Elementary Particles Questions and Answers

Q: The following nuclear reaction\[ \pi^{+} + n \rightarrow K^{\circ} + P \]is examined on the basis of conservation laws of charge, Baryon number, strangeness and third component of Isospin. It is observed that the reaction cannot be induced due to non-conservation of the :

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.

Q: The following particles\(K^{+}, K^{-}, \pi^{+}, \pi^{0}, \pi^{-}, K^{0}\)are broadly classified as :

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.

Q: Considering U, d, S Quarks, the quark content of proton and neutron are respectively :

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.

Q: The violation of CP invariance theorem was observed in the :

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.

Q: The following nuclear reaction\(\text{P} + \text{P} \rightarrow \text{K}^{+} + \Sigma^{+}\)is not allowed due to non-conservation of :

The reaction violates conservation of baryon number and isospin.Check each quantity on both sides of \(p + p \rightarrow K^{+} + \Sigma^{+}\):Quantityp + pK+ + Sigma+Charge Q+2+1 + 1 = +2Baryon number B20 + 1 = 1Strangeness S0+1 - 1 = 0Isospin \(I_{3}\)+1+1/2 + 1 = +3/2Charge 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.

Q: The quark content of baryon \(\Xi^{0}\) with strangeness S = -2 is given by :

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.

Q: Based on the additive quantum numbers such as strangeness (S), baryon number (B), charge (Q), and isospin (I), indicate for which among the four options given below, the following nuclear reaction cannot be induced\[ \Pi^{+} + n \rightarrow \Pi^{o} + K^{+} \]

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 \(\Pi\)):QBSI3π+ + n1 + 0 = 10 + 1 = 101 - 1/2 = 1/2π0 + K+0 + 1 = 10 + 0 = 00 + 1 = 10 + 1/2 = 1/2Charge 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.

Q: Parity is not conserved in :

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.

Q: The quark content of the subatomic particle K° with strangeness S = +1 and isospin I =1/2 is given by :

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 \(\eta\).

Q: What are the appropriate values for the corresponding lifetimes of hadronic decay, electromagnetic decay and weak decay ?

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.

11.

The following nuclear reaction

\[ \pi^{+} + n \rightarrow K^{\circ} + P \]

is examined on the basis of conservation laws of charge, Baryon number, strangeness and third component of Isospin. It is observed that the reaction cannot be induced due to non-conservation of the :

A.Charge and strangeness
B.Baryon number and charge
C.Third component of Isospin and Baryon number
D.Strangeness and third component of Isospin

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

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.

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12.

The following particles

\(K^{+}, K^{-}, \pi^{+}, \pi^{0}, \pi^{-}, K^{0}\)

are broadly classified as :

A.Leptons
B.Quarks
C.Baryons
D.Mesons

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

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.

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13.

Considering U, d, S Quarks, the quark content of proton and neutron are respectively :

A.UUS and ddS
B.UdS and dSS
C.UUS and USS
D.UUd and Udd

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

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.

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14.

The violation of CP invariance theorem was observed in the :

A.Decay of beta particles
B.Decay of alpha particles
C.Decay of heavy ions
D.Decay of K° meson

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

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.

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15.

The following nuclear reaction

\(\text{P} + \text{P} \rightarrow \text{K}^{+} + \Sigma^{+}\)

is not allowed due to non-conservation of :

A.Baryon number and isospin
B.Strangeness and baryon number
C.Charge and strangeness
D.Isospin and charge

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The reaction violates conservation of baryon number and isospin.

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

Quantity	p + p	K+ + Sigma+
Charge Q	+2	+1 + 1 = +2
Baryon number B	2	0 + 1 = 1
Strangeness S	0	+1 - 1 = 0
Isospin \(I_{3}\)	+1	+1/2 + 1 = +3/2

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.

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16.

The quark content of baryon \(\Xi^{0}\) with strangeness S = -2 is given by :

A.\(u\bar{s}\bar{s}\)
B.\(d\,s\,s\)
C.\(d\bar{s}\bar{s}\)
D.\(u\,s\,s\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

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.

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17.

Based on the additive quantum numbers such as strangeness (S), baryon number (B), charge (Q), and isospin (I), indicate for which among the four options given below, the following nuclear reaction cannot be induced

\[ \Pi^{+} + n \rightarrow \Pi^{o} + K^{+} \]

A.Q, B, S are conserved, but \(I_{3}\) is not conserved
B.Q and B are conserved, but S and \(I_{3}\) are not conserved
C.Q and \(I_{3}\) are conserved, but B and S are not conserved
D.B, S and \(I_{3}\) are conserved, but Q is not conserved

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

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 \(\Pi\)):

	Q	B	S	I₃
π⁺ + n	1 + 0 = 1	0 + 1 = 1	0	1 - 1/2 = 1/2
π⁰ + K⁺	0 + 1 = 1	0 + 0 = 0	0 + 1 = 1	0 + 1/2 = 1/2

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.

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18.

Parity is not conserved in :

A.strong nuclear interactions
B.gravitational interactions
C.weak interactions
D.electromagnetic interactions

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

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.

Discuss in Forum

19.

The quark content of the subatomic particle K° with strangeness S = +1 and isospin I =1/2 is given by :

A.\(d\bar{s}\)
B.\(\bar{d}s\)
C.\(d\bar{d}\)
D.\(u\bar{u}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

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 \(\eta\).

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20.

What are the appropriate values for the corresponding lifetimes of hadronic decay, electromagnetic decay and weak decay ?

A.10\(^{-9}\) sec, 10\(^{-6}\) sec, 10\(^{-3}\) sec
B.10\(^{-12}\) sec, 10\(^{-9}\) sec, 10\(^{-6}\) sec
C.10\(^{-15}\) sec, 10\(^{-13}\) sec, 10\(^{-6}\) sec
D.10\(^{-23}\) sec, 10\(^{-18}\) sec, 10\(^{-10}\) sec

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

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.

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