Particle Physics: Question 5

Worked solution

Part (a): The nuclide equation for $\beta^-$ decay

In $\beta^-$ decay, a neutron in the nucleus transforms into a proton, releasing an electron (the beta particle) and an electron antineutrino. Since a proton is gained, the proton number of the daughter nuclide is one more than that of the parent, while the nucleon number is unchanged: $^{24}_{11}\text{Na} \rightarrow {}^{24}_{12}\text{Mg} + {}^{0}_{-1}e + {}^{0}_{0}\bar\nu_e$

Here, the beta particle is written as $^{0}_{-1}e$ (zero nucleon number, charge $-1$) and the antineutrino as $^{0}_{0}\bar\nu_e$ (zero nucleon number, zero charge).

Part (b): Proton number, nucleon number and conservation

The daughter nuclide of magnesium has proton number $12$ and nucleon number $24$.

Checking nucleon number conservation (top numbers): $24 = 24 + 0 + 0 \checkmark$

Checking charge/proton number conservation (bottom numbers): $11 = 12 + (-1) + 0 \checkmark$

Both the total nucleon number and the total charge are the same before and after the decay, so both conservation laws hold.

Part (c): The quark-level change and its charge conservation

A neutron has quark composition $udd$ and a proton has quark composition $uud$. Since the decay converts a neutron into a proton, exactly one down quark must change into an up quark: $d \rightarrow u + e^{-} + \bar\nu_e$

Checking charge conservation, using quark and lepton charges in units of $e$ (down quark $=-\frac13$, up quark $=+\frac23$, electron $=-1$, antineutrino $=0$):

Left-hand side: $-\frac13$

Right-hand side: $+\frac23 + (-1) + 0 = -\frac13$

The left- and right-hand sides are equal, so electric charge is conserved in this quark-level equation, consistent with the nuclide-level equation found in part (a).

Final answers

• (a) $^{24}_{11}\text{Na} \rightarrow {}^{24}_{12}\text{Mg} + {}^{0}_{-1}e + {}^{0}_{0}\bar\nu_e$
• (b) Daughter nuclide: proton number $= \boxed{12}$, nucleon number $= \boxed{24}$; nucleon number ($24=24$) and charge ($11=12-1$) are both conserved.
• (c) $d \rightarrow u + e^{-} + \bar\nu_e$; charge conserved since $-\frac13 = \frac23 - 1$.