd$. Proton = $uud$.

Calculate charge on the left

$$Q_{\text{LHS}} = +\frac{2}{3} - \frac{1}{3} - \frac{1}{3} = 0$$

Calculate charge on the right

Proton: $+\frac{2}{3} + \frac{2}{3} - \frac{1}{3} = +1$

Electron: $-1$. Anti-neutrino: $0$.

$$Q_{\text{RHS}} = +1 + (-1) + 0 = 0$$

Conclusion

$Q_{\text{LHS}} = Q_{\text{RHS}} = 0$. Charge is conserved.

Answer

Both sides have total $charge = 0$, so charge is conserved.

Examiner Tips and Tricks

Quantum numbers for exotic particles (kaons, lambdas, sigmas) will be given in the question.
You need to know the quantum numbers for protons, neutrons, electrons, neutrinos and pions from the data sheet.

Related:Particle Interactions