Question 26.4: Decay of the Particle The lambda (Λ) particle (Table 26.4) i...
Decay of the Particle
The lambda (Λ) particle (Table 26.4) is an uncharged, heavy baryon that can decay via the reactions \Lambda \rightarrow p^{+}+\pi^{-} \text {and } \Lambda \rightarrow n+\pi^{0}. Examine both these reactions from the standpoint of conservation laws.
| TABLE 26.4 Table of Hadrons | |||||||||
| Particle name | Symbol | Anti- particle | Rest energy (MeV) | Mean lifetime (s) | Main decay modes | Spin | Baryon number B | Strangeness number S | Charm number C |
| Mesons | |||||||||
| Pion | \pi^{-}\\ \pi^{0} | \pi^{+} Self | 140
135 |
2.6 \times 10^{-8}\\8.4 \times 10^{-17} | \mu^{+} v_{\mu}\\ 2 \gamma | 0
0 |
0
0 |
0
0 |
0
0 |
| Kaon | K ^{+}\\K _{S}^{0}\\K _{L}^{0} | K ^{-}\\\overline{ K }_{S}^{0}\\ \overline{ K }_{L}^{0} | 494
498 498 |
1.2 \times 10^{-8}\\9.0 \times 10^{-11}\\ 5.1 \times 10^{-8} | \mu^{+} \nu_{\mu}, \pi^{+} \pi^{0}\\\pi^{+} \pi^{-}, 2 \pi^{0}\\\pi^{\pm} e^{\mp} \nu_{e}, 3 \pi^{0}\\\pi^{\pm} \mu^{\mp} \nu_{\mu}\pi^{+} \pi^{-} \pi^{0} | 0
0 0 |
0
0 0 |
1
1 1 |
0
0 0 |
| Eta | \eta^{0} | Self | 548 | 5 \times 10^{-19} | 2 \gamma, 3 \pi^{0}\\\pi^{+} \pi^{-} \pi^{0} | 0 | 0 | 0 | 0 |
| Charmed D’s | D ^{+}\\D ^{0}\\D _{S}^{+} | D ^{-} \\ \overline{ D }^{0}\\ \overline{ D }_{S}^{-} | 1870
1865 1968 |
1.0 \times 10^{-12}\\ 4.1 \times 10^{-13}\\ 5.0 \times 10^{-13} | e^{+}, K ^{\pm}, K ^{0} \overline{ K }^{0}+\text { anything }\\\text { Same as } D ^{+} | 0
0 0 |
0
0 0 |
0
0 1 |
1
1 1 |
| Bottom B’s | B ^{+} \\ B ^{0} | B ^{-}\\ \overline{ B }^{0} | 5280
5280 |
1.6 \times 10^{-12}\\ 1.5 \times 10^{-12} | Various | 0
0 |
0
0 |
0
0 |
0
0 |
| J/Psi | J / \psi | Self | 3097 | 10^{-20} | Various | 0 | 0 | 0 | 0 |
| Upsilon | \Upsilon(\text { IS }) | Self | 9460 | 10^{-20} | Various | 0 | 0 | 0 | 0 |
| Baryons | |||||||||
| Proton | p | \bar{p} | 938.3 | Stable (?) | \frac{1}{2} | 1 | 0 | 0 | |
| Neutron | n | \bar{n} | 939.6 | 886 | p e^{-}\bar{\nu}_{e} | \frac{1}{2} | 1 | 0 | |
| Lambda | Λ | \bar{\Lambda} | 1116 | 2.6 \times 10^{-10} | p \pi^{-}, n \pi^{0} | \frac{1}{2} | 1 | 0 | 0 |
| Sigmas | \Sigma^{+} \\ \Sigma^{0} \\ \Sigma^{-} | \bar{\Sigma}^{-}\\ \bar{\Sigma}^{0}\\ \bar{\Sigma}^{+} | 1189
1193 1197 |
8.0 \times 10^{-11}\\ 7.4 \times 10^{-20}\\ 1.5 \times 10^{-10} | p \pi^{0}, n \pi^{+}\\\Lambda \gamma \\ n \pi^{-} | \frac{1}{2}\\ \frac{1}{2}\\ \frac{1}{2} | 1
1 1 |
-1
-1 -1 |
0
0 0 |
| Xi | \Xi^{0}\\ \Xi^{-} | \bar{\Xi}^{0}\\\Xi^{+} | 1315
1321 |
2.9 \times 10^{-10} \\ 1.6 \times 10^{-10} | \Lambda \pi^{0}\\ \Lambda \pi^{-} | \frac{1}{2}\\ \frac{1}{2} | 1
1 |
-2
-2 |
0 |
| Omega | \Omega^{-} | \Omega^{+} | 1672 | 0.82 \times 10^{-10} | \Lambda K ^{-}, \Xi^{0} \pi^{-} | \frac{3}{2} | 1 | -3 | 0 |
| Charmed lambda | \Lambda_{C}^{-} | \bar{\Lambda}_{C}^{-} | 2286 | 2.0 \times 10^{-13} | Various | \frac{1}{2} | 1 | 0 | 1 |
It’s best to consider the conservation laws one at a time. For energy conservation, note that a lambda particle’s mass is larger than the total mass of a proton and pion or neutron and pion. Therefore, a lambda particle has enough rest energy for these decays. Charge is conserved in both decays, because the net charge of the proton and \pi^{-} \text {is } e-e=0 . Similarly, the neutron and neutral pion are both uncharged. Baryon number is conserved in each decay, because the lambda has baryon number 1, and so do the proton and neutron. The pions are mesons, with baryon number zero. Thus the net baryon number is 1 on each side of both reactions. None of the particles involved carry charm, so there’s no issue with conservation of charm. The lambda particle has strangeness -1, but none of the decay products do. Thus, strangeness is not conserved. This indicates that the weak force is involved in this decay, because strangeness need not be conserved in weak-force reactions.
REFLECT Note that these are the only two likely decay modes for the lambda particle. Baryon conservation dictates that the lambda must decay into another baryon, which must have less mass than the lambda in order to conserve energy. This leaves only the proton and neutron. With the rest energy of the proton or neutron committed, the only mesons light enough to form from the leftover energy are pions.