Elementary Particles Questions and Answers

Answer: D

The reaction fails on strangeness and the third component of isospin.

Check each quantity, initial against final, for \( \pi^{+} + n \rightarrow K^{0} + p \):

• Charge: \( +1 + 0 = 0 + 1 \), conserved.
• Baryon number: \( 0 + 1 = 0 + 1 \), conserved.
• Strangeness: \( 0 + 0 \ne +1 + 0 \), violated.
• Isospin \( I_{3} \): \( +1 - \frac{1}{2} = +\frac{1}{2} \) against \( -\frac{1}{2} + \frac{1}{2} = 0 \), violated.

Since charge and baryon number are both conserved, every option naming either of them is wrong. A strong reaction must conserve both strangeness and I₃, so it cannot occur.

Answer: D

The particles are Mesons.

Kaons and pions are bound states of a quark and an antiquark, which is the definition of a meson. For example the positive pion is an up quark with an anti-down quark, and the positive kaon is an up quark with an anti-strange quark. All have integer spin and are bosons.

Leptons, such as electrons and neutrinos, are not made of quarks. Quarks are the constituents, not composite particles. Baryons, such as protons and neutrons, contain three quarks.

Answer: D

The quark contents are UUd and Udd.

The proton is up-up-down: charges 2/3 + 2/3 - 1/3 = +1. The neutron is up-down-down: 2/3 - 1/3 - 1/3 = 0. Neither contains a strange quark, which is why their strangeness is zero.

Every option with an S quark is wrong for nucleons. Combinations such as uus are the Sigma-plus baryon and dds the Sigma-minus, both strange particles.

Answer: D

CP violation was observed in the decay of the neutral K meson (printed K°).

In 1964 Christenson, Cronin, Fitch and Turlay found that the long-lived neutral kaon, \(K_{L}^{0}\), which should decay only into three pions if CP were conserved, occasionally decays into two pions \(\pi^{+}\pi^{-}\), a CP-violating mode, about twice in a thousand decays. Cronin and Fitch received the 1980 Nobel Prize for it.

Beta decay was where parity (P) violation was found, by Wu in 1957, but CP appeared conserved there. Alpha decay and heavy-ion decay are strong or electromagnetic processes, which conserve CP.

Answer: A

The reaction violates conservation of baryon number and isospin.

Check each quantity on both sides of \(p + p \rightarrow K^{+} + \Sigma^{+}\):

Charge and strangeness balance, but baryon number falls from 2 to 1 and \(I_{3}\) changes from 1 to 3/2, which a strong interaction cannot do. So options naming charge or strangeness are wrong.

Answer: D

The quark content of \(\Xi^{0}\) is u s s.

A baryon is made of three quarks (not antiquarks). Strangeness -2 requires two strange quarks, each with S = -1. The third quark must give total charge zero:

\[ Q = +\tfrac{2}{3} - \tfrac{1}{3} - \tfrac{1}{3} = 0 \]

so it is an up quark, giving uss.

d s s is the charged \(\Xi^{-}\), with charge \(-\frac{1}{3} - \frac{1}{3} - \frac{1}{3} = -1\). The options with anti-strange quarks have strangeness +2 and contain antiquarks mixed with quarks, which is not a baryon at all.

Answer: C

The reaction cannot occur because Q and \(I_{3}\) are conserved, but B and S are not.

Tabulate the additive quantum numbers (the paper writes the pion as \(\Pi\)):

Charge and \(I_{3}\) balance, but the baryon number falls from 1 to 0 and strangeness rises from 0 to 1. Baryon number is absolutely conserved, so the reaction cannot be induced, and strangeness is also conserved in strong interactions. The other options misstate which quantities fail.

Answer: C

Parity is not conserved in weak interactions.

In 1956 T.D. Lee and C.N. Yang pointed out that there was no experimental evidence for parity conservation in weak processes, and in 1957 C.S. Wu's experiment on the beta decay of polarised cobalt-60 showed that it is violated. The weak interaction acts only on left-handed particles (and right-handed antiparticles), so its mirror image is not a possible process.

The strong and electromagnetic interactions conserve parity to high precision, which is why parity is a good quantum number for nuclear and atomic states. Gravity also respects parity, so the weak interaction is the only one of the four that violates it.

Answer: A

The quark content is \(d\bar{s}\).

Strangeness is \(-1\) for each \(s\) quark and \(+1\) for each \(\bar{s}\) antiquark. A \(K^{0}\) with \(S = +1\) must contain an \(\bar{s}\); being neutral, it pairs it with a \(d\) quark (charge \(-\frac{1}{3} + \frac{1}{3} = 0\)). The \(d\) gives isospin \(\frac{1}{2}\), and \(K^{0}\) and \(K^{+} = u\bar{s}\) form an isospin doublet.

\(\bar{d}s\) is the \(\bar{K}^{0}\), with \(S = -1\). \(d\bar{d}\) and \(u\bar{u}\) have zero strangeness and appear in the \(\pi^{0}\) and \(\eta\).

Answer: D

The lifetimes are about 10^-23 s, 10^-18 s and 10^-10 s.

Decay rates scale with the strength of the interaction responsible:

• Strong (hadronic) decays, such as resonances like the \(\Delta\), take about the time for light to cross a nucleon, \(10^{-23}\) s.
• Electromagnetic decays, such as \(\pi^{0} \to \gamma\gamma\) (about \(10^{-17}\) s) and \(\Sigma^{0} \to \Lambda\gamma\), are around \(10^{-20}\) to \(10^{-16}\) s.
• Weak decays, such as \(\Lambda \to p\pi^{-}\) (about \(10^{-10}\) s), are much slower.

The other options give lifetimes far too long for strong and electromagnetic decays.