Elementary Particles and Conservation Laws

Threshold Energy for Particle Production

Conditions: A projectile particle of mass m₁ and total energy E incident on a stationary target particle of mass m₂ to produce new particles with total rest mass ∑M.

• Write the relativistic energy-momentum four-vector invariant for the initial state in the lab frame: (E + m₂c²)² - p²c² = E_cm².
• Express the square of the total center-of-mass energy using E² - p²c² = m₁²c⁴: E_cm² = m₁²c⁴ + m₂²c⁴ + 2Em₂c².
• At threshold, the final products are produced at rest relative to each other in the center-of-mass frame, so the minimum center-of-mass energy is E_cm,min = (∑M)c².
• Equating the invariant mass-energy squares yields the threshold energy condition for the incident particle.

Result: $E_{th} = \frac{(\sum M)^2 c^2 - m_1^2 c^2 - m_2^2 c^2}{2 m_2}$

Q-Value of a Nuclear or Particle Reaction

Conditions: Reaction of the form A + B \rightarrow C + D at non-relativistic or relativistic limits.

• Write the conservation of total relativistic energy: E_A + $E_B = E_C + E_D$.
• Expand each total energy into its rest mass energy and kinetic energy: (m_A c^2 + K_A) + (m_B c^2 + $K_B) = (m_C c^2 + K_C) + (m_D c^2 + K_D)$.
• Define the Q-value as the difference between the total kinetic energy after and before the reaction: $Q = (K_C + K_D) - (K_A + K_B)$.
• Substitute back to express Q purely in terms of rest mass energies.

Result: $Q = [m_A + m_B - (m_C + m_D)] c^2$

Bubble Chamber Particle Tracks

Draw: Curved tracks of charged particles in a uniform magnetic field showing spirals, vertices, and missing neutral particle paths.

Interpret: Radius of curvature $r = p / (qB) allows determination of momentum$, charge sign (from direction of curvature), and lifetime of unstable intermediate states.

Exam use: Questions identifying particle decay modes and sign of charge from magnetic deflection.

Binding Energy per Nucleon Curve

Draw: A plot of Binding Energy per Nucleon (B.E./A) vs Mass Number (A) rising sharply to a peak around A = 56 and slowly decreasing.

Interpret: Highlights nuclear stability regions and explains energy release in nuclear fission and fusion.

Exam use: Direct conceptual questions on nuclear stability and energetics.

Scattering Cross-Section vs Energy

Draw: A resonance peak curve showing sharp spikes in cross-section at specific energy thresholds.

Interpret: Indicates the formation of short-lived unstable excited hadron states (resonances) during particle collisions.

Exam use: Interpreting resonance peaks and unstable particle generation.

Determine whether the following reaction is allowed by conservation laws: p + p \rightarrow p + n + \pi^+.

• Check Baryon Number (B): Initial $B = 1 + 1 = 2$. Final $B = 1 (proton) + 1 (neutron) + 0 (pion) = 2$. Baryon number is conserved.
• Check Electric Charge (Q): Initial $Q = 1 + 1 = 2$. Final $Q = 1 (proton) + 0 (neutron) + 1 (pion) = 2$. Charge is conserved.
• Check Lepton Number (L): Initial L = 0, Final L = 0. Lepton number is conserved.

Answer: Allowed by conservation of charge, baryon number, and lepton number.

Check the validity of the reaction K^- + p \rightarrow \Lambda^0 + \pi^0 using conservation of strangeness, baryon number, and charge.

• Write quantum numbers for K^-: B = 0, $Q = -1$, $S = -1$. For proton (p): B = 1, Q = 1, S = 0. Initial totals: B = 1, Q = 0, $S = -1$.
• Write quantum numbers for \Lambda^0: B = 1, Q = 0, $S = -1$. For \pi^0: B = 0, Q = 0, S = 0. Final totals: B = 1, Q = 0, $S = -1$.
• Compare initial and final totals: $B_initial = B_final (1 = 1)$, $Q_initial = Q_final (0 = 0)$, $S_initial = S_final (-1 = -1)$.

Answer: Allowed in strong interactions as strangeness, baryon number, and charge are all conserved.

Calculate the threshold kinetic energy of a proton incident on a stationary proton to produce an antiproton via the reaction p + p \rightarrow p + p + p + \bar{p}.

• Identify masses: $m_1 = m_2 = m_3 = m_4 = m_p = m$. Total rest mass of final products ∑M = 4m.
• Use the threshold energy formula: $E_{th} = \frac{(\sum M)^2 c^2 - m_1^2 c^2 - m_2^2 c^2}{2 m_2}$.
• Substitute values: $E_{th} = \frac{(4m)^2 c^2 - m^2 c^2 - m^2 c^2}{2 m c^2} = \frac{16m^2 c^2 - 2m^2 c^2}{2mc^2} = 7mc^2$.
• Since total incident energy $E = K + mc^2$, the threshold kinetic energy is $K_{th} = E_{th} - mc^2 = 7mc^2 - mc^2 = 6mc^2$.

Answer: 6mc^2 (or 6 times the rest mass energy of a proton, approximately 5.63 GeV)