KAHIBARO
Table of Contents
Why this quantity matters
Binding energy per nucleon tells us how strongly, on average, the nucleons in a nucleus are held together. A nucleus contains protons and neutrons, and the total binding energy is the energy needed to separate the whole nucleus into free nucleons. When we divide that total by the number of nucleons, we get the binding energy per nucleon.
If a nucleus has mass number $A$, then
$$
\text{binding energy per nucleon} = \frac{B}{A}
$$
where $B$ is the total binding energy.
This quantity is useful because nuclei can have very different sizes. Comparing only total binding energy can be misleading, since a large nucleus usually has more total binding energy simply because it has more nucleons. Dividing by $A$ gives a fairer comparison of how tightly packed and stable the nucleons are on average.
The binding energy per nucleon is
$$
\frac{B}{A}
$$
It is the average energy required to remove one nucleon from a nucleus, in an approximate sense.
Physical meaning
A larger binding energy per nucleon usually means a more tightly bound nucleus. That means the nucleons are, on average, harder to pull apart. So this quantity is closely related to nuclear stability.
However, it is important to understand the word average. Not every proton or neutron in a nucleus is bound by exactly the same amount. Some nucleons may be more tightly bound than others, especially those near the surface of the nucleus. So $\frac{B}{A}$ is a summary number, not a detailed description of each individual nucleon.
For light nuclei, the binding energy per nucleon tends to be smaller. As nuclei become larger, this quantity generally rises, reaches a maximum for medium mass nuclei, and then slowly decreases for very heavy nuclei.
How to calculate it
To find binding energy per nucleon, first find the total binding energy $B$, then divide by the mass number $A$.
If the mass defect is $\Delta m$, then
$$
B = \Delta m c^2
$$
and therefore
$$
\frac{B}{A} = \frac{\Delta m c^2}{A}
$$
If $B$ is given in MeV, then $\frac{B}{A}$ is usually also expressed in MeV per nucleon.
If total binding energy is known, use
$$
\frac{B}{A} = \frac{\text{total binding energy}}{\text{mass number}}
$$
If mass defect is known, use
$$
\frac{B}{A} = \frac{\Delta m c^2}{A}
$$
Example
Consider a nucleus with mass number $A = 4$ and total binding energy $B = 28.3\ \text{MeV}$.
Then
$$
\frac{B}{A} = \frac{28.3\ \text{MeV}}{4} = 7.08\ \text{MeV per nucleon}
$$
This means that, on average, each nucleon contributes about $7.08\ \text{MeV}$ to the binding of the nucleus.
Now consider a larger nucleus with $A = 56$ and total binding energy about $492\ \text{MeV}$.
$$
\frac{B}{A} = \frac{492\ \text{MeV}}{56} \approx 8.79\ \text{MeV per nucleon}
$$
This larger value tells us that this nucleus is more tightly bound on average than the first one.
Trend across nuclei
The binding energy per nucleon does not stay constant across all elements. Its general pattern is very important in nuclear physics.
For very light nuclei, $\frac{B}{A}$ is relatively small. As $A$ increases, $\frac{B}{A}$ rises quickly. It reaches its highest values for nuclei in the middle of the periodic table, especially near iron and nickel. For heavier nuclei, it decreases slowly.
This trend explains why energy can be released in two different ways. Light nuclei can release energy by joining together, because the product nucleus often has a higher binding energy per nucleon. Very heavy nuclei can release energy by splitting apart, because the fragments often have higher binding energy per nucleon than the original heavy nucleus.
Nuclei near iron and nickel have the highest binding energy per nucleon and are among the most stable nuclei.
Simple comparison table
| Nucleus | Mass number $A$ | Approximate total binding energy $B$ | Approximate $B/A$ |
|---|---|---|---|
| Deuterium | 2 | $2.2\ \text{MeV}$ | $1.1\ \text{MeV/nucleon}$ |
| Helium 4 | 4 | $28.3\ \text{MeV}$ | $7.1\ \text{MeV/nucleon}$ |
| Iron 56 | 56 | $492\ \text{MeV}$ | $8.8\ \text{MeV/nucleon}$ |
| Uranium 238 | 238 | about $1800\ \text{MeV}$ | about $7.6\ \text{MeV/nucleon}$ |
This table shows the rise, peak, and slow decline of binding energy per nucleon.
Stability and energy release
Suppose one nuclear reaction turns a less tightly bound arrangement into a more tightly bound one. Then the final nuclei have a larger binding energy per nucleon, and energy is released.
A useful idea is this. Systems tend to move toward lower total energy. A nucleus with higher binding energy per nucleon is generally in a more strongly bound, lower energy state.
That is why fusion of light nuclei and fission of very heavy nuclei can both release energy. In both cases, the products move toward the region of larger $\frac{B}{A}$.
A visual idea
The graph of binding energy per nucleon versus mass number has a characteristic shape. It rises steeply at small $A$, reaches a broad maximum around medium mass nuclei, and then falls slowly for large $A$.
This drawing is only schematic. It shows the general behavior, not exact numerical values.
Important caution
Binding energy per nucleon is a very helpful guide, but it does not tell everything about a nucleus. Two nuclei with similar values of $\frac{B}{A}$ may still behave differently in nuclear reactions. It is best used as a broad measure of average nuclear stability.
A higher binding energy per nucleon usually means a more stable, more tightly bound nucleus, but it is an average measure, not a complete description of nuclear behavior.
Summary
Binding energy per nucleon is the total binding energy divided by the number of nucleons:
$$
\frac{B}{A}
$$
It measures how tightly bound a nucleus is on average. Higher values usually mean greater stability. This quantity increases from light nuclei to a maximum near iron and nickel, then decreases slowly for very heavy nuclei. That simple trend helps explain why fusion of light nuclei and fission of heavy nuclei can both release energy.