Energy Released in Fission and Fusion: HSC Physics Guide

Suppose one deuterium nucleus fuses with one tritium nucleus. The reactants have a total mass of \(5.030151\ \text{u}\), while the products have a total mass of only \(5.011268\ \text{u}\).

That difference is just \(0.018883\ \text{u}\). Yet it corresponds to about \(17.6\ \text{MeV}\) of released energy.

So where did the missing mass go?

It wasn’t destroyed. The system’s rest-mass energy decreased, and that energy appears in other forms, such as kinetic energy of the products and electromagnetic radiation. The calculation is a direct application of mass-energy equivalence.

01Start with the whole reaction

The first useful model is a set of scales.

Imagine putting everything that enters the reaction on one scale and everything that leaves on another. You cannot compare just the original nucleus with one of the products. Every particle must be counted.

If the products have less total rest mass than the reactants, predict what happens to the difference.

Energy is released.

If the products have more total rest mass, energy must be supplied.

The simple rule is therefore:

Products lighter than reactants → energy released.

Products heavier than reactants → energy absorbed.

The analogy has one important limit. The missing mass does not literally peel off the nucleus and fly away. Mass and energy are two parts of the same energy accounting. The lower rest mass of the products means there is more energy available in other forms.

02The reaction-energy equation

The formal quantity is called the Q-value, or reaction energy:

\[
Q=(m_\text{reactants}-m_\text{products})c^2
\]

Here:

• \(Q\) is the reaction energy, in joules or electronvolts,
• \(m_\text{reactants}\) is the total rest mass of everything before the reaction,
• \(m_\text{products}\) is the total rest mass of everything after the reaction,
• \(c\) is the speed of light, \(3.00\times10^8\ \text{m s}^{-1}\).

For an energy-releasing reaction, \(Q\) is positive.

For an energy-absorbing reaction, \(Q\) is negative.

Notice the order of subtraction. It is:

\[
\boxed{\text{reactants} – \text{products}}
\]

If you reverse that order, you reverse the sign.

Why such a tiny mass gives so much energy

The factor \(c^2\) is enormous:

\[
c^2=(3.00\times10^8)^2
=9.00\times10^{16}\ \text{m}^2\text{s}^{-2}
\]

So even a tiny mass difference produces a measurable amount of energy.

Nuclear physics usually measures mass in atomic mass units, written as \(\text{u}\), and energy in MeV.

Using

\[
1\ \text{u}=1.66\times10^{-27}\ \text{kg}
\]

and \(E=mc^2\),

\[
1\ \text{u}\,c^2\approx931.5\ \text{MeV}
\]

That gives the HSC-friendly shortcut:

\[
\boxed{Q(\text{MeV})=\Delta m(\text{u})\times931.5}
\]

where

\[
\Delta m=m_\text{reactants}-m_\text{products}
\]

is the mass difference in atomic mass units.

If you need the deeper explanation of why bound nuclei have less mass than their separated nucleons, see Mass defect: how missing mass becomes nuclear binding energy.

03Why both fission and fusion can release energy

At first, fission and fusion seem like opposites.

• Fusion joins small nuclei together.
• Fission splits a large nucleus apart.

Yet both can move the products towards a more tightly bound state.

For light nuclei, combining nuclei can produce a nucleus with greater binding energy per nucleon. For very heavy nuclei, splitting the nucleus into medium-sized fragments can do the same.

Greater binding means a lower-energy bound state. Since energy and mass are related by \(E=mc^2\), the more tightly bound products have a lower total rest mass.

That is the physical reason behind the mass difference.

The graph is useful for understanding why energy can be released. It is not a substitute for the calculation. To calculate the actual energy for a particular reaction, compare the masses of every reactant and every product.

04The calculation method

For almost every HSC question of this type, use the same sequence.

1. Write the balanced nuclear equation.
2. Add the masses of all reactants.
3. Add the masses of all products.
4. Calculate \(\Delta m=m_\text{reactants}-m_\text{products}\).
5. Multiply by \(931.5\) to get MeV if the masses are in u.
6. Convert to joules if required.

There is a particularly important trap in step 3.

If a fission reaction produces three neutrons, those three neutrons belong in the product mass. They are not “extra energy” that can be ignored.

05Atomic masses: one detail that can save you a mistake

Nuclear data tables often give atomic masses, which include the electrons surrounding a neutral atom.

For the D-T reaction, there are two electrons on the reactant side and two electrons on the product side:

• deuterium + tritium: \(1+1=2\) electrons,
• helium + neutron: \(2+0=2\) electrons.

So the electron masses cancel.

The same happens in the uranium fission example:

• uranium: 92 electrons,
• barium + krypton: \(56+36=92\) electrons.

Therefore, for these reactions, you can use the listed atomic masses directly.

The rule is not “atomic masses are always fine”. The rule is that the electron contribution must be accounted for consistently. This becomes especially important when dealing with beta decay.

06The most tempting mistakes

One more useful check: if your calculated energy is wildly larger than the usual nuclear scale of a few MeV to hundreds of MeV per reaction, go back and check whether you forgot a neutron or mixed up the units.

07Questions and solutions

Question 1

A deuterium-deuterium fusion reaction follows

\[
{}^{2}_{1}\mathrm{H}+{}^{2}_{1}\mathrm{H}
\rightarrow
{}^{3}_{1}\mathrm{H}+{}^{1}_{1}\mathrm{H}
\]

Use these atomic masses:

• deuterium: \(2.014102\ \text{u}\)
• tritium: \(3.016049\ \text{u}\)
• hydrogen-1: \(1.007825\ \text{u}\)

Calculate the energy released per reaction in MeV and joules.

Solution 1

The reaction releases \(4.03\ \text{MeV}\), or approximately \(6.46\times10^{-13}\ \text{J}\) per reaction.

First find the total mass before and after the reaction:

\[
m_\text{reactants}
=2(2.014102)
=4.028204\ \text{u}
\]

\[
m_\text{products}
=3.016049+1.007825
=4.023874\ \text{u}
\]

Therefore,

\[
\Delta m
=4.028204-4.023874
=0.004330\ \text{u}
\]

Convert this to energy:

\[
Q=(0.004330)(931.5)
=4.033\ \text{MeV}
\]

Then

\[
Q=(4.033)(1.602\times10^{-13})
=6.46\times10^{-13}\ \text{J}
\]

The positive mass difference tells us the products have lower rest mass, so the reaction releases energy.

Question 2

A model fission reaction has a total reactant mass of \(236.052589\ \text{u}\). The two heavy fission fragments have a combined mass of \(232.840600\ \text{u}\), and three free neutrons are produced. Each neutron has a mass of \(1.008665\ \text{u}\).

Calculate the reaction energy in MeV.

A student ignores the three neutrons because they are moving away carrying energy. Calculate the student’s incorrect Q-value, and explain exactly why it is wrong.

Solution 2

The correct reaction energy is \(173.3\ \text{MeV}\). The student’s result of about \(2.99\times10^3\ \text{MeV}\) is wrong because the three neutrons are still particles with rest mass and must be included among the products.

The complete product mass is

\[
m_\text{products}
=232.840600+3(1.008665)
\]

\[
m_\text{products}
=235.866595\ \text{u}
\]

So

\[
\Delta m
=236.052589-235.866595
=0.185994\ \text{u}
\]

and

\[
Q=(0.185994)(931.5)
=173.25\ \text{MeV}
\]

The student’s calculation leaves the neutrons out:

\[
\Delta m_\text{wrong}
=236.052589-232.840600
=3.211989\ \text{u}
\]

which gives

\[
Q_\text{wrong}
=(3.211989)(931.5)
\approx2992\ \text{MeV}
\]

The trap is thinking that because the neutrons carry kinetic energy, they should be counted as energy rather than mass. They carry kinetic energy and have rest mass. The Q-value compares the rest masses of everything before and after the reaction.

Question 3

Compare the D-T fusion reaction from the worked example with the uranium fission reaction from the worked example.

The fusion reaction has

\[
\Delta m=0.018883\ \text{u}
\]

and total reactant mass \(5.030151\ \text{u}\).

The fission reaction has

\[
\Delta m=0.185994\ \text{u}
\]

and total reactant mass \(236.052589\ \text{u}\).

Calculate the percentage of the initial rest mass represented by the mass difference for each reaction.

A student then says, “Fission must convert more of its starting mass into released energy because it releases \(173\ \text{MeV}\), compared with only \(17.6\ \text{MeV}\) for fusion.” Is that conclusion correct?

Solution 3

Fusion has the larger fractional rest-mass change, about \(0.375\%\), compared with \(0.0788\%\) for fission. The student’s conclusion is therefore incorrect because energy per reaction and fractional mass change are different comparisons.

For fusion,

\[
\frac{\Delta m}{m_\text{reactants}}\times100
=
\frac{0.018883}{5.030151}\times100
\]

\[
=0.375\%
\]

For fission,

\[
\frac{\Delta m}{m_\text{reactants}}\times100
=
\frac{0.185994}{236.052589}\times100
\]

\[
=0.0788\%
\]

Fission releases more energy per individual reaction in these examples because a fission event involves far more nucleons and therefore a much larger absolute mass difference.

But the fusion reaction starts with only about \(5\ \text{u}\) of reactant mass, compared with about \(236\ \text{u}\) for the fission reaction. The percentage comparison shows that a larger Q-value does not automatically mean a larger fraction of the initial rest energy has been released.

This is also why “energy per reaction” should not be confused with “energy per kilogram of fuel”.

Question 4

A nuclear reaction has a Q-value of \(12.0\ \text{MeV}\). After the reaction, \(7.5\ \text{MeV}\) is measured as kinetic energy of the products and \(4.5\ \text{MeV}\) is emitted as gamma radiation.

Calculate the mass difference between the reactants and products in:

• atomic mass units,
• kilograms.

A student calculates the mass difference using only the \(7.5\ \text{MeV}\) of kinetic energy because gamma photons have zero rest mass. Explain why this reasoning is incorrect.

Solution 4

The mass difference is \(0.0129\ \text{u}\), or approximately \(2.14\times10^{-29}\ \text{kg}\). The student is wrong because a photon can have zero rest mass while still carrying energy.

The entire Q-value comes from the change in rest-mass energy:

\[
Q=12.0\ \text{MeV}
\]

Using \(1\ \text{u}\,c^2=931.5\ \text{MeV}\),

\[
\Delta m
=\frac{12.0}{931.5}
=0.01288\ \text{u}
\]

Convert to kilograms:

\[
\Delta m
=(0.01288)(1.66\times10^{-27})
\]

\[
\Delta m
=2.14\times10^{-29}\ \text{kg}
\]

The energy accounting is

\[
12.0\ \text{MeV}
=
7.5\ \text{MeV}
+
4.5\ \text{MeV}
\]

Both forms of energy came from the reaction’s mass difference.

The mistake is treating “zero rest mass” as “zero energy”. A gamma photon has no rest mass, but it carries electromagnetic energy. So the \(4.5\ \text{MeV}\) must be included when determining the total Q-value.

Question 5

A deuterium-tritium fusion reaction occurs with the initial nuclei effectively at rest:

\[
{}^{2}_{1}\mathrm{H}+{}^{3}_{1}\mathrm{H}
\rightarrow
{}^{4}_{2}\mathrm{He}+{}^{1}_{0}\mathrm{n}
\]

From the mass difference, the reaction releases approximately \(17.6\ \text{MeV}\).

Assume:

• no gamma photon is emitted,
• both products are in their ground states,
• all \(17.6\ \text{MeV}\) becomes kinetic energy of the helium nucleus and neutron,
• the products can be treated using the non-relativistic equation \(K=p^2/(2m)\).

Determine the kinetic energy of each product.

A student predicts that the \(17.6\ \text{MeV}\) should split equally, giving \(8.8\ \text{MeV}\) to each particle. Decide whether that can be correct.

Solution 5

The neutron carries about \(14.1\ \text{MeV}\), while the helium nucleus carries about \(3.54\ \text{MeV}\). The energy cannot be split equally because momentum conservation requires the two particles to have equal and opposite momenta.

The initial system is at rest, so its total momentum is zero. With only two final particles and no photon, momentum conservation requires

\[
p_\mathrm{n}=p_\mathrm{He}
\]

in magnitude, with the particles travelling in opposite directions.

For non-relativistic kinetic energy,

\[
K=\frac{p^2}{2m}
\]

Since both particles have the same momentum magnitude,

\[
\frac{K_\mathrm{n}}{K_\mathrm{He}}
=
\frac{m_\mathrm{He}}{m_\mathrm{n}}
\]

Using the particle masses,

\[
\frac{K_\mathrm{n}}{K_\mathrm{He}}
=
\frac{4.002603}{1.008665}
\approx3.97
\]

So the neutron has almost four times the kinetic energy of the helium nucleus.

Let

\[
K_\mathrm{He}=x
\]

Then

\[
K_\mathrm{n}=3.97x
\]

All the released energy becomes kinetic energy:

\[
x+3.97x=17.6
\]

\[
4.97x=17.6
\]

\[
x\approx3.54\ \text{MeV}
\]

Therefore,

\[
K_\mathrm{He}\approx3.54\ \text{MeV}
\]

and

\[
K_\mathrm{n}\approx14.1\ \text{MeV}
\]

The equal split of \(8.8\ \text{MeV}\) each sounds reasonable if you only think about energy conservation. But it fails the other constraint: momentum conservation. Equal kinetic energies would give the heavier helium nucleus a larger momentum than the neutron, so the final momentum would not add to zero.

That is the deeper lesson: the mass difference tells you how much energy is available. Conservation laws and the details of the reaction determine where that energy goes.

08What this lets you understand next

Once you can calculate a Q-value, a nuclear reaction becomes an energy ledger rather than a mysterious equation: compare the total rest masses, find the difference, and then track where the released energy goes.

The next useful step is to connect that calculation to mass defect and binding energy. That explains why particular nuclei sit at lower-energy states in the first place. From there, fission adds another layer: the released energy is linked to the fission products and the neutrons that can trigger further reactions.