KAHIBARO
Table of Contents
Core Idea
A nucleus is made of protons and neutrons, which are together called nucleons. If we simply add the masses of all the separate nucleons, we might expect that this sum would equal the mass of the nucleus they form. But experiment shows something surprising. The mass of a bound nucleus is actually smaller than the total mass of its free protons and neutrons.
This missing mass is called the mass defect.
The mass defect is one of the clearest signs that mass and energy are deeply connected in nuclear physics. When nucleons bind together, energy is released, and the mass of the final nucleus becomes smaller by an amount corresponding to that released energy.
Definition of Mass Defect
Consider a nucleus with atomic number $Z$ and mass number $A$. It contains $Z$ protons and $A - Z$ neutrons.
If $m_p$ is the mass of a proton, $m_n$ is the mass of a neutron, and $m_{\text{nucleus}}$ is the actual measured mass of the nucleus, then the mass defect is
$$
\Delta m = Z m_p + (A - Z)m_n - m_{\text{nucleus}}
$$
This quantity is positive for a bound nucleus.
The mass defect is the difference between the sum of the masses of the separated nucleons and the actual mass of the bound nucleus:
$$
\Delta m = Z m_p + (A - Z)m_n - m_{\text{nucleus}}
$$
A bound nucleus has less mass than its separated parts.
Why Mass Is Missing
The mass is not really lost. Instead, it has been converted into energy released when the nucleus was formed. This is a direct application of Einstein's relation
$$
E = mc^2
$$
When protons and neutrons come together and form a stable nucleus, nuclear forces bind them. During this process, energy is given off, often as gamma radiation or as kinetic energy of emitted particles. Because energy leaves the system, the mass of the final bound nucleus is reduced.
So the mass defect corresponds to the binding energy of the nucleus, which is the energy required to completely separate the nucleus back into free nucleons.
Mass defect and binding energy are connected by
$$
E_b = \Delta m \, c^2
$$
where $E_b$ is the binding energy.
A Simple Physical Picture
Imagine bringing several nucleons together from far apart. As they bind, the system moves to a lower energy state. A lower energy state means a lower total mass-energy. The nucleus is therefore lighter than the unbound collection of particles.
This is similar in spirit to many bound systems in physics. A bound system has less total energy than its separated parts. In nuclei, because the energy changes are very large, the effect on mass is measurable.
Using Nuclear Masses and Atomic Masses
In practice, physicists often use tabulated masses. There is an important detail here. A neutral atom includes electrons, while a nucleus does not. So when calculating mass defect, you must be careful to use masses consistently.
There are two common approaches. One uses nuclear masses directly. The other uses atomic masses, but then all electron masses must be handled properly. For many nuclear calculations, the atomic mass unit is especially convenient, but the main idea of mass defect stays the same.
The exact bookkeeping with electrons belongs to broader nuclear mass calculations, but the key point here is consistency. You must compare like with like.
When calculating a mass defect, always use a consistent set of masses. Do not mix nuclear masses and atomic masses carelessly.
Example Structure of a Calculation
Suppose a nucleus contains $Z$ protons and $A-Z$ neutrons.
First, calculate the mass of the separated nucleons:
$$
m_{\text{separate}} = Zm_p + (A-Z)m_n
$$
Then subtract the measured nuclear mass:
$$
\Delta m = m_{\text{separate}} - m_{\text{nucleus}}
$$
Finally, convert this mass defect into energy:
$$
E_b = \Delta m c^2
$$
If masses are expressed in atomic mass units, then the conversion
$$
1\,\text{u} \approx 931.5\,\text{MeV}/c^2
$$
is very useful, so
$$
E_b(\text{MeV}) \approx \Delta m(\text{u}) \times 931.5
$$
Useful conversion:
$$
1\,\text{u} \approx 931.5\,\text{MeV}/c^2
$$
Therefore,
$$
E_b(\text{MeV}) \approx \Delta m(\text{u}) \times 931.5
$$
Numerical Illustration
Take a simple symbolic example. Suppose the sum of the free nucleon masses is
$$
m_{\text{separate}} = 10.085\,\text{u}
$$
and the actual nuclear mass is
$$
m_{\text{nucleus}} = 10.012\,\text{u}
$$
Then the mass defect is
$$
\Delta m = 10.085 - 10.012 = 0.073\,\text{u}
$$
The corresponding binding energy is
$$
E_b \approx 0.073 \times 931.5 \,\text{MeV}
$$
$$
E_b \approx 68.0\,\text{MeV}
$$
This means that $68.0\,\text{MeV}$ of energy would be required to pull the nucleus completely apart into separate nucleons.
Sign of the Mass Defect
For a stable bound nucleus, the actual nuclear mass is less than the sum of the masses of its free nucleons, so the mass defect is positive.
If there were no binding, there would be no mass defect. A positive mass defect tells us that the nucleons are in a lower-energy bound state.
This is why heavier nuclei are not just simple sums of their parts. Their internal energy matters.
Mass Defect and Nuclear Stability
Mass defect is closely related to nuclear stability, because a larger binding energy usually means a more strongly bound nucleus. However, the best way to compare nuclei of different sizes is not just by total binding energy, but by binding energy per nucleon, which is treated separately.
For the present chapter, the important idea is this: mass defect is the measurable mass difference that reveals how much energy is associated with nuclear binding.
Summary Table
| Quantity | Meaning | Formula |
|---|---|---|
| Sum of free nucleon masses | Mass before binding | $Zm_p + (A-Z)m_n$ |
| Nuclear mass | Actual bound mass | $m_{\text{nucleus}}$ |
| Mass defect | Missing mass due to binding | $\Delta m = Zm_p + (A-Z)m_n - m_{\text{nucleus}}$ |
| Binding energy | Energy equivalent of mass defect | $E_b = \Delta m c^2$ |
Final Insight
Mass defect shows that a nucleus is not just a pile of protons and neutrons. It is a bound system with lower total mass-energy than its separated components. The difference in mass is small in ordinary units, but it corresponds to a very large amount of energy on the nuclear scale.
A nucleus has less mass than the sum of its free nucleons because energy is released when the nucleus forms.
$$
\Delta m > 0 \quad \Rightarrow \quad E_b = \Delta m c^2
$$
Mass defect is direct evidence of nuclear binding.