Why Nuclear Binding Energy Is Positive Despite Lower Energy

A free proton and a free neutron have a greater combined mass than the deuterium nucleus they form. The difference corresponds to approximately \(2.224\ \mathrm{MeV}\) of energy.

But here is the apparent contradiction: the bound nucleus has lower energy, yet its binding energy is quoted as positive \(2.224\ \mathrm{MeV}\).

There is no contradiction. Binding energy is defined as the positive amount of energy you must supply to separate a nucleus into its individual protons and neutrons. The bound nucleus’s energy is lower than the separated particles’ energy by exactly this amount.

In other words, the binding energy is positive, but the energy change when the nucleus forms is negative. Understanding which quantity you’re describing makes the signs straightforward.

01Why binding energy is positive when the nucleus has lower energy

The first step is to choose a reference level.

Imagine two particles that are completely separated and at rest. We assign their combined energy a reference value of zero for the purpose of comparing their nuclear binding.

When they form a bound nucleus, energy is released. The nucleus moves to a lower energy state.

For deuterium, the difference is approximately \(2.224\ \mathrm{MeV}\).

We can express this as:

\[
E_{\text{bound}}-E_{\text{separated}}=-B
\]

Here:

• \(E_{\text{bound}}\) is the energy of the bound nucleus relative to the chosen reference.
• \(E_{\text{separated}}\) is the energy of the separated nucleons.
• \(B\) is the binding energy, measured in mega-electronvolts (MeV).

The negative sign tells us that the bound nucleus has lower energy.

So why do we report \(B\) as positive?

Because binding energy describes the amount of energy required to reverse the process.

For a nucleus with binding energy \(B\):

When the nucleus forms, its energy decreases by \(B\), and energy is released to the surroundings.

When the nucleus is separated, its energy increases by \(B\), and energy must be supplied.

Both processes involve the same energy difference. Only the direction changes.

A useful analogy: a ball in a valley

Imagine a ball resting at the bottom of a valley. Choose the height of the surrounding ridge as zero gravitational potential energy.

The ball has negative gravitational potential energy relative to that reference. However, the height you must lift it to reach the ridge is positive.

You wouldn’t describe the required lifting energy as negative just because the ball starts below the reference level.

Nuclear binding energy works in a similar way. It is the positive energy needed to move from the lower-energy bound state to the separated-particle reference.

The analogy is about energy levels and reference points, not the forces involved. Nuclear binding comes from nuclear interactions, not gravity.

The important distinction: A positive binding energy does not mean the nucleus has positive energy relative to its separated constituents. It means the nucleus is bound by an energy gap of that positive magnitude.

The nucleus’s actual total rest energy is still positive. The negative value applies only to its energy relative to the chosen separated-nucleon reference.

02How physicists made the mass difference measurable

There is a practical reason this distinction matters. The difference between the mass of a bound system and the combined mass of its constituents is extremely small.

Physicists needed accurate measurements of atomic masses before they could make reliable comparisons.

An important development came from Francis Aston’s mass spectrograph, developed in 1919.

The original measurement problem

At the time, an important question was whether atoms of the same element always had exactly the same mass.

Chemical measurements could provide average atomic masses, but they could not directly distinguish the masses of individual atoms.

Aston wanted to separate atoms according to their mass-to-charge ratios.

His apparatus used positively charged ions and electromagnetic fields.

How Aston’s mass spectrograph worked

The apparatus followed a sequence.

1. A gas sample entered a low-pressure discharge tube. An electrical discharge removed electrons from some atoms, producing positive ions.

2. An electric field accelerated the positive ions along the tube.

3. The ions passed through narrow slits, producing a more tightly defined beam.

4. The beam entered an electric field between parallel plates, which deflected the ions.

5. A diaphragm selected a group of ions before they entered a magnetic field.

6. The magnetic field bent their paths in the opposite direction to the electric deflection.

7. The ions eventually reached a photographic plate, where their positions were recorded.

What should the photographic plate show?

If all atoms of an element had the same mass and charge state, their ions would produce a single mass line.

If the element contained isotopes with different masses, separate lines would appear.

Aston’s measurements revealed separate mass lines for different isotopes. Neon, for example, showed components with masses close to 20 and 22 atomic mass units.

This supported the interpretation that neon contained different isotopes rather than one type of atom with an unusual average mass.

Why the original design needed improvement

An earlier arrangement, known as the parabola method, could distinguish particles, but it was not sufficiently accurate for the close mass comparisons Aston wanted.

In his 1922 Nobel lecture, Aston noted that the best existing apparatus was reliable to only about one per cent.

He revised the arrangement using narrow parallel slits and an electromagnetic focusing principle. This produced sharper mass lines and improved the accuracy of the measurements.

His work on mass spectra and isotopes was recognised with the 1922 Nobel Prize in Chemistry.

You can read the historical account in Aston’s Nobel lecture, Mass Spectra and Isotopes, and see the Science Museum Group record of his first mass spectrograph.

What did these measurements establish?

A mass spectrograph measures the mass-to-charge ratio of ions. It does not directly measure nuclear binding energy.

However, accurate atomic masses allow physicists to compare the mass of a bound nucleus with the combined masses of its separate nucleons.

That comparison reveals the mass defect, which can then be converted into energy using Einstein’s mass-energy relationship.

For perspective, the mass difference in deuterium is only about \(0.118\%\) of the combined mass of a neutral hydrogen atom and a neutron.

That is approximately one part in 845.

A very small mass difference can still represent a substantial amount of energy because the conversion involves the square of the speed of light.

03Keep nuclear mass, atomic mass, and binding energy separate

These quantities are related, but they describe different things.

The distinction between nuclear mass and atomic mass is particularly important in calculations.

Nuclear mass

A nucleus contains protons and neutrons, collectively called nucleons.

For a nucleus written as

\[
{}^{A}_{Z}X
\]

• \(A\) is the nucleon number.
• \(Z\) is the proton number.
• \(N=A-Z\) is the neutron number.

If you are given the nuclear mass, use the proton and neutron masses to calculate the mass defect:

\[
\Delta m=Zm_p+Nm_n-m_{\text{nucleus}}
\]

Here, \(m_p\) is the mass of a proton, \(m_n\) is the mass of a neutron, and \(m_{\text{nucleus}}\) is the measured nuclear mass.

The subtraction order is important. The combined mass of the separate nucleons comes first.

Atomic mass

An atomic mass includes the nucleus and its surrounding electrons.

For a neutral atom containing \(Z\) protons, there are also \(Z\) electrons.

Therefore, approximately:

\[
m_{\text{atom}}=m_{\text{nucleus}}+Zm_e
\]

where \(m_e\) is the electron mass.

If you are given neutral atomic masses, use the mass of a neutral hydrogen atom instead of the bare proton mass:

\[
\Delta m=Zm_H+Nm_n-m_{\text{atom}}
\]

Here, \(m_H\) is the mass of a neutral hydrogen atom.

The electron contributions cancel because both sides contain the same number of electrons. Small differences in electronic binding energies are neglected at normal HSC calculation precision.

A common mistake is to combine proton masses with neutral atomic masses. This introduces an electron-mass error into the calculation.

For a more detailed treatment of the calculation methods, see How to Calculate Binding Energy per Nucleon from Nuclear Mass.

04A nucleus’s binding energy is not the same as a reaction’s released energy

So far, we’ve considered one nucleus and the energy needed to separate its nucleons.

A nuclear reaction is a different calculation. It involves comparing the complete initial and final systems.

The energy released by a reaction is commonly represented by \(Q\):

\[
Q=(m_{\text{initial}}-m_{\text{final}})c^2
\]

Here:

• \(m_{\text{initial}}\) is the total initial rest mass.
• \(m_{\text{final}}\) is the total final rest mass.
• \(Q\) is the energy released, using the convention that positive \(Q\) means energy leaves the reaction’s rest-mass energy.

Therefore:

• \(Q>0\): Energy is released.
• \(Q<0\): Energy must be supplied.

For reactions in which the total numbers of protons and neutrons are separately conserved, we can also write:

\[
Q=B_{\text{final}}-B_{\text{initial}}
\]

This is because the total masses of the free constituent nucleons cancel when we compare the two sides.

The important point is that every nucleus can have a positive binding energy while the overall reaction still requires an energy input.

05The contrasting case: A bound nucleus can absorb energy

Consider the reverse of nuclear binding.

A deuterium nucleus can be separated into a proton and a neutron through photodisintegration:

\[
{}^{2}_{1}\mathrm{H}+\gamma
\rightarrow
{}^{1}_{1}\mathrm{p}
+
{}^{1}_{0}\mathrm{n}
\]

Here, \(\gamma\) represents a gamma-ray photon.

Check conservation:

For nucleon number:

\[
2=1+1
\]

For nuclear charge:

\[
1=1+0
\]

Both quantities are conserved. The photon carries neither nucleon number nor electric charge.

The deuterium nucleus has a binding energy of \(2.224\ \mathrm{MeV}\).

To separate it completely, energy must be supplied to overcome this binding.

In the simplified energy-level model, the change in the nuclear system’s energy is:

\[
\Delta E_{\text{system}}=+2.224\ \mathrm{MeV}
\]

The energy transfer is into the nuclear system.

For an actual photon interacting with a stationary deuteron, the threshold photon energy is slightly greater than the binding energy because the outgoing particles must also carry momentum. The \(2.224\ \mathrm{MeV}\) value describes the binding-energy gap in the simplified energy accounting.

This example exposes the misconception that positive binding energy means energy is released whenever a nucleus changes.

It does not.

The positive number tells you the energy required to undo the binding. The direction of the process determines whether energy is released or absorbed.

06Questions and solutions

Question 1

A neutral tritium atom, \({}^{3}_{1}\mathrm{H}\), has an atomic mass of \(3.016049\ \mathrm{u}\).

Use the following data:

• Hydrogen-1 atomic mass: \(1.007825\ \mathrm{u}\)
• Neutron mass: \(1.008665\ \mathrm{u}\)
• \(1\ \mathrm{u}\,c^2=931.5\ \mathrm{MeV}\)

Calculate the total binding energy of the tritium nucleus. Explain why the result is positive.

Solution 1

The total binding energy of tritium is approximately \(8.48\ \mathrm{MeV}\). It is positive because this is the energy required to separate its one proton and two neutrons completely.

Step 1: Determine the number of protons and neutrons.

For tritium:

\[
A=3,\qquad Z=1
\]

Therefore:

\[
N=3-1=2
\]

Step 2: Select the correct mass expression.

The question gives an atomic mass, so:

\[
\Delta m=m_H+2m_n-m_T
\]

Step 3: Substitute the supplied values.

\[
\begin{aligned}
\Delta m
&=1.007825+2(1.008665)\\
&\quad-3.016049\\
&=0.009106\ \mathrm{u}
\end{aligned}
\]

Step 4: Convert the mass defect into energy.

\[
\begin{aligned}
B&=0.009106\times931.5\\
&=8.4822\ \mathrm{MeV}
\end{aligned}
\]

Therefore:

\[
B\approx8.48\ \mathrm{MeV}
\]

The positive result means energy must be supplied to completely separate the tritium nucleus.

The signed energy change when its free nucleons form tritium is instead approximately \(-8.48\ \mathrm{MeV}\).

Question 2

The binding energy of a deuterium nucleus is \(+2.224\ \mathrm{MeV}\).

A second source describes the energy of the bound deuterium nucleus as \(-2.224\ \mathrm{MeV}\), relative to a separated proton and neutron at rest.

A student claims that the two descriptions cannot both be correct because the same nucleus cannot have both positive and negative energy.

Explain why the two descriptions are consistent.

Then state:

(a) The signed energy change of the nuclear system when a free proton and neutron form deuterium.

(b) The energy transferred to the surroundings during formation.

(c) The signed energy change of the nuclear system when deuterium is completely separated.

(d) The direction of energy transfer during separation.

Assume that the only energy change being considered is the nuclear binding-energy change.

Solution 2

Both descriptions are correct. The binding energy is the positive magnitude of the energy required for separation, whereas the negative value describes the bound system’s energy relative to the separated-particle reference.

Step 1: Identify the reference energy.

The separated proton and neutron have a relative energy of zero:

\[
E_{\text{separated}}=0
\]

The bound deuterium nucleus has:

\[
E_{\text{bound}}=-2.224\ \mathrm{MeV}
\]

The negative sign means its energy is lower than the reference level.

Step 2: Consider formation.

The change in the nuclear system’s energy is:

\[
\begin{aligned}
\Delta E_{\text{system}}
&=E_{\text{bound}}-E_{\text{separated}}\\
&=-2.224-0\\
&=-2.224\ \mathrm{MeV}
\end{aligned}
\]

Therefore, the nuclear system loses \(2.224\ \mathrm{MeV}\).

That energy is transferred to the surroundings:

\[
E_{\text{released}}=+2.224\ \mathrm{MeV}
\]

Step 3: Reverse the process.

Complete separation takes the nucleus from \(-2.224\ \mathrm{MeV}\) back to zero.

\[
\begin{aligned}
\Delta E_{\text{system}}
&=0-(-2.224)\\
&=+2.224\ \mathrm{MeV}
\end{aligned}
\]

The nuclear system gains energy, so energy must be supplied.

Step 4: Identify the student’s mistake.

The student has confused the positive magnitude of the binding energy with the signed energy of the bound state.

These are different quantities.

The binding energy remains positive because it measures the size of the energy gap. The relative energy of the bound state is negative because it lies below the chosen reference level.

Question 3

A neutral helium-4 atom has an atomic mass of \(4.002603\ \mathrm{u}\).

Use the following rounded values:

A student calculates the mass defect using:

\[
\Delta m=2m_p+2m_n-m_{\text{He atom}}
\]

(a) Explain the mistake.

(b) Calculate the correct mass defect and total binding energy of helium-4.

(c) Calculate the binding energy obtained using the student’s incorrect expression.

(d) Explain the difference between the two answers.

Solution 3

The correct binding energy is approximately \(28.30\ \mathrm{MeV}\). The student’s method gives approximately \(27.27\ \mathrm{MeV}\) because it mixes bare proton masses with the mass of a neutral atom.

Step 1: Identify the error.

The supplied helium-4 mass is an atomic mass, so it includes two electrons.

However, the student’s expression uses the masses of two bare protons, which contain no electrons.

The electron contributions therefore do not cancel.

Step 2: Use the correct atomic-mass expression.

Helium-4 contains:

\[
Z=2,\qquad N=2
\]

Therefore:

\[
\Delta m=2m_H+2m_n-m_{\text{He atom}}
\]

Step 3: Calculate the correct mass defect.

\[
\begin{aligned}
\Delta m
&=2(1.007825)+2(1.008665)\\
&\quad-4.002603\\
&=4.032980-4.002603\\
&=0.030377\ \mathrm{u}
\end{aligned}
\]

Step 4: Calculate the correct binding energy.

\[
\begin{aligned}
B&=0.030377\times931.5\\
&=28.2962\ \mathrm{MeV}
\end{aligned}
\]

Therefore:

\[
B\approx28.30\ \mathrm{MeV}
\]

Step 5: Calculate the student’s incorrect result.

Using the incorrect expression:

\[
\begin{aligned}
\Delta m_{\text{wrong}}
&=2(1.007276)+2(1.008665)\\
&\quad-4.002603\\
&=0.029279\ \mathrm{u}
\end{aligned}
\]

The corresponding binding energy is:

\[
\begin{aligned}
B_{\text{wrong}}
&=0.029279\times931.5\\
&=27.2734\ \mathrm{MeV}
\end{aligned}
\]

Therefore:

\[
B_{\text{wrong}}\approx27.27\ \mathrm{MeV}
\]

Step 6: Explain the difference.

The error is:

\[
\begin{aligned}
\Delta B
&=28.2962-27.2734\\
&=1.0228\ \mathrm{MeV}
\end{aligned}
\]

This corresponds to the mass-energy equivalent of approximately two electron masses.

The student’s answer is too small because the two electron masses included in the neutral helium atom were not included in the constituent-mass total.

The correct approach is to use either:

• Bare proton and neutron masses with the bare nuclear mass, or
• Hydrogen atom and neutron masses with the neutral atomic mass.

Never mix the two methods.

Question 4

Consider the nuclear reaction:

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

Use the following rounded neutral atomic masses:

(a) Verify conservation of nucleon number and nuclear charge.

(b) Calculate the reaction’s \(Q\)-value.

(c) Determine whether energy is released or absorbed.

(d) Explain how the reaction can require energy even though both deuterium product nuclei have positive binding energies.

Solution 4

The reaction requires approximately \(3.27\ \mathrm{MeV}\) of energy input. Both deuterium nuclei are individually bound, but their combined binding energy is smaller than the initial helium-3 nucleus’s binding energy.

Step 1: Check nucleon number conservation.

The initial nucleon number is:

\[
A_i=3+1=4
\]

The final nucleon number is:

\[
A_f=2+2=4
\]

Therefore, nucleon number is conserved.

Step 2: Check nuclear charge conservation.

The initial nuclear charge is:

\[
Z_i=2+0=2
\]

The final nuclear charge is:

\[
Z_f=1+1=2
\]

Nuclear charge is also conserved.

Step 3: Calculate the total initial mass.

\[
\begin{aligned}
m_i&=3.016029+1.008665\\
&=4.024694\ \mathrm{u}
\end{aligned}
\]

Step 4: Calculate the total final mass.

There are two deuterium atoms:

\[
\begin{aligned}
m_f&=2(2.014102)\\
&=4.028204\ \mathrm{u}
\end{aligned}
\]

Step 5: Calculate the mass difference.

\[
\begin{aligned}
\Delta m&=m_i-m_f\\
&=4.024694-4.028204\\
&=-0.003510\ \mathrm{u}
\end{aligned}
\]

Notice that the mass difference is negative.

The products have greater total rest mass than the reactants.

Step 6: Calculate the reaction energy.

\[
\begin{aligned}
Q&=-0.003510\times931.5\\
&=-3.2696\ \mathrm{MeV}
\end{aligned}
\]

Therefore:

\[
Q\approx-3.27\ \mathrm{MeV}
\]

The negative sign means approximately \(3.27\ \mathrm{MeV}\) must be supplied.

Step 7: Explain why positive product binding energies do not guarantee energy release.

Each deuterium nucleus has a binding energy of approximately:

\[
B_D=2.224\ \mathrm{MeV}
\]

The total binding energy of the two products is:

\[
B_f=2(2.224)=4.448\ \mathrm{MeV}
\]

The initial helium-3 nucleus has a total binding energy of approximately \(7.718\ \mathrm{MeV}\).

The neutron has zero nuclear binding energy.

Therefore:

\[
\begin{aligned}
Q&=B_f-B_i\\
&=4.448-7.718\\
&=-3.270\ \mathrm{MeV}
\end{aligned}
\]

This agrees with the mass calculation.

The tempting mistake is to reason that two bound product nuclei must mean energy has been released.

However, the initial helium-3 nucleus was more tightly bound overall than the two deuterium nuclei combined.

The reaction therefore moves to a higher total energy state and requires an energy input.

Question 5

Consider the following hypothetical nuclear reaction:

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

The excited helium-3 nucleus subsequently emits a gamma photon and returns to its ground state:

\[
{}^{3}_{2}\mathrm{He}^{*}
\rightarrow
{}^{3}_{2}\mathrm{He}
+
\gamma
\]

Use these values:

• Ground-state binding energy of deuterium: \(2.224\ \mathrm{MeV}\) per nucleus.
• Ground-state binding energy of helium-3: \(7.718\ \mathrm{MeV}\).
• Excitation energy of the hypothetical helium-3 state: \(4.00\ \mathrm{MeV}\).

For this problem, assume the hypothetical excited helium-3 state is a genuine bound state below every particle-emission threshold. This is an illustrative energy-level model, not a claim about an experimentally established helium-3 excited level.

Ignore the small recoil energy associated with gamma emission.

Calculate:

(a) The binding energy of the excited helium-3 nucleus.

(b) The total initial binding energy of the two deuterium nuclei.

(c) The \(Q\)-value for the formation of the excited helium-3 nucleus and neutron.

(d) The energy released when the excited helium-3 nucleus emits its gamma photon.

(e) The net energy released across both stages.

Finally, explain how the first reaction can require energy even though the excited helium-3 nucleus has a positive binding energy.

Solution 5

The first stage absorbs \(0.730\ \mathrm{MeV}\), while the gamma emission releases \(4.00\ \mathrm{MeV}\). The complete two-stage process releases a net \(3.27\ \mathrm{MeV}\).

The excited helium-3 nucleus still has positive binding energy, but that does not guarantee that forming it from two already-bound deuterium nuclei releases energy.

Step 1: Check conservation of nucleon number and nuclear charge.

For nucleon number:

\[
2+2=3+1=4
\]

For nuclear charge:

\[
1+1=2+0=2
\]

Both quantities are conserved.

Step 2: Calculate the binding energy of excited helium-3.

The excited state is \(4.00\ \mathrm{MeV}\) above the ground state.

Its energy is therefore higher, meaning its binding energy is lower by the same amount.

\[
\begin{aligned}
B_{\mathrm{He3}^{*}}
&=B_{\mathrm{He3}}-E_{\mathrm{exc}}\\
&=7.718-4.000\\
&=3.718\ \mathrm{MeV}
\end{aligned}
\]

The excited nucleus still has positive binding energy:

\[
B_{\mathrm{He3}^{*}}=3.718\ \mathrm{MeV}
\]

Under the stated assumption, it remains bound.

Its rest mass is also greater than the ground-state mass by an amount equivalent to \(4.00\ \mathrm{MeV}\).

Step 3: Calculate the initial total binding energy.

There are two deuterium nuclei, each with a binding energy of \(2.224\ \mathrm{MeV}\).

Therefore:

\[
\begin{aligned}
B_i&=2(2.224)\\
&=4.448\ \mathrm{MeV}
\end{aligned}
\]

The incoming neutron has zero binding energy.

Step 4: Calculate the final total binding energy after the first reaction.

The products are excited helium-3 and a free neutron.

Therefore:

\[
B_f=3.718+0
\]

\[
B_f=3.718\ \mathrm{MeV}
\]

Step 5: Calculate the first reaction’s \(Q\)-value.

Using:

\[
Q=B_f-B_i
\]

We obtain:

\[
\begin{aligned}
Q_1&=3.718-4.448\\
&=-0.730\ \mathrm{MeV}
\end{aligned}
\]

The first reaction is endothermic.

It requires \(0.730\ \mathrm{MeV}\) of additional energy, which must come from the available energy of the incoming system or an external source.

This is the important result.

Although the excited helium-3 nucleus has positive binding energy, the complete product arrangement has less total binding energy than the two incoming deuterium nuclei.

Step 6: Calculate the energy released during gamma emission.

The excited helium-3 nucleus returns to its ground state.

The energy difference between these states is:

\[
E_{\gamma}=4.00\ \mathrm{MeV}
\]

Therefore, approximately \(4.00\ \mathrm{MeV}\) is released as gamma radiation.

Step 7: Calculate the net energy released.

The first stage absorbs \(0.730\ \mathrm{MeV}\), while the second stage releases \(4.00\ \mathrm{MeV}\).

The net energy released is:

\[
\begin{aligned}
Q_{\text{net}}
&=Q_1+Q_2\\
&=-0.730+4.000\\
&=+3.270\ \mathrm{MeV}
\end{aligned}
\]

Therefore:

\[
Q_{\text{net}}=+3.27\ \mathrm{MeV}
\]

We can check this result by considering the final ground-state products directly:

\[
\begin{aligned}
Q_{\text{net}}
&=B_{\mathrm{He3}}-2B_D\\
&=7.718-2(2.224)\\
&=3.270\ \mathrm{MeV}
\end{aligned}
\]

Both approaches agree.

Step 8: Explain the conceptual trap.

The tempting argument is:

“The excited helium-3 nucleus has positive binding energy, so forming it must release energy.”

But positive binding energy only tells us that the nucleus is at a lower energy than its completely separated constituent nucleons.

It does not tell us whether it is lower in energy than the particular reactants used to form it.

The initial system already contains \(4.448\ \mathrm{MeV}\) of binding energy. The excited product contains only \(3.718\ \mathrm{MeV}\).

Forming that excited product therefore requires energy.

When it subsequently emits a \(4.00\ \mathrm{MeV}\) gamma photon, it releases the excitation energy it was carrying.

The net result is a release of \(3.27\ \mathrm{MeV}\).

This is why you must distinguish three things:

• The positive binding energy of an individual nucleus.
• The signed energy change when that nucleus forms or separates.
• The net energy released by an entire nuclear reaction.

They are related, but they are not interchangeable.

07Where to go next

Once the sign convention is clear, the next step is comparing the total binding energies of different nuclei.

This is the reasoning behind the nuclear binding-energy curve. It explains why some fusion and fission reactions release energy while others require an input.

The key decision remains the same: compare the total binding energy before and after the change, rather than assuming that a positive binding energy automatically means energy will be released.

Continue with How to Read the Nuclear Binding-Energy Curve and Iron Limit.