Mass defect and binding energy

A bound system weighs less

Pull a spring apart and its energy goes up. A bound system needs energy to take it apart, so it has less energy than its pieces apart. Einstein told us that energy has mass. So a nucleus must weigh less than the protons and neutrons it is made of.

That is exactly what we find. Take helium-4. Its nucleus has $2$ protons and $2$ neutrons. We use atomic masses, because the mass of a bare nucleus is hard to measure. The $2$ electrons of the helium atom are balanced by the electrons of two hydrogen-1 atoms, so we can use $m_H = 1.007825\ \text{u}$ for each proton.

1. The parts: $2 \times 1.007825\ \text{u} + 2 \times 1.008665\ \text{u} = 4.032980\ \text{u}$.
2. The helium-4 atom: $4.002603\ \text{u}$.
3. The difference: $4.032980\ \text{u} - 4.002603\ \text{u} = 0.030377\ \text{u}$.

The missing $0.030377\ \text{u}$ is the mass defect $\Delta m$. In general, for a nuclide with $Z$ protons and $N$ neutrons and atomic mass $M$,

Mass and energy: $E = mc^2$

Einstein's relation says that a mass $m$ is equivalent to an energy $E = mc^2$. The speed of light is $c = 2.998 \times 10^8$ m/s, which is $3.00 \times 10^8$ m/s in most calculations. Take one atomic mass unit:

Nuclear physics uses a much smaller energy unit, the MeV. One electronvolt is the energy of an electron that goes through one volt, $1\ \text{eV} = 1.602177 \times 10^{-19}$ J, and $1\ \text{MeV} = 10^6\ \text{eV} = 1.602177 \times 10^{-13}$ J. So

The unit $\text{MeV}/c^2$ is a unit of mass. People often say that one u is "931.494 MeV". They mean the energy that goes with it.

What NCERT prints. NCERT and many A-Level and IB data booklets round this to $931.5\ \text{MeV}/c^2$ (and some exam questions say $931\ \text{MeV}$). We use the accurate value $931.494$ MeV, as modern tables do. The difference between $931.494$ and $931.5$ is less than $0.001\ \%$, so your answer is the same to four figures.

The binding energy

The binding energy $B$ of a nucleus is the energy needed to pull it apart into separate protons and neutrons, far from one another and at rest. It is the energy that goes with the mass defect:

For helium-4, $B = 0.030377\ \text{u} \times 931.494\ \text{MeV/u} = 28.30\ \text{MeV}$. To break one helium nucleus into four nucleons we must supply $28.3$ MeV. Making one from four nucleons gives back the same amount.

The binding energy per nucleon is $B/A$. It tells us how tightly a typical nucleon is held. For helium-4 it is $\dfrac{28.30\ \text{MeV}}{4} = 7.07\ \text{MeV}$. The larger it is, the more tightly bound the nucleus is.

Worked example: oxygen-16

Oxygen-16 has $8$ protons and $8$ neutrons, and the atomic mass is $15.994915\ \text{u}$.

1. The parts. $8 \times 1.007825\ \text{u} + 8 \times 1.008665\ \text{u} = 8.062600\ \text{u} + 8.069320\ \text{u} = 16.131920\ \text{u}$.
2. The mass defect. $16.131920\ \text{u} - 15.994915\ \text{u} = 0.137005\ \text{u}$.
3. The binding energy. $B = 0.137005\ \text{u} \times 931.494\ \text{MeV/u} = 127.6\ \text{MeV}$.
4. Per nucleon. $\dfrac{B}{A} = \dfrac{127.6\ \text{MeV}}{16} = 7.98\ \text{MeV}$.

Check. The answer is a little more than the $7.07$ MeV of helium-4. We shall see in the next lesson that this is what the curve shows.

Why nuclear energy is so large

When a carbon atom burns, it releases about $4\ \text{eV}$. In making helium-4 the nucleons give up $7\ \text{MeV}$ each. A nuclear change releases about a million times more energy for each particle than a chemical change, so a small mass of nuclear fuel goes a very long way. It also shows why a nuclear change can change the mass noticeably: a mass defect of $0.03\ \text{u}$ is $0.75\ \%$ of the mass of the parts, while a chemical change alters the mass by far less than one part in a billion.

Electrons in the atom are bound too, by $13.6\ \text{eV}$ for hydrogen. Compared with MeV this is so small that we ignore it in nuclear calculations.

Set the mass defect to $0.030$ u. This is close to helium-4. What energy do you read? Now double the mass defect. Does the energy double?

Practise with new numbers

Try it in Python

The program finds the mass defect, the binding energy and the binding energy per nucleon of four nuclei, from their atomic masses.

Common mistakes

• Using the mass of the nucleus and the mass of the atom together. Use atomic masses with $m_H$ for each proton, or nuclear masses with $m_p$. Never mix them.
• Forgetting the sign. The mass defect is the parts minus the nucleus. It is positive.
• Dividing by $A$ twice, or not at all. The binding energy per nucleon is $B/A$. The total is $B$.
• Leaving the answer in u. A mass defect in u is not an energy. Multiply by $931.494$ MeV/u.
• Thinking mass is lost. The mass is not destroyed. It is carried away as energy when the nucleus forms.

Remember these

• Mass defect: $\Delta m = Zm_H + Nm_n - M$, using atomic masses.
• $E = mc^2$, and $1\ \text{u}$ is equivalent to $931.494$ MeV.
• Binding energy: $B = \Delta m \times 931.494\ \text{MeV/u}$. It is the energy needed to pull the nucleus apart.
• Binding energy per nucleon: $B/A$. Helium-4 has $7.07$ MeV and oxygen-16 has $7.98$ MeV.
• Nuclear changes involve MeV. Chemical changes involve eV.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. The nucleus of deuterium has a mass defect of $0.002388$ u. What is its binding energy in MeV?

2. A nucleus has $A = 56$ and a binding energy of $492$ MeV. What is its binding energy per nucleon?

3. Why is the mass of a stable nucleus less than the sum of the masses of its protons and neutrons?

4. A mass defect of $1$ u appears when a nucleus forms. How much energy is released?

References

• NCERT Physics Part II (Class 12, rationalised edition), Chapter 13 Nuclei, on mass-energy and nuclear binding energy.
• OpenStax University Physics Volume 3, Chapter 10 Nuclear Physics, on nuclear binding and nuclear forces.
• AME2020 atomic mass evaluation (Wang and others, Chinese Physics C, 2021), for the atomic masses.