Why Total and Per-Nucleon Binding Energy Give Different Rankings

A uranium-238 nucleus has a total binding energy of approximately 1,802 MeV. An iron-56 nucleus has a total binding energy of only about 492 MeV. Yet iron-56 has the higher binding energy per nucleon: approximately 8.79 MeV per nucleon, compared with uranium-238’s 7.57 MeV per nucleon.

Both rankings are correct. The difference is that total binding energy measures the energy needed to separate every nucleon in a nucleus, while binding energy per nucleon measures the average energy per nucleon. A large nucleus can have a greater total binding energy simply because it contains many more nucleons, even if its nucleons are less tightly bound on average.

01The two rankings measure different things

Imagine comparing two basketball teams. One team has 12 players and scores 120 points. Another has five players and scores 75 points.

The first team has the higher total score, but the second has the higher average score per player.

Nuclear binding energy works in a similar way.

• Total binding energy, \(B\), measures the energy required to completely separate a nucleus into its individual protons and neutrons.
• Binding energy per nucleon, \(B/A\), measures the average binding energy associated with each nucleon.

Here, \(A\) is the mass number, which is the total number of protons and neutrons.

The relationship is:

\[ \boxed{\text{Binding energy per nucleon}=\frac{B}{A}} \]

The analogy is useful for understanding the mathematics, but nucleons are not independent players. They interact through the nuclear force, and their average binding depends on the structure and composition of the nucleus.

A comparison of real nuclei

Consider the following approximate binding energies for four isotopes.

These are approximate reference values, rounded to illustrate the differences. The values are calculated from measured isotope masses, rather than average atomic weights. Reference isotope masses are available from NIST’s atomic mass tables.

Now rank the nuclei using each measurement.

Ranking by total binding energy:

\[ \mathrm{U\text{-}238 > Fe\text{-}56 > C\text{-}12 > He\text{-}4} \]

Ranking by binding energy per nucleon:

\[ \mathrm{Fe\text{-}56 > C\text{-}12 > U\text{-}238 > He\text{-}4} \]

Notice the difference.

Uranium-238 has a much greater total binding energy than iron-56. However, each nucleon in iron-56 is more tightly bound on average.

This is exactly why the two measurements can produce different rankings.

There is one useful mathematical observation here.

If two nuclei have the same mass number, their total binding energy and binding energy per nucleon must give the same ranking. Both values are divided by the same number.

The rankings can reverse when comparing nuclei with different mass numbers.

02How physicists discovered the mass differences

Before physicists could compare nuclear binding energies, they needed a way to measure the masses of different isotopes accurately.

This was not straightforward.

The mass of an atom is close to a whole-number multiple of the atomic mass unit, but the measured values do not follow that pattern exactly. Why?

The answer became clearer through the development of the mass spectrograph.

Aston’s mass spectrograph

In 1919, physicist Francis Aston developed a mass spectrograph that could separate isotopes according to their mass-to-charge ratios.

The apparatus worked in a sequence.

1. An ion source produced positively charged ions from a sample.
2. An electric field accelerated the ions into a narrow beam.
3. Slits restricted the beam so that the ions travelled along controlled paths.
4. Electric and magnetic fields deflected the ions.
5. The ions reached a photographic plate, where their positions produced separate lines.

For ions with the same charge, differences in mass produce differences in their trajectories. Lighter ions are deflected more than heavier ions under equivalent conditions.

What should the instrument show if every atom of an element has exactly the same mass?

It should produce one mass line.

Instead, Aston observed separate lines for different isotopes. This provided experimental evidence that elements could contain atoms with different masses.

The first instrument had limited resolving power. Aston’s later work, published in 1927, described a more powerful instrument designed to improve the resolution of heavier isotope masses and measure small departures from whole-number masses.

You can read his original account in Nature’s 1927 publication, Atoms and their Packing Fractions.

What did the mass measurements reveal?

The measured mass of a nucleus was smaller than the combined mass of its separate protons and neutrons.

This difference is called the mass defect.

Einstein’s mass-energy relationship,

\[ E=mc^2 \]

provided the explanation.

When separate protons and neutrons combine to form a bound nucleus, energy is released. The resulting bound system has less rest mass than the original collection of separated particles.

The mass difference corresponds to the energy released during formation.

The reverse process requires the same amount of energy: completely separating the nucleus into its individual nucleons.

The experiment measured isotope masses, not binding energy directly. Binding energy was calculated from those measurements using mass-energy equivalence.

That distinction is important when interpreting experimental evidence.

03From nuclear mass to total binding energy

To understand why the rankings differ, we need to be clear about what each mass represents.

A nucleus contains protons and neutrons, collectively called nucleons.

For a nucleus written as:

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

• \(A\) is the mass number (total nucleons).
• \(Z\) is the atomic number (protons).
• \(N\) is the number of neutrons.

The number of neutrons is:

\[ N=A-Z \]

There are four related quantities that you should keep separate.

Calculating mass defect from nuclear mass

If a question gives you the mass of the nucleus itself, compare it with the combined masses of its separate protons and neutrons.

The mass defect is:

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

Where:

• \(m_p\) is the mass of one proton, in u.
• \(m_n\) is the mass of one neutron, in u.
• \(m_{\mathrm{nucleus}}\) is the measured nuclear mass, in u.

Our sign convention makes the mass defect positive for a bound nucleus.

Calculating mass defect from atomic mass

There is an important difference if the question supplies the mass of a neutral atom rather than its nucleus.

A neutral atom contains \(Z\) electrons. We therefore need to account for their masses.

A convenient approach is to use the mass of a neutral hydrogen atom instead of a proton.

The mass of a neutral hydrogen atom includes one proton and one electron. Using \(Z\) hydrogen atoms therefore accounts for the same number of electrons as the neutral atom being investigated.

The expression becomes:

\[ \Delta m=Zm_H+Nm_n-M_{\mathrm{atom}} \]

Where:

• \(m_H\) is the mass of one neutral hydrogen atom.
• \(M_{\mathrm{atom}}\) is the measured atomic mass of the isotope.

At HSC calculation precision, small differences in electronic binding energies are neglected.

Never combine neutral atomic masses with bare proton masses without correcting for the electrons.

Converting mass defect into binding energy

Once the mass defect is known, use Einstein’s mass-energy relationship:

\[ B=\Delta mc^2 \]

If the mass defect is in atomic mass units, the convenient conversion is:

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

Therefore:

\[ \boxed{B=\Delta m\times931.5} \]

The result is the total binding energy in MeV per nucleus.

To calculate binding energy per nucleon, divide by the mass number:

\[ \boxed{\frac{B}{A}=\frac{\Delta m\times931.5}{A}} \]

The result is measured in MeV per nucleon.

For a more detailed treatment of choosing the correct mass formula, see How to Calculate Binding Energy per Nucleon from Nuclear Mass.

04Why does binding energy per nucleon peak around iron?

There is another reason why the two rankings behave differently.

The total binding energy generally increases as the number of nucleons increases. However, the average binding energy per nucleon does not increase indefinitely.

To understand this, consider the two important interactions inside a nucleus.

The strong nuclear force

The strong nuclear force attracts nucleons and holds the nucleus together.

It is extremely powerful, but it acts over a very short range. Each nucleon interacts strongly with nearby nucleons rather than gaining an unlimited amount of additional attraction from every nucleon in the nucleus.

As a nucleus grows, its total binding energy generally increases because it contains more nucleons.

However, the average contribution per nucleon approaches a relatively limited value.

Electrostatic repulsion

Protons are positively charged, so they repel one another through the electrostatic force.

As a nucleus becomes larger, the number of protons increases. The electrostatic repulsion becomes increasingly important.

This reduces the average binding energy per nucleon in very heavy nuclei.

The result is the characteristic binding-energy-per-nucleon curve:

• Very light nuclei generally have relatively low binding energy per nucleon.
• The average rises as nuclei become heavier.
• It reaches a broad maximum around iron and nickel.
• It gradually decreases for very heavy nuclei.

This explains the uranium-238 and iron-56 comparison.

Uranium-238 contains 238 nucleons, so its total binding energy is enormous.

However, its average binding energy per nucleon is lower than iron-56’s because electrostatic repulsion reduces the average binding in such a heavy nucleus.

Iron-56 has fewer nucleons, but each nucleon is more tightly bound on average.

A higher binding energy per nucleon is therefore a useful indicator of stronger average nuclear binding. It is not, however, a complete predictor of radioactive stability or half-life. Other possible decay processes must also be considered.

06Binding energy versus energy released in a nuclear reaction

So far, we have compared individual nuclei.

But what happens when two nuclei combine, or one nucleus splits into smaller nuclei?

This introduces another quantity: released energy, usually represented by \(Q\).

The energy released in a nuclear reaction depends on the difference between the total rest mass of the reactants and the total rest mass of the products.

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

Our sign convention is:

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

There is also a useful way to calculate released energy using total binding energies.

When the reaction conserves the total number of protons and neutrons separately, the free nucleon masses cancel.

Therefore:

\[ \boxed{Q=B_{\mathrm{products}}-B_{\mathrm{reactants}}} \]

This is a comparison of the total binding energies on both sides, not their binding energies per nucleon.

There is an important qualification: conserving mass number and charge does not always mean that the numbers of protons and neutrons are separately conserved. Beta decay, for example, converts a neutron into a proton or vice versa. In those reactions, the proton and neutron rest-mass difference must also be considered.

Check charge and nucleon number first

Before calculating reaction energy, balance the nuclear equation.

For every nuclear reaction:

\[ \sum A_{\mathrm{reactants}} = \sum A_{\mathrm{products}} \]

And:

\[ \sum Z_{\mathrm{reactants}} = \sum Z_{\mathrm{products}} \]

Here, \(A\) represents nucleon number and \(Z\) represents nuclear charge.

A neutron contributes one to the nucleon number but zero to the nuclear charge.

These checks are essential before using the binding-energy relationship.

07A quick decision rule for HSC questions

The most important skill is recognising what the question is actually asking you to compare.

One final trap: a high binding energy per nucleon does not automatically mean that a nucleus cannot decay.

It is an average measure of binding, not a complete description of every possible nuclear transformation.

08Questions and solutions

The following questions build from straightforward comparisons to more demanding applications involving mass measurements, fusion, fission, and beta decay.

Question 1

Two hypothetical nuclei have the following properties:

(a) Calculate the binding energy per nucleon of each nucleus.

(b) Identify which nucleus has the greater total binding energy.

(c) Identify which nucleus has the greater binding energy per nucleon.

(d) Would the rankings be able to differ if both nuclei had the same mass number? Explain.

Solution 1

Nucleus Q has the greater total binding energy, but nucleus P has the greater binding energy per nucleon.

Part (a): Calculate binding energy per nucleon.

For nucleus P:

\[ \begin{aligned} \frac{B}{A}&=\frac{192}{24}\\ &=8.00\ \mathrm{MeV/nucleon} \end{aligned} \]

For nucleus Q:

\[ \begin{aligned} \frac{B}{A}&=\frac{360}{48}\\ &=7.50\ \mathrm{MeV/nucleon} \end{aligned} \]

Part (b): Compare total binding energies.

\[ 360\ \mathrm{MeV}>192\ \mathrm{MeV} \]

Nucleus Q has the greater total binding energy.

Part (c): Compare binding energies per nucleon.

\[ 8.00>7.50 \]

Nucleus P has the greater binding energy per nucleon.

Part (d): Consider nuclei with the same mass number.

No, their rankings cannot differ if both nuclei have the same \(A\).

Dividing both total binding energies by the same positive number cannot reverse their order.

The reversal occurs because the two nuclei have different numbers of nucleons.

Question 2

A student is comparing iron-56 and uranium-238 using measured mass data.

The following rounded values are supplied:

Use the following particle masses:

• Proton: \(1.007276\ \mathrm{u}\)
• Neutron: \(1.008665\ \mathrm{u}\)
• Neutral hydrogen atom: \(1.007825\ \mathrm{u}\)

Use \(1\ \mathrm{u}\,c^2=931.5\ \mathrm{MeV}\).

(a) Calculate the total binding energy of each nucleus.

(b) Calculate the binding energy per nucleon of each nucleus.

(c) Explain why the two nuclei have different rankings when compared using total binding energy and binding energy per nucleon.

Assume electronic binding-energy corrections are negligible.

Solution 2

Uranium-238 has the greater total binding energy, but iron-56 has the greater binding energy per nucleon.

The important step is choosing the correct mass formula for each isotope.

Part (a): Calculate the total binding energies.

Start with iron-56.

Iron-56 contains:

\[ Z=26 \]

\[ N=56-26=30 \]

The question gives the nuclear mass, so use proton and neutron masses directly.

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

Substituting:

\[ \begin{aligned} \Delta m &=26(1.007276)\\ &\quad+30(1.008665)-55.920673\\ &=56.449126-55.920673\\ &=0.528453\ \mathrm{u} \end{aligned} \]

Convert the mass defect into binding energy:

\[ \begin{aligned} B&=0.528453\times931.5\\ &=492.25\ \mathrm{MeV} \end{aligned} \]

Now consider uranium-238.

It contains:

\[ Z=92 \]

\[ N=238-92=146 \]

This time, the question supplies atomic mass. We must use the neutral hydrogen atom mass.

\[ \Delta m=Zm_H+Nm_n-M_{\mathrm{atom}} \]

Substituting:

\[ \begin{aligned} \Delta m &=92(1.007825)\\ &\quad+146(1.008665)-238.050788\\ &=239.984990-238.050788\\ &=1.934202\ \mathrm{u} \end{aligned} \]

Convert into binding energy:

\[ \begin{aligned} B&=1.934202\times931.5\\ &\approx1801.7\ \mathrm{MeV} \end{aligned} \]

Part (b): Calculate binding energy per nucleon.

For iron-56:

\[ \begin{aligned} \frac{B}{A}&=\frac{492.25}{56}\\ &=8.79\ \mathrm{MeV/nucleon} \end{aligned} \]

For uranium-238:

\[ \begin{aligned} \frac{B}{A}&=\frac{1801.7}{238}\\ &=7.57\ \mathrm{MeV/nucleon} \end{aligned} \]

Part (c): Explain the different rankings.

Uranium-238 has a much greater total binding energy because it contains substantially more nucleons.

However, iron-56 has the greater average binding energy per nucleon.

The larger total binding energy of uranium-238 does not mean that each of its nucleons is more tightly bound. The two quantities measure different properties.

The calculation also demonstrates why using the correct mass convention matters. Using proton masses directly with a neutral atomic mass would give an incorrect mass defect.

Question 3

Two deuterium nuclei undergo a fusion reaction:

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

The total binding energies are:

• Deuterium: \(2.225\ \mathrm{MeV}\) per nucleus.
• Helium-3: \(7.718\ \mathrm{MeV}\) per nucleus.
• Free neutron: \(0\ \mathrm{MeV}\).

(a) Verify conservation of nucleon number and charge.

(b) Calculate the energy released per reaction.

(c) A student calculates the difference between the binding energy per nucleon of helium-3 and deuterium and claims this is the energy released. Explain why this method is incorrect.

Solution 3

The reaction releases approximately \(3.27\ \mathrm{MeV}\) per event. The correct calculation compares the total binding energies of all reactants and products.

Part (a): Check conservation.

Nucleon number:

\[ A_{\mathrm{initial}}=2+2=4 \]

\[ A_{\mathrm{final}}=3+1=4 \]

Charge:

\[ Z_{\mathrm{initial}}=1+1=2 \]

\[ Z_{\mathrm{final}}=2+0=2 \]

Both quantities are conserved.

The reaction also conserves the separate numbers of protons and neutrons:

• Initially: two protons and two neutrons.
• Finally: two protons and two neutrons.

Therefore, we can use the difference in total binding energies.

Part (b): Calculate energy released.

The total binding energy of the reactants is:

\[ \begin{aligned} B_{\mathrm{reactants}} &=2(2.225)\\ &=4.450\ \mathrm{MeV} \end{aligned} \]

The total binding energy of the products is:

\[ \begin{aligned} B_{\mathrm{products}} &=7.718+0\\ &=7.718\ \mathrm{MeV} \end{aligned} \]

Therefore:

\[ \begin{aligned} Q&=B_{\mathrm{products}}-B_{\mathrm{reactants}}\\ &=7.718-4.450\\ &=3.268\ \mathrm{MeV} \end{aligned} \]

Thus:

\[ \boxed{Q=3.27\ \mathrm{MeV}} \]

Part (c): Explain the incorrect method.

The student is comparing average binding energies per nucleon, not the total binding energies involved in the reaction.

The binding energy per nucleon of deuterium is:

\[ \frac{2.225}{2}=1.1125\ \mathrm{MeV/nucleon} \]

For helium-3:

\[ \frac{7.718}{3}=2.573\ \mathrm{MeV/nucleon} \]

Subtracting these values gives approximately:

\[ 2.573-1.1125=1.461\ \mathrm{MeV/nucleon} \]

This is not the energy released by the reaction.

There are two deuterium nuclei involved, and their total initial binding energy must be included.

The correct reaction energy is the difference between the combined binding energies of the entire initial and final systems.

Question 4

A hypothetical uranium-236 nucleus undergoes fission according to:

\[ {}^{236}_{92}\mathrm{U} \rightarrow {}^{140}_{54}\mathrm{Xe} + {}^{94}_{38}\mathrm{Sr} + 2{}^{1}_{0}\mathrm{n} \]

The following illustrative total binding energies are supplied:

Assume the supplied values are for the nuclear ground states and ignore any additional initial excitation energy.

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

(b) Calculate the energy released by the fission reaction.

(c) Calculate the binding energy per nucleon of the parent nucleus and each of the two fragments.

(d) Calculate the average binding energy per nucleon of the complete final system, including the two free neutrons.

(e) A student calculates the ordinary arithmetic mean of the two fragment binding energies per nucleon and claims this is the average for the products. Explain why this is incorrect.

Solution 4

The reaction releases \(160\ \mathrm{MeV}\). The combined products have a higher average binding energy per nucleon than the original uranium nucleus.

Part (a): Verify conservation.

Nucleon number:

\[ 236=140+94+2 \]

\[ 236=236 \]

Nuclear charge:

\[ 92=54+38+0 \]

\[ 92=92 \]

Both quantities are conserved.

The separate proton and neutron counts are also conserved in this reaction.

Part (b): Calculate energy released.

The total binding energy of the reactants is:

\[ B_{\mathrm{reactants}}=1790\ \mathrm{MeV} \]

The total binding energy of the products is:

\[ \begin{aligned} B_{\mathrm{products}} &=1160+790+2(0)\\ &=1950\ \mathrm{MeV} \end{aligned} \]

Therefore:

\[ \begin{aligned} Q&=B_{\mathrm{products}}-B_{\mathrm{reactants}}\\ &=1950-1790\\ &=160\ \mathrm{MeV} \end{aligned} \]

The reaction releases:

\[ \boxed{Q=160\ \mathrm{MeV}} \]

Part (c): Calculate the binding energy per nucleon.

For uranium-236:

\[ \begin{aligned} \frac{B}{A} &=\frac{1790}{236}\\ &=7.585\ \mathrm{MeV/nucleon} \end{aligned} \]

For xenon-140:

\[ \begin{aligned} \frac{B}{A} &=\frac{1160}{140}\\ &=8.286\ \mathrm{MeV/nucleon} \end{aligned} \]

For strontium-94:

\[ \begin{aligned} \frac{B}{A} &=\frac{790}{94}\\ &=8.404\ \mathrm{MeV/nucleon} \end{aligned} \]

Both fragments have a higher average binding energy per nucleon than the original uranium nucleus.

Part (d): Calculate the average for the complete final system.

There are 236 nucleons in the complete final system:

\[ A_{\mathrm{final}}=140+94+2=236 \]

The two free neutrons contribute zero binding energy.

Therefore:

\[ \begin{aligned} \left(\frac{B}{A}\right)_{\mathrm{final}} &=\frac{1950}{236}\\ &=8.263\ \mathrm{MeV/nucleon} \end{aligned} \]

The initial average was approximately \(7.585\ \mathrm{MeV/nucleon}\), while the final average is approximately \(8.263\ \mathrm{MeV/nucleon}\).

The average binding energy per nucleon has increased.

Part (e): Explain why an ordinary arithmetic mean is incorrect.

The ordinary arithmetic mean of the two fragment values is:

\[ \frac{8.286+8.404}{2} =8.345\ \mathrm{MeV/nucleon} \]

This is not the average for the complete final system.

Xenon-140 and strontium-94 contain different numbers of nucleons. Their contributions must therefore be weighted according to their mass numbers.

The two free neutrons must also be included in the total nucleon count, even though they contribute zero binding energy.

The correct average is:

\[ \frac{140(8.286)+94(8.404)+2(0)}{236} \]

\[ \approx8.263\ \mathrm{MeV/nucleon} \]

The key point is that an average across different-sized groups must account for the number of members in each group.

Question 5

Beryllium-10 undergoes beta-minus decay to form boron-10:

\[ {}^{10}_{4}\mathrm{Be} \rightarrow {}^{10}_{5}\mathrm{B} + {}^{0}_{-1}\mathrm{e} + \bar{\nu}_e \]

The following rounded neutral atomic masses are supplied:

Use:

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

Assume electronic binding-energy corrections and the antineutrino rest mass are negligible.

(a) Verify conservation of nucleon number and charge.

(b) Calculate the total binding energy of each nucleus.

(c) Calculate the binding energy per nucleon of each nucleus.

(d) Calculate the energy released by the decay using the atomic masses.

(e) Explain how the decay can release energy even though the daughter nucleus has a lower total binding energy and a lower binding energy per nucleon.

Solution 5

The decay releases approximately \(0.557\ \mathrm{MeV}\), even though the daughter nucleus has a slightly lower binding energy per nucleon.

This is possible because beta-minus decay changes a neutron into a proton. The proton and neutron rest masses therefore do not cancel when calculating the reaction energy.

Part (a): Check conservation.

Nucleon number:

\[ 10=10+0+0 \]

Charge:

\[ 4=5-1+0 \]

Both quantities are conserved.

The parent contains four protons and six neutrons.

The daughter contains five protons and five neutrons.

One neutron has effectively transformed into a proton, an electron, and an electron antineutrino.

Part (b): Calculate the total binding energy of each nucleus.

Both masses are neutral atomic masses, so use:

\[ \Delta m=Zm_H+Nm_n-M_{\mathrm{atom}} \]

For beryllium-10:

\[ Z=4,\qquad N=6 \]

Therefore:

\[ \begin{aligned} \Delta m_{\mathrm{Be}} &=4(1.007825)\\ &\quad+6(1.008665)-10.0135347\\ &=10.083290-10.0135347\\ &=0.0697553\ \mathrm{u} \end{aligned} \]

Convert into binding energy:

\[ \begin{aligned} B_{\mathrm{Be}} &=0.0697553\times931.5\\ &=64.977\ \mathrm{MeV} \end{aligned} \]

For boron-10:

\[ Z=5,\qquad N=5 \]

Therefore:

\[ \begin{aligned} \Delta m_{\mathrm{B}} &=5(1.007825)\\ &\quad+5(1.008665)-10.0129369\\ &=10.082450-10.0129369\\ &=0.0695131\ \mathrm{u} \end{aligned} \]

Convert into binding energy:

\[ \begin{aligned} B_{\mathrm{B}} &=0.0695131\times931.5\\ &=64.751\ \mathrm{MeV} \end{aligned} \]

Part (c): Calculate binding energy per nucleon.

Both nuclei contain ten nucleons.

For beryllium-10:

\[ \begin{aligned} \frac{B}{A} &=\frac{64.977}{10}\\ &=6.498\ \mathrm{MeV/nucleon} \end{aligned} \]

For boron-10:

\[ \begin{aligned} \frac{B}{A} &=\frac{64.751}{10}\\ &=6.475\ \mathrm{MeV/nucleon} \end{aligned} \]

The daughter nucleus has a slightly lower total binding energy and a slightly lower binding energy per nucleon.

Part (d): Calculate the energy released.

For beta-minus decay, using neutral atomic masses, the energy released is calculated directly from the parent and daughter atomic masses.

\[ Q=(M_{\mathrm{parent}}-M_{\mathrm{daughter}})c^2 \]

Substitute:

\[ \begin{aligned} Q&=(10.0135347-10.0129369)\\ &\quad\times931.5\\ &=0.0005978\times931.5\\ &=0.5569\ \mathrm{MeV} \end{aligned} \]

Therefore:

\[ \boxed{Q\approx0.557\ \mathrm{MeV}} \]

The decay releases energy.

This is consistent with the evaluated decay energy of approximately \(0.556\ \mathrm{MeV}\) for this transition, as reported by the TUNL Nuclear Data Project.

Part (e): Explain the apparent contradiction.

This is the important conceptual step.

A tempting approach is to use:

\[ Q=B_{\mathrm{products}}-B_{\mathrm{reactants}} \]

That would give:

\[ Q=64.751-64.977 \]

\[ Q=-0.226\ \mathrm{MeV} \]

This incorrectly suggests that energy must be supplied.

The problem is that the shortcut only works when the total numbers of protons and neutrons are separately conserved.

Here, a neutron transforms into a proton during beta-minus decay.

The rest mass of a neutron is different from the combined rest mass of a proton and an electron. This difference must be included.

Using neutral atomic masses, the relevant additional contribution is:

\[ (m_n-m_H)c^2 \]

Substituting:

\[ \begin{aligned} (m_n-m_H)c^2 &=(1.008665-1.007825)\\ &\quad\times931.5\\ &=0.782\ \mathrm{MeV} \end{aligned} \]

The complete energy calculation can therefore be understood as:

\[ Q=(m_n-m_H)c^2 +(B_{\mathrm{daughter}}-B_{\mathrm{parent}}) \]

\[ Q=0.782-0.226 \]

\[ \boxed{Q\approx0.556\ \mathrm{MeV}} \]

The small difference between the rounded results comes from rounding the supplied masses and binding energies.

The additional contribution from the neutron-to-proton transformation is larger than the reduction in nuclear binding energy.

Consequently, the overall reaction still releases energy.

This example exposes an important limitation of using binding energy per nucleon as a standalone stability test. Even though beryllium-10 has the higher average binding energy, it can undergo an energetically favourable beta decay.

09Where to go next

When comparing nuclei, keep three questions separate:

• How much energy is needed to completely separate one nucleus? Calculate total binding energy.
• How tightly bound are its nucleons on average? Calculate binding energy per nucleon.
• How much energy does a nuclear reaction release? Compare the complete initial and final systems, accounting for mass, charge, nucleon number, and any emitted particles.

Once these distinctions are clear, the next useful step is analysing why nuclear fusion and fission release energy. Both processes can move nuclei towards higher binding energy per nucleon, but the actual energy released must still be calculated using the total energy change of the reaction.

Course note (November 2026): This explanation follows the established NSW Physics Stage 6 nuclear-energy treatment. NESA schedules implementation of the Physics 11-12 Syllabus (2025) for Year 11 in 2027 and Year 12 in Term 4 of 2027, with the first new-syllabus HSC examination in 2028. Check the NESA syllabus information for the requirements applying to later cohorts.