Specification: AQA A-level Physics (7408) 3.2.1.3 | Paper: 1 | Data and formulae booklet: rest energies and E = hf = hc/λ given in the exam; learn the 2 × rest energy rule
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Rest energy: the energy equivalent of a particle’s mass when it is at rest, E = mc². The data booklet gives rest energies in MeV, for example 0.510999 MeV for the electron and 938.257 MeV for the proton.
Annihilation: when a particle and its antiparticle meet, they are both destroyed and their total energy (rest energy plus kinetic energy) is converted into two photons. Two photons moving in opposite directions are needed so that momentum is conserved.
Pair production: a photon converts into a particle and its antiparticle, usually near a nucleus. The minimum photon energy is 2 × the rest energy of the particle; any extra energy becomes the kinetic energy of the pair.
The equation
rest energy = mass × (speed of light)² E = mc² and minimum photon energy for pair production = 2 × rest energy hfmin = 2E0
| Quantity | Symbol | Unit |
|---|---|---|
| rest energy of one particle | E0 (or E) | joules (J), or mega-electronvolts (MeV) |
| rest mass | m | kilograms (kg) |
| speed of light in a vacuum | c | metres per second (m s⁻¹) |
| minimum photon energy | hfmin | joules (J), or MeV |
| photon frequency | f | hertz (Hz) |
| photon wavelength | λ | metres (m) |
Rearranged: m = E ÷ c² | λmax = hc ÷ (2E0) for pair production | each annihilation photon (particles at rest) has E = E0, so λ = hc ÷ E0 | MeV → J: × 1.60 × 10−13; J → MeV: ÷ 1.60 × 10−13 | Always convert MeV to J before using E = hf or E = hc ÷ λ.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) Calculate the rest energy of an electron (a) in joules, to 3 significant figures, and (b) in MeV, to 3 significant figures. (me = 9.11 × 10−31 kg; c = 3.00 × 108 m s⁻¹; 1 MeV = 1.60 × 10−13 J)
Show solution
F: E = mc²
I: E = 9.11 × 10−31 × (3.00 × 108)²
F: E is already the subject
A: (a) 8.199 × 10−14 J = 8.20 × 10−14 J (3 s.f.)
(b) 8.199 × 10−14 ÷ 1.60 × 10−13 = 0.5124… MeV = 0.512 MeV (3 s.f.), in line with the booklet value of 0.510999 MeV
Q2 (Core) Use the booklet rest energy of the electron, 0.510999 MeV, to find the minimum energy of a photon that can produce an electron–positron pair, to 3 significant figures.
Show solution
F: hfmin = 2E0
I: hfmin = 2 × 0.510999
F: hfmin is already the subject
A: 1.021998 MeV = 1.02 MeV (3 s.f.)
Q3 (Core) Convert the minimum photon energy for electron–positron pair production, 1.021998 MeV, into joules, then calculate the photon’s frequency. Give both to 3 significant figures. (1 MeV = 1.60 × 10−13 J; h = 6.63 × 10−34 J s)
Show solution
Convert first: 1.021998 MeV × 1.60 × 10−13 = 1.635… × 10−13 J = 1.64 × 10−13 J (3 s.f.)
F: E = hf
I: 1.635… × 10−13 = 6.63 × 10−34 × f
F: f = 1.635… × 10−13 ÷ 6.63 × 10−34
A: 2.466… × 1020 Hz = 2.47 × 1020 Hz (3 s.f.), a gamma-ray photon
Q4 (Core) The rest energy of a proton is 938.257 MeV. Calculate the minimum energy of a photon that can produce a proton–antiproton pair, in MeV and in joules, to 3 significant figures. (1 MeV = 1.60 × 10−13 J)
Show solution
F: hfmin = 2E0
I: hfmin = 2 × 938.257
F: hfmin is already the subject
A: 1876.514 MeV = 1880 MeV (3 s.f.)
Convert: 1876.514 × 1.60 × 10−13 = 3.002… × 10−10 J = 3.00 × 10−10 J (3 s.f.)
Watch out: the photon has to make two particles, so double the rest energy. Using 938 MeV on its own is a common error.
Q5 (Exam) An electron and a positron, both effectively at rest, annihilate. (a) Explain why two photons are produced rather than one. (b) Calculate the wavelength of each photon, to 3 significant figures. (Electron rest energy 0.510999 MeV; 1 MeV = 1.60 × 10−13 J; h = 6.63 × 10−34 J s; c = 3.00 × 108 m s⁻¹)
Show solution
(a) the total momentum before is zero. A single photon always carries momentum, so momentum would not be conserved. Two photons of equal energy moving in opposite directions have a total momentum of zero.
Step 1 – energy of each photon
the total rest energy 2 × 0.510999 MeV is shared equally, so each photon has 0.510999 MeV
Convert first: 0.510999 × 1.60 × 10−13 = 8.175… × 10−14 J
Step 2 – (b) wavelength
F: E = hc ÷ λ
I: 8.175… × 10−14 = 6.63 × 10−34 × 3.00 × 108 ÷ λ
F: λ = 1.989 × 10−25 ÷ 8.175… × 10−14
A: 2.432… × 10−12 m = 2.43 × 10−12 m (3 s.f.)
Watch out: each photon gets one rest energy, not two. Using 1.022 MeV for each photon halves the wavelength.
Q6 (Exam) A proton and an antiproton, both effectively at rest, annihilate to produce two photons. Calculate the wavelength of each photon, to 3 significant figures. (Proton rest energy 938.257 MeV; 1 MeV = 1.60 × 10−13 J; h = 6.63 × 10−34 J s; c = 3.00 × 108 m s⁻¹)
Show solution
Convert first: each photon has E = 938.257 MeV = 938.257 × 1.60 × 10−13 = 1.501… × 10−10 J
F: E = hc ÷ λ
I: 1.501… × 10−10 = 1.989 × 10−25 ÷ λ
F: λ = 1.989 × 10−25 ÷ 1.501… × 10−10
A: 1.324… × 10−15 m = 1.32 × 10−15 m (3 s.f.), about the size of a nucleus
Q7 (Exam) Calculate the maximum wavelength of a photon that can produce an electron–positron pair, to 3 significant figures. (Electron rest energy 0.510999 MeV; 1 MeV = 1.60 × 10−13 J; h = 6.63 × 10−34 J s; c = 3.00 × 108 m s⁻¹)
Show solution
Step 1 – minimum energy
hfmin = 2 × 0.510999 = 1.021998 MeV = 1.635… × 10−13 J
Step 2 – maximum wavelength
F: E = hc ÷ λ
I: 1.635… × 10−13 = 1.989 × 10−25 ÷ λmax
F: λmax = 1.989 × 10−25 ÷ 1.635… × 10−13
A: 1.216… × 10−12 m = 1.22 × 10−12 m (3 s.f.)
Watch out: a minimum energy means a maximum wavelength, because E is inversely proportional to λ. Any photon with a longer wavelength cannot make the pair.
Q8 (Exam) A gamma-ray photon has a frequency of 3.00 × 1020 Hz. Show whether it could produce (a) an electron–positron pair and (b) a proton–antiproton pair. (h = 6.63 × 10−34 J s; 1 MeV = 1.60 × 10−13 J; rest energies: electron 0.510999 MeV, proton 938.257 MeV)
Show solution
F: E = hf
I: E = 6.63 × 10−34 × 3.00 × 1020
F: E = 1.989 × 10−13 J; ÷ 1.60 × 10−13 = 1.243… MeV
A: photon energy = 1.24 MeV (3 s.f.)
(a) 1.24 MeV > 2 × 0.510999 = 1.02 MeV, so yes, it can produce an electron–positron pair
(b) 1.24 MeV is far less than 2 × 938.257 = 1880 MeV, so no, it cannot produce a proton–antiproton pair
Q9 (Stretch) A photon of energy 1.50 MeV produces an electron–positron pair near a nucleus. Assume the nucleus gains negligible energy and the kinetic energy is shared equally. Calculate the kinetic energy of the electron (a) in MeV, to 3 significant figures, and (b) in joules, to 3 significant figures. (Electron rest energy 0.510999 MeV; 1 MeV = 1.60 × 10−13 J)
Show solution
Step 1 – energy left after making the pair
energy conservation: hf = 2E0 + total kinetic energy
total kinetic energy = 1.50 − 2 × 0.510999 = 1.50 − 1.021998 = 0.478002 MeV
Step 2 – (a) share for each particle
Ek = 0.478002 ÷ 2 = 0.239001 MeV
A: 0.239 MeV (3 s.f.)
Step 3 – (b) in joules
0.239001 × 1.60 × 10−13 = 3.824… × 10−14 J
A: 3.82 × 10−14 J (3 s.f.)
Watch out: subtract the rest energy of both particles before sharing out the kinetic energy. The nucleus is needed so that momentum can be conserved.
Q10 (Stretch) A high-energy photon of wavelength 5.00 × 10−16 m produces a proton–antiproton pair. Assume the kinetic energy is shared equally between the two particles. Calculate (a) the photon energy in MeV, to 4 significant figures, (b) the kinetic energy of each particle in MeV, to 3 significant figures, and (c) the total kinetic energy in joules, to 3 significant figures. (Proton rest energy 938.257 MeV; 1 MeV = 1.60 × 10−13 J; h = 6.63 × 10−34 J s; c = 3.00 × 108 m s⁻¹)
Show solution
Step 1 – (a) photon energy
F: E = hc ÷ λ
I: E = 6.63 × 10−34 × 3.00 × 108 ÷ 5.00 × 10−16 = 3.978 × 10−10 J
F: E = 3.978 × 10−10 ÷ 1.60 × 10−13 = 2486.25 MeV
A: 2486 MeV (4 s.f.), which is more than 2 × 938.257 = 1876.514 MeV, so the pair can be made
Step 2 – (b) kinetic energy of each particle
total Ek = 2486.25 − 1876.514 = 609.736 MeV; each = 609.736 ÷ 2 = 304.868 MeV
A: 305 MeV (3 s.f.)
Step 3 – (c) total kinetic energy in joules
609.736 × 1.60 × 10−13 = 9.755… × 10−11 J
A: 9.76 × 10−11 J (3 s.f.)
Watch out: do not round the photon energy to 2490 MeV before subtracting. Taking away two large, similar numbers magnifies any rounding error.
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