How many joules is one MeV?

One MeV is 1.6 times ten to the minus thirteenth joules, so 200 MeV is only 3.2 times ten to the minus eleventh joules.

Why look at binding energy per nucleon?

A higher value means tighter binding, and iron-56 peaks, which is why both fission and fusion release energy.

S6 Physics Nuclear Energy, Paper 3: binding energy and unit conversion

2026-11-02

Ten graded items turning mass defect and binding energy into joules, then efficiency, fission rate, fuel use and MeV conversion.

Focus　Keep the MeV and joule scales apart, since one MeV is 1.6 times ten to the minus thirteenth joules, and divide electrical output by efficiency for thermal power.

01　Paper details (check answers afterwards)

Item	Detail
Questions	10
Suggested time	25
Difficulty mix	Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1
Chapter	Chapter: S6 Physics: Nuclear Energy
Level	Four basic, three advanced, two challenge and one error-repair item

02　Questions (10 items)

FoundationHow much energy is released when 3 kg of mass is converted completely, with c as 3.0 x 10^8 m/s?
FoundationA reaction has a mass defect of 0.5 u. Find the energy released, with 1 u as 931 MeV.
FoundationA mass defect of 5 g releases how much energy?
FoundationA plant delivers 600 MW of electricity at 40 per cent efficiency. Find its thermal power.
IntermediateEach fission gives 3.0 x 10^-11 J. How many fissions per second supply a 300 W lamp?
Intermediate1 kg of uranium-235 holds about 2.56 x 10^24 nuclei and each fission gives 3.0 x 10^-11 J. Find the total energy.
IntermediateIron-56 has 56 nucleons and a binding energy of 8.8 MeV per nucleon. Find the total binding energy.
ChallengeThe Sun converts 4.0 x 10^9 kg of mass into energy every second. Find its power output.
ChallengeA plant delivers 900 MW of electricity at 33 per cent efficiency. Find the thermal power and the daily uranium use at 8.2 x 10^13 J per kg.
Error redoEach fission releases 200 MeV and a student writes 200 J. Find the correct value in joules and the factor by which he is wrong.

03　Answers and one-line notes

#	Answer	One-line note
1	2.7 x 10^17 J	Three kilograms times nine times ten to the sixteenth joules per kilogram gives 2.7 times ten to the seventeenth joules.
2	465.5 MeV	Half an atomic mass unit at 931 MeV each gives 465.5 MeV.
3	4.5 x 10^14 J	Five grams is 0.005 kg, so the energy is 4.5 times ten to the fourteenth joules.
4	1.5 x 10^9 W	Dividing the electrical output by the efficiency gives 1.5 gigawatts of thermal power.
5	1.0 x 10^13 per second	Three hundred joules per second divided by three times ten to the minus eleventh gives ten to the thirteenth fissions.
6	About 7.7 x 10^13 J	Multiplying nuclei by energy per fission gives about 7.7 times ten to the thirteenth joules per kilogram.
7	492.8 MeV	Multiplying the binding energy per nucleon by the nucleon number gives 492.8 MeV for iron-56.
8	About 3.6 x 10^26 W	Four billion kilograms per second times nine times ten to the sixteenth gives about 3.6 times ten to the twenty-sixth watts.
9	About 2.7 x 10^9 W and 0.95 kg per day	Thermal power is about 2.7 GW and the daily electricity demand of 7.8 times ten to the thirteenth joules needs about 0.95 kg of fuel.
10	3.2 x 10^-11 J, so he is 6.25 x 10^12 times too large	Two hundred MeV equals 3.2 times ten to the minus eleventh joules, so treating it as 200 J is 6.25 times ten to the twelfth too large.

04　What to do after marking

After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
Repeat the same item two days later; move to the next paper only when both attempts are correct.
Parents only need the answer and note columns; there is no need to explain every item.

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