Mass-Energy Equivalence and Nuclear Units
For an isolated process, total relativistic energy is conserved. A rest-mass difference Delta m corresponds to rest-energy difference Delta E = Delta m c^2; 1 u c^2 is approximately 931.5 MeV. Unit conversion must not insert c^2 twice.
Why this shows up in the exam
Converting mass loss to released energy · Estimating fuel consumption at a stated power · Converting MeV, joules, and kilowatt-hours
Learn the idea
A change in rest mass corresponds to energy through E = mc², with atomic mass units converting conveniently to MeV. Nuclear energy can be viewed as a tiny measurable difference between initial and final rest masses. Multiplying that difference by c squared turns the small mass change into a large energy.
🧠 Memory hook: Mass in u times 931.5 gives energy in MeV.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- E0 = m c² — rest energy of mass m
- 1 u c² approximately 931.5 MeV — atomic-mass-unit energy conversion
- P = Delta E / Delta t — power produced when energy Delta E is released over time Delta t
How to approach it
- 1Choose either the SI route or the u-to-MeV route
- 2Track per-nucleus versus total-sample quantities
- 3Perform the final unit conversion only once
Common slip-ups that cost marks
- •Multiplying by c squared after already using 931.5 MeV per u
- •Confusing energy with power
- •Using grams where kilograms are required in SI
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
In hydrogen, an electron transitions from n = 2 to n = 1. Using E_n = -13.6/n^2 eV, find the emitted photon energy.
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