Physics • Year 12 • Module 7 • Lesson 14
Relativistic Mass, Momentum and Energy
Build HSC Band 5–6 extended-response technique on relativistic dynamics, mass–energy equivalence, and the failure of classical mechanics.
1. Data + scenario: proton in the Australian Synchrotron (Band 5–6)
8 marks Band 5–6
Scenario. A proton ($m_p c^2 = 938$ MeV, $e = 1.60 \times 10^{-19}$ C) is accelerated from rest through a total potential difference of 3.0 GV in a particle accelerator. The accelerator operator records the values in the table below.
| Quantity | Recorded value |
|---|---|
| Potential difference | $3.0 \times 10^9$ V |
| Proton rest mass energy $m_p c^2$ | 938 MeV |
| Elementary charge $e$ | $1.60 \times 10^{-19}$ C |
| Speed of light $c$ | $3.00 \times 10^8$ m/s |
Q1. Analyse the data above to fully characterise the relativistic state of the proton after acceleration. In your response you must:
- Calculate the proton's kinetic energy in MeV and in joules.
- Calculate the Lorentz factor $\gamma$ and hence the proton's speed as a decimal fraction of $c$ (4 significant figures).
- Calculate the proton's total energy and its relativistic momentum in MeV/$c$.
- Use the energy-momentum relation $E^2 = (pc)^2 + (mc^2)^2$ to verify your momentum value.
- Compare the relativistic kinetic energy to the classical prediction $E_k^{class} = \frac{1}{2}mv^2$ and explain quantitatively why they differ.
2. Experimental design — testing $E = mc^2$ with electron-positron annihilation (Band 5–6)
7 marks Band 5–6
Context. When an electron ($m_e c^2 = 0.511$ MeV) meets a positron at rest, they annihilate and produce two gamma-ray photons. A nuclear physics student wants to use this to test Einstein's mass-energy equivalence $E = mc^2$. The student has access to a positron emitter (sodium-22 source), a scintillator detector, an energy spectrometer, a coincidence counter, and standard lab equipment.
Constraints: You may not use equipment beyond what is listed. Assume both the electron and positron are approximately at rest at the moment of annihilation.
Q2. Design the investigation to test whether the energy of each photon produced equals $m_e c^2 = 0.511$ MeV. Present your answer under the headings below.
- Prediction: state the expected energy (in MeV) of each photon and the angle between the two photons, with justification.
- Procedure: describe at least four numbered steps explaining how to measure photon energy and confirm conservation of energy.
- How the data would confirm or refute the prediction.
- One significant source of systematic error and one improvement.
- Explain why conservation of both energy and momentum requires two photons, not one.
Q1 — Sample Band 6 response (8 marks), annotated
Step 1 — Kinetic energy: $E_k = qV = (1.60 \times 10^{-19})(3.0 \times 10^9) = 4.80 \times 10^{-10}$ J [1]. In MeV: $E_k = 4.80 \times 10^{-10} / (1.60 \times 10^{-13}) = 3000$ MeV [1].
Step 2 — Lorentz factor and speed: $\gamma - 1 = E_k / (m_p c^2) = 3000/938 = 3.1982$, so $\gamma = 4.198$. Then $v/c = \sqrt{1 - 1/\gamma^2} = \sqrt{1 - 1/17.62} = \sqrt{0.9432} = 0.9712$ [1 for $\gamma$; 1 for $v/c$].
Step 3 — Total energy and momentum: $E_{total} = \gamma m_p c^2 = 4.198 \times 938 = 3938$ MeV [1]. Using $pc = \sqrt{E_{total}^2 - (m_p c^2)^2} = \sqrt{3938^2 - 938^2} = \sqrt{15508 - 880} \times 10^3 = \sqrt{14308 \times 10^3}$. Wait, computing: $3938^2 = 15507844$; $938^2 = 879844$; difference $= 14628000$; $pc = \sqrt{14628000} = 3824$ MeV/$c$ [1]. Alternatively, $p = \gamma m_p v = 4.198 \times 938 \times 0.9712 = 3824$ MeV/$c$ (consistent) [1 for verification].
Step 4 — Classical comparison: $E_k^{class} = \frac{1}{2}(v/c)^2 m_p c^2 = 0.5 \times (0.9712)^2 \times 938 = 0.5 \times 0.9432 \times 938 = 442$ MeV. The relativistic value is 3000 MeV vs classical 442 MeV — a discrepancy of 2558 MeV (about 579% or $\sim$6.8 times). The classical formula fails because it does not account for the relativistic increase in inertia (the growing $\gamma$ factor) as $v \to c$; it treats mass as constant [1 for calculation; 1 for explanation].
Marking criteria (8 marks): 1 = correct $E_k$ in joules; 1 = correct $E_k$ in MeV (3000 MeV); 1 = correct $\gamma$ (4.198 $\pm$ rounding); 1 = correct $v/c$ (0.971 $\pm$ 0.001); 1 = correct $E_{total}$ (3938 MeV); 1 = correct $pc$ via energy-momentum relation; 1 = consistent verification of $p$ via $\gamma m_p v$; 1 = correct classical $E_k$ with quantitative comparison and explanation of divergence.
Q2 — Sample Band 6 response (7 marks), annotated
Prediction: If $E = mc^2$ holds, each photon should carry energy equal to the rest energy of one lepton: $E_\gamma = m_e c^2 = 0.511$ MeV. Because the electron and positron are both at rest, the total momentum before annihilation is zero. By conservation of momentum, the two photons must be emitted in exactly opposite directions (180° to each other) with equal energies [1 — prediction with correct justification including momentum conservation].
Procedure: (1) Position two scintillator-spectrometer detectors on opposite sides of the sodium-22 source (180° apart). Connect both detectors to a coincidence counter that registers an event only when both fire simultaneously. (2) Allow the source to emit positrons; each will annihilate with a nearby electron in the surrounding material. (3) For coincident events, record the energy deposited in each detector. Repeat for $\geq 500$ annihilation events to build a statistical distribution. (4) Plot a histogram of measured photon energies; check whether the peak occurs at 0.511 MeV. Check that both detectors fire in coincidence (confirming back-to-back emission at 180°) [1 for four clear steps including coincidence method].
Confirmation / refutation: If the energy histogram peaks at $0.511 \pm \delta$ MeV and the coincidence rate is high with detectors at 180° (dropping significantly when detectors are moved off 180°), the data confirm mass-energy equivalence. If photon energies are significantly different from 0.511 MeV or not emitted back-to-back, the prediction is refuted [1].
Systematic error: Some positrons will annihilate while still in motion (not at rest), adding kinetic energy to the photons and shifting their energy above 0.511 MeV (Doppler broadening the peak). Improvement: use a moderating medium (e.g. aluminium block) to thermalise positrons before annihilation, reducing their kinetic energy contribution [1 error; 1 improvement].
Why two photons are required: Before annihilation, the electron and positron are at rest, so the total momentum of the system is zero. A single photon would carry non-zero momentum $p = E/c \neq 0$, violating conservation of momentum. Two back-to-back photons with equal energy carry equal and opposite momenta, so the total momentum remains zero, satisfying both conservation of energy and conservation of momentum [1].
Marking criteria (7 marks): 1 = prediction: 0.511 MeV per photon with correct justification. 1 = prediction: 180° back-to-back emission with momentum conservation argument. 1 = procedure includes coincidence counting to confirm simultaneous emission. 1 = procedure includes energy measurement and $\geq 4$ steps. 1 = clear statement of how data would confirm or refute. 1 = one specific systematic error (positron kinetic energy or Compton scattering) with improvement. 1 = conservation of momentum argument requiring two photons.