---
title: "Energy from D-T fusion and U-235 fission"
subject: "Energy Resources"
chapter: 7
key: "semester_6/energy_resources:fusion-fission-energy-mass-defect"
source: https://www.bctengineer.com/q/semester_6/energy_resources/fusion-fission-energy-mass-defect
site: BCT Engineering
---

# Energy from D-T fusion and U-235 fission

Energy Resources · Chapter 7: Nuclear energy

- Asked in: year not given, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/energy_resources/fusion-fission-energy-mass-defect
- Whole chapter: https://www.bctengineer.com/md/semester_6/energy_resources/chapter-7

## Question

Using atomic masses D = 2.014102 u, T = 3.016049 u, He-4 = 4.002603 u and neutron = 1.008665 u (1 u = 931.5 MeV), find the energy released in the D-T fusion reaction. Compare the energy released per kg of fuel with that from fission of U-235, assuming 200 MeV per fission. ($N_A = 6.022\times10^{23}$, 1 MeV = $1.602\times10^{-13}$ J.) How many kg of coal of calorific value 25 MJ/kg are equivalent to 1 kg of U-235?

## Short answer

Mass defect Δm = 0.018883 u; energy = 17.59 MeV. Energy/kg D-T = (1000/5.030)×N_A × 17.59 MeV = 3.37×10^14 J/kg. U-235 yields 8.21×10^13 J/kg, so fusion ~4.1 times greater. 1 kg U-235 equals ~3.28×10^6 kg coal (3280 tonnes).

## Answer

#### Mass defect of D-T fusion
Reaction: $D + T \rightarrow He + n$

$$
\begin{aligned}
m_{in} &= 2.014102 + 3.016049 = 5.030151\ \text{u} \\
m_{out} &= 4.002603 + 1.008665 = 5.011268\ \text{u} \\
\Delta m &= 0.018883\ \text{u}
\end{aligned}
$$

$$
E = \Delta m \times 931.5 = 0.018883 \times 931.5 = 17.59\ \text{MeV}
$$

#### Energy per kg of D-T fuel
One reaction uses 5.030 u of fuel, i.e. 1 kg holds $\dfrac{1000\ \text{g}}{5.030\ \text{g/mol}} \times N_A = 1.197\times10^{26}$ reactions.

$$
E_{fus} = 1.197\times10^{26} \times 17.59 \times 1.602\times10^{-13} = 3.37\times10^{14}\ \text{J/kg}
$$

#### Energy per kg of U-235
Atoms in 1 kg: $\dfrac{1000}{235}\times 6.022\times10^{23} = 2.563\times10^{24}$.

$$
E_{fis} = 2.563\times10^{24} \times 200 \times 1.602\times10^{-13} = 8.21\times10^{13}\ \text{J/kg}
$$

#### Comparison and coal equivalent
$$
\frac{E_{fus}}{E_{fis}} = \frac{3.37\times10^{14}}{8.21\times10^{13}} = 4.1
$$

Coal equivalent of 1 kg U-235:

$$
\frac{8.21\times10^{13}}{25\times10^{6}} = 3.28\times10^{6}\ \text{kg} \approx 3280\ \text{tonnes}
$$

**Answer:** D-T fusion releases 17.59 MeV; fusion gives $3.37\times10^{14}$ J/kg, about 4.1 times the $8.21\times10^{13}$ J/kg of fission; 1 kg of U-235 is equal to about 3.3 million kg (3280 tonnes) of coal.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the energy of D-T fusion found?
   A. Sum of reactant masses times 931.5
   B. Mass of reactants minus products (0.018883 u) times 931.5 MeV/u
   C. Mass of He only times 931.5
   D. Mass of neutron times 200
   Answer: B. The mass defect is converted to energy: 0.018883 x 931.5 = 17.59 MeV. (Key point: Mass defect method)

2. How is the number of fusion reactions in 1 kg of fuel found?
   A. (1000 g / 5.030 g/mol) x N_A
   B. (1000 / 235) x N_A
   C. 1000 x 17.59
   D. 1000 / N_A
   Answer: A. One reaction consumes a D and a T, total 5.030 u, so moles are 1000/5.030. (Key point: Reactions per kg)

3. How does fusion compare with fission per kg of fuel?
   A. Fusion gives about 4.1 times more
   B. They are equal
   C. Fission gives about 4 times more
   D. Fusion gives 100 times more
   Answer: A. 3.37 x 10^14 J/kg compared with 8.21 x 10^13 J/kg is about 4.1. (Key point: Fusion vs fission per kg)

4. How is the coal equivalent of 1 kg U-235 found?
   A. Divide 8.21 x 10^13 J by 25 x 10^6 J/kg, about 3.3 x 10^6 kg
   B. Multiply 8.21 x 10^13 by 25 x 10^6
   C. Divide 25 x 10^6 by 8.21 x 10^13
   D. Use 200 MeV x 25
   Answer: A. Energy per kg of U-235 divided by coal calorific value gives about 3280 tonnes. (Key point: Coal equivalent)

## Related in Energy Resources

- [Nuclear fission, fusion and fusion requirements](https://www.bctengineer.com/question.md?key=semester_6%2Fenergy_resources%3Anuclear-fission-fusion-requirements) — Energy Resources, chapter 7
- [Fuel cell voltage, efficiency and hydrogen use](https://www.bctengineer.com/question.md?key=semester_6%2Fenergy_resources%3Afuel-cell-stack-numerical) — Energy Resources, chapter 6

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