Specification: AQA A-level Physics (7408) 3.12.3.5 Turning points option | Paper: 3 Section B | Data and formulae booklet: E = mc² = m₀c²/√(1 − v²/c²) is given in the exam; kinetic energy = (m − m₀)c² is not, so learn it
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Mass–energy equivalence: mass and energy are equivalent, E = mc². A particle’s rest mass m₀ is its mass measured at rest, and its rest energy is m₀c². Its total energy at speed v is E = mc², where m = m₀/√(1 − v²/c²) is its relativistic mass.
Kinetic energy: kinetic energy = total energy − rest energy, Ek = (m − m₀)c². For a charge accelerated from rest through a pd V, the work done eV equals this kinetic energy. ½m₀v² is only a good approximation when v is much less than c.
Bertozzi’s experiment: electrons accelerated to known kinetic energies were timed over a fixed distance. Their speed approached but never exceeded c as the kinetic energy increased, which is direct evidence that Ek = ½mv² fails at high speeds and that the relativistic equation is correct.
The equation
total energy = mass × (speed of light)² = rest mass × (speed of light)² ÷ √(1 − v²/c²) E = mc² = m₀c²/√(1 − v²/c²) and kinetic energy Ek = (m − m₀)c²
| Quantity | Symbol | Unit |
|---|---|---|
| total energy | E | joules (J), often MeV (1 MeV = 1.60 × 10−13 J) |
| relativistic mass at speed v | m | kilograms (kg) |
| rest mass | m₀ | kilograms (kg) |
| speed of the particle | v | metres per second (m s⁻¹) |
| speed of light in a vacuum, 3.00 × 108 | c | metres per second (m s⁻¹) |
| kinetic energy | Ek | joules (J) |
Rearranged: m = E ÷ c² | m ÷ m₀ = E ÷ m₀c² = 1/√(1 − v²/c²) | v = c√(1 − (m₀/m)²) | E = m₀c² + Ek, and for an accelerating pd, Ek = eV | Convert MeV to J (× 1.60 × 10−13) and kV or MV to V before you substitute.
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 a proton, in J and in MeV, each to 3 significant figures. (c = 3.00 × 108 m s⁻¹, mp = 1.673 × 10−27 kg, e = 1.60 × 10−19 C)
Show solution
F: E = m₀c²
I: E = 1.673 × 10−27 × (3.00 × 108)²
F: E is already the subject
A: 1.5057 × 10−10 J = 1.51 × 10−10 J (3 s.f.); ÷ 1.60 × 10−13 = 941.0… MeV = 941 MeV (3 s.f.)
Q2 (Core) An electron moves at 0.600c. Calculate its total energy, to 3 significant figures. (c = 3.00 × 108 m s⁻¹, me = 9.11 × 10−31 kg, e = 1.60 × 10−19 C)
Show solution
F: E = m₀c²/√(1 − v²/c²)
I: E = 9.11 × 10−31 × (3.00 × 108)² ÷ √(1 − 0.600²) = 8.199 × 10−14 ÷ 0.800
F: E is already the subject
A: 1.024… × 10−13 J = 1.02 × 10−13 J (3 s.f.)
Q3 (Core) A proton in an accelerator has a total energy of 3.00 × 10−10 J. Calculate (a) its relativistic mass and (b) the ratio of its mass to its rest mass, each to 3 significant figures. (c = 3.00 × 108 m s⁻¹, mp = 1.673 × 10−27 kg, e = 1.60 × 10−19 C)
Show solution
F: E = mc²
I: 3.00 × 10−10 = m × (3.00 × 108)²
F: m = E ÷ c² = 3.00 × 10−10 ÷ 9.00 × 1016
A: (a) 3.333… × 10−27 kg = 3.33 × 10−27 kg (3 s.f.); (b) m ÷ m₀ = 3.333… × 10−27 ÷ 1.673 × 10−27 = 1.99 (3 s.f.)
Q4 (Core) An electron has a kinetic energy of 1.00 MeV. Calculate (a) its total energy in J and (b) its relativistic mass, each to 3 significant figures. (c = 3.00 × 108 m s⁻¹, me = 9.11 × 10−31 kg, e = 1.60 × 10−19 C)
Show solution
Convert first: 1.00 MeV = 1.00 × 106 × 1.60 × 10−19 = 1.60 × 10−13 J
Step 1 – (a) total energy
rest energy m₀c² = 9.11 × 10−31 × (3.00 × 108)² = 8.199 × 10−14 J
E = m₀c² + Ek = 8.199 × 10−14 + 1.60 × 10−13 = 2.4199 × 10−13 J = 2.42 × 10−13 J (3 s.f.)
Step 2 – (b) relativistic mass
F: E = mc², so m = E ÷ c²
I: m = 2.4199 × 10−13 ÷ 9.00 × 1016
A: 2.688… × 10−30 kg = 2.69 × 10−30 kg (3 s.f.), nearly three times the rest mass
Watch out: kinetic energy is not the total energy. Add the rest energy before using E = mc².
Q5 (Exam) An electron travels at 0.800c. (a) Calculate its kinetic energy using Ek = (m − m₀)c². (b) Calculate the value given by ½m₀v² and compare the two. Give answers to 3 significant figures. (c = 3.00 × 108 m s⁻¹, me = 9.11 × 10−31 kg, e = 1.60 × 10−19 C)
Show solution
Step 1 – (a) relativistic kinetic energy
m = m₀/√(1 − 0.800²) = m₀ ÷ 0.600, so Ek = (m − m₀)c² = m₀c²(1 ÷ 0.600 − 1) = 8.199 × 10−14 × 0.6666…
A: 5.466 × 10−14 J = 5.47 × 10−14 J (3 s.f.)
Step 2 – (b) Newtonian value
v = 0.800 × 3.00 × 108 = 2.40 × 108 m s⁻¹; ½m₀v² = 0.5 × 9.11 × 10−31 × (2.40 × 108)² = 2.623… × 10−14 J
A: 2.62 × 10−14 J (3 s.f.). The relativistic value is 2.08 times bigger, so ½m₀v² badly underestimates Ek at this speed.
Q6 (Exam) A proton is accelerated from rest through a pd of 2.00 MV. (a) Calculate the percentage increase in its mass, to 2 significant figures. (b) Calculate its speed using the relativistic equations and using ½m₀v² = eV, each to 3 significant figures, and comment. (c = 3.00 × 108 m s⁻¹, mp = 1.673 × 10−27 kg, e = 1.60 × 10−19 C)
Show solution
Convert first: 2.00 MV = 2.00 × 106 V; Ek = eV = 1.60 × 10−19 × 2.00 × 106 = 3.20 × 10−13 J
Step 1 – (a) mass increase
rest energy m₀c² = 1.673 × 10−27 × (3.00 × 108)² = 1.5057 × 10−10 J; m ÷ m₀ = (m₀c² + Ek) ÷ m₀c² = 1.5089 × 10−10 ÷ 1.5057 × 10−10 = 1.00212…
A: 0.21% increase (2 s.f.)
Step 2 – (b) speeds
relativistic: v = c√(1 − (m₀/m)²) = 3.00 × 108 × √(1 − (1 ÷ 1.00212…)²) = 3.00 × 108 × 0.06509… = 1.95 × 107 m s⁻¹ (3 s.f.)
Newtonian: v = √(2Ek ÷ m₀) = √(2 × 3.20 × 10−13 ÷ 1.673 × 10−27) = 1.96 × 107 m s⁻¹ (3 s.f.)
A: The two agree to within 0.2%, because 2 MeV is tiny compared with the proton’s rest energy of 941 MeV. Relativistic effects only matter when Ek is comparable with m₀c².
Q7 (Exam) (a) Calculate the speed at which an electron’s relativistic mass is twice its rest mass, in m s⁻¹ to 3 significant figures. (b) Calculate the accelerating pd needed to give an electron this speed, to 3 significant figures. (c = 3.00 × 108 m s⁻¹, me = 9.11 × 10−31 kg, e = 1.60 × 10−19 C)
Show solution
Step 1 – (a) speed
F: m = m₀/√(1 − v²/c²)
I: 2m₀ = m₀/√(1 − v²/c²), so √(1 − v²/c²) = 0.5
F: 1 − v²/c² = 0.25, so v/c = √0.75 = 0.8660…
A: v = 0.8660… × 3.00 × 108 = 2.598… × 108 m s⁻¹ = 2.60 × 108 m s⁻¹ (3 s.f.)
Step 2 – (b) accelerating pd
Ek = (m − m₀)c² = (2m₀ − m₀)c² = m₀c² = 9.11 × 10−31 × (3.00 × 108)² = 8.199 × 10−14 J
eV = Ek, so V = 8.199 × 10−14 ÷ 1.60 × 10−19 = 512 437.5 V
A: 512 kV (3 s.f.)
Q8 (Exam) In Bertozzi’s experiment, electrons with a kinetic energy of 1.5 MeV were timed over a flight path of 8.4 m. (a) Use the relativistic equations to calculate their speed, to 3 significant figures, and the flight time, to 2 significant figures. (b) Calculate the flight time predicted by ½m₀v² = Ek, to 2 significant figures, and comment. (c = 3.00 × 108 m s⁻¹, me = 9.11 × 10−31 kg, e = 1.60 × 10−19 C)
Show solution
Convert first: 1.5 MeV = 1.5 × 1.60 × 10−13 = 2.40 × 10−13 J
Step 1 – (a) relativistic speed
E = m₀c² + Ek = 8.199 × 10−14 + 2.40 × 10−13 = 3.2199 × 10−13 J; m ÷ m₀ = 3.2199 × 10−13 ÷ 8.199 × 10−14 = 3.927…
v = c√(1 − (m₀/m)²) = 3.00 × 108 × √(1 − (1 ÷ 3.927…)²) = 3.00 × 108 × 0.9670… = 2.901… × 108 m s⁻¹ = 2.90 × 108 m s⁻¹ (3 s.f.)
flight time = 8.4 ÷ 2.901… × 108 = 2.895… × 10−8 s = 29 ns (2 s.f.)
Step 2 – (b) Newtonian prediction
v = √(2Ek ÷ m₀) = √(2 × 2.40 × 10−13 ÷ 9.11 × 10−31) = 7.258… × 108 m s⁻¹; time = 8.4 ÷ 7.258… × 108 = 1.157… × 10−8 s = 12 ns (2 s.f.)
A: The Newtonian speed is more than twice c, which is impossible. Bertozzi’s measured times matched the relativistic prediction, with speeds always just below c.
Watch out: never use ½mv² for electrons of MeV energies. If your speed comes out bigger than 3.00 × 108 m s⁻¹, you have used the wrong equation.
Q9 (Stretch) An electron is accelerated from rest through a pd of 500 kV. (a) Calculate its total energy and its relativistic mass, each to 3 significant figures. (b) Calculate its speed, to 3 significant figures. (c) Calculate the speed predicted by ½m₀v² = eV, to 3 significant figures, and describe how a graph of kinetic energy against speed differs between the two theories. (c = 3.00 × 108 m s⁻¹, me = 9.11 × 10−31 kg, e = 1.60 × 10−19 C)
Show solution
Convert first: 500 kV = 5.00 × 105 V; Ek = eV = 1.60 × 10−19 × 5.00 × 105 = 8.00 × 10−14 J
Step 1 – (a) total energy and mass
E = m₀c² + Ek = 8.199 × 10−14 + 8.00 × 10−14 = 1.6199 × 10−13 J = 1.62 × 10−13 J (3 s.f.)
m = E ÷ c² = 1.6199 × 10−13 ÷ 9.00 × 1016 = 1.799… × 10−30 kg = 1.80 × 10−30 kg (3 s.f.)
Step 2 – (b) speed
m ÷ m₀ = 1.799… × 10−30 ÷ 9.11 × 10−31 = 1.975…; v = c√(1 − (m₀/m)²) = 3.00 × 108 × √(1 − 0.2561…) = 3.00 × 108 × 0.8624…
A: 2.587… × 108 m s⁻¹ = 2.59 × 108 m s⁻¹ (3 s.f.)
Step 3 – (c) Newtonian prediction and graphs
v = √(2eV ÷ m₀) = √(2 × 8.00 × 10−14 ÷ 9.11 × 10−31) = √(1.756… × 1017) = 4.19 × 108 m s⁻¹ (3 s.f.), faster than light
A: Newtonian Ek = ½m₀v² is a parabola that passes c with a finite kinetic energy. The relativistic curve matches it at low speeds but rises ever more steeply, becoming infinite as v approaches c, so c can never be reached.
Q10 (Stretch) A linear accelerator gives electrons a kinetic energy of 25.0 MeV. (a) Calculate the relativistic mass of an electron and the ratio m ÷ m₀, each to 3 significant figures. (b) Calculate the percentage by which its speed is less than c, to 2 significant figures. (c) The electrons then enter a uniform magnetic field of flux density 0.500 T at right angles to their motion. Using r = mv ÷ Be with the relativistic mass, calculate the radius of their path, to 3 significant figures, and compare it with the radius predicted using the rest mass at the same speed. (c = 3.00 × 108 m s⁻¹, me = 9.11 × 10−31 kg, e = 1.60 × 10−19 C)
Show solution
Convert first: 25.0 MeV = 25.0 × 1.60 × 10−13 = 4.00 × 10−12 J
Step 1 – (a) mass
E = m₀c² + Ek = 8.199 × 10−14 + 4.00 × 10−12 = 4.08199 × 10−12 J
m = E ÷ c² = 4.08199 × 10−12 ÷ 9.00 × 1016 = 4.535… × 10−29 kg = 4.54 × 10−29 kg (3 s.f.); m ÷ m₀ = 49.78… = 49.8 (3 s.f.)
Step 2 – (b) speed
v ÷ c = √(1 − (1 ÷ 49.78…)²) = √(1 − 4.034… × 10−4) = 0.999798…; (1 − 0.999798…) × 100 = 0.0201…%
A: 0.020% less than c (2 s.f.), so v = 2.999… × 108 m s⁻¹
Step 3 – (c) radius in the field
F: r = mv ÷ Be
I: r = 4.535… × 10−29 × 2.999… × 108 ÷ (0.500 × 1.60 × 10−19)
A: 0.1700… m = 0.170 m (3 s.f.). Using the rest mass gives 9.11 × 10−31 × 2.999… × 108 ÷ (0.500 × 1.60 × 10−19) = 3.42 × 10−3 m, about 50 times too small.
Watch out: at this energy almost all the energy is kinetic, but v hardly changes. Increasing Ek increases the mass, not the speed.
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