Nuclear Mass, Binding Energy and Strong Nuclear Force — AP Physics 2 Study Guide

For: AP Physics 2 candidates sitting AP Physics 2.

Covers: Mass defect calculation, mass-energy equivalence for binding energy, binding energy per nucleon, properties of the strong nuclear force, and analysis of nuclear stability via the binding energy per nucleon curve.

You should already know: 1. Mass-energy equivalence ($E=m{c}^2$) from modern physics. 2. Basic nuclear notation for mass number, atomic number, and neutron number. 3. General properties of Coulomb repulsion between charged particles.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 2 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Nuclear Mass, Binding Energy and Strong Nuclear Force?

This core topic in AP Physics 2 Unit 7 makes up a significant portion of the 20-25% exam weight assigned to Unit 7, and questions involving these concepts appear in both multiple-choice (MCQ) and free-response (FRQ) sections. Nuclear mass refers to the measured mass of a neutral atom (or its nucleus), and we use atomic mass units (u) for most nuclear calculations, where 1 u is defined as 1/12 the mass of a neutral carbon-12 atom. Binding energy is the total energy required to completely separate a bound nucleus into its individual constituent protons and neutrons (collectively called nucleons). The strong nuclear force is the fundamental short-range interaction that binds nucleons together to form stable nuclei, overcoming the Coulomb repulsion between positively charged protons. Synonyms you may encounter include mass deficit (for mass defect) and specific binding energy (for binding energy per nucleon). This topic connects mass-energy conversion to nuclear stability, and is the required foundation for all further nuclear physics topics on the exam.

2. Mass Defect and Mass-Energy Conversion

The measured mass of any stable bound nucleus is always less than the sum of the masses of its individual free protons and neutrons. This missing mass is called the mass defect, $\Delta m$. The reason for this difference is that when nucleons bind together to form a nucleus, some mass is converted to binding energy (per mass-energy equivalence $E=m{c}^2$) that holds the nucleus together.

For calculation convenience on the AP exam, we almost always use neutral atomic masses instead of bare nuclear masses, because the electron masses cancel out automatically in the calculation. The formula for mass defect becomes: $$\Delta m=Z{m}_{\text{H}}+N{m}_n-m_{\text{atom}}$$ where $Z$ is atomic number (number of protons/electrons), $N=A-Z$ is neutron number, $m_{\text{H}}$ is the mass of a neutral hydrogen atom, $m_n$ is the mass of a free neutron, and $m_{\text{atom}}$ is the mass of the neutral atom in question. A very useful conversion for nuclear calculations is that $1\text{u}=931.5\text{MeV}/c^2$, so binding energy can be calculated directly as $E_b(\text{MeV})=\Delta m(\text{u})\times 931.5$, no extra $c^2$ term needed.

Worked Example

Calculate the mass defect and total binding energy of lithium-7 (${}^7\text{Li}$), given $m_{\text{atom}}{(}^7\text{Li})=7.01600\text{u}$, $m_{\text{H}}=1.00783\text{u}$, $m_n=1.00866\text{u}$.

1. Identify $Z$ and $N$: Lithium has $Z=3$, so $N=7-3=4$.
2. Calculate the total mass of free constituents: $Z{m}_{\text{H}}+N{m}_n=3(1.00783)+4(1.00866)=3.02349+4.03464=7.05813\text{u}$.
3. Calculate mass defect by subtracting the bound atomic mass: $\Delta m=7.05813-7.01600=0.04213\text{u}$.
4. Convert mass defect to binding energy: $E_b=0.04213\times 931.5\approx 39.2\text{MeV}$.

Exam tip: Always use atomic masses for AP calculations, the electron masses cancel out automatically — you do not need to subtract electron masses from the atomic mass to get a bare nuclear mass.

3. Binding Energy Per Nucleon and Nuclear Stability

Total binding energy tells you how much energy it takes to split an entire nucleus into free nucleons, but it does not tell you how stable the nucleus is, because larger nuclei always have higher total binding energy simply because they have more nucleons. To compare stability between different-sized nuclei, we use binding energy per nucleon, defined as $\text{BE}/A=E_b/A$, where $A$ is the total number of nucleons. Higher binding energy per nucleon means the nucleus is more tightly bound and more stable.

When we plot binding energy per nucleon against mass number $A$, we get a characteristic curve: it rises sharply for light nuclei ($A<20$), peaks at $A\approx 56$ (iron-56 and nickel-62 are the most stable nuclei), then slowly decreases for heavier nuclei ($A>56$). This curve explains why energy is released in fusion of light nuclei and fission of heavy nuclei: when light nuclei fuse, they form a heavier nucleus closer to the peak with higher $\text{BE}/A$, and when heavy nuclei split, they form smaller daughter nuclei also closer to the peak with higher $\text{BE}/A$. The excess energy is released as kinetic energy and radiation.

Worked Example

Fusion of two deuterium nuclei (${}^2\text{H}$) produces one helium-3 nucleus and one neutron. The total binding energy of two deuterium nuclei is $4.46\text{MeV}$, and the binding energy of helium-3 is $7.72\text{MeV}$. How much energy is released in this reaction?

1. Recall that energy released equals the increase in total binding energy of products compared to reactants.
2. Total binding energy of reactants (two deuterium) is given as $4.46\text{MeV}$. The free neutron has 0 binding energy, so total binding energy of products is just the binding energy of helium-3: $7.72\text{MeV}$.
3. Calculate energy released: $\Delta E={\text{BE}}_{\text{products}}-{\text{BE}}_{\text{reactants}}=7.72-4.46=3.26\text{MeV}$.
4. Confirm with mass defect logic: products have less total mass than reactants, so the missing mass is converted to released energy, matching our result.

Exam tip: When asked to justify whether energy is released in a nuclear reaction, you can use either the binding energy difference or the mass difference — both are acceptable on the AP exam, as long as your sign is correct.

4. The Strong Nuclear Force

The nucleus is made of positively charged protons and neutral neutrons. Protons repel each other via the long-range Coulomb electromagnetic force, so there must be an attractive force strong enough to overcome this repulsion to hold the nucleus together: this is the strong nuclear force. AP Physics 2 requires you to remember four key properties of the strong nuclear force:

1. It is very short-range: it only acts between adjacent nucleons, with a range of ~1-2 femtometers ($1\text{fm}=10^{-15}\text{m}$). Beyond 2 fm, it drops to nearly zero.
2. It is ~100 times stronger than the Coulomb repulsion at short (1 fm) distances.
3. It is repulsive at distances less than ~0.5 fm, which prevents the nucleus from collapsing into a point.
4. It is charge-independent: it acts the same between any pair of nucleons (proton-proton, proton-neutron, neutron-neutron).

The short-range property explains the shape of the binding energy per nucleon curve: in large nuclei, each nucleon only interacts with its immediate neighbors via the strong force, so adding more nucleons does not increase the strong attraction per nucleon. But Coulomb repulsion is long-range, so every proton in the nucleus repels every other proton, and cumulative repulsion increases as the nucleus gets larger. This is why binding energy per nucleon decreases for heavy nuclei, making them unstable. Heavy stable nuclei require more neutrons than protons because neutrons add strong attraction without adding Coulomb repulsion.

Worked Example

Tin-120 is a stable heavy nucleus with $Z=50$ and $N=70$. Explain why it has many more neutrons than protons, rather than an equal number of each.

1. Recall the range properties: strong nuclear force is short-range, Coulomb repulsion between protons is long-range.
2. In a large nucleus with 50 protons, every proton experiences repulsive Coulomb force from all 49 other protons in the entire nucleus, leading to a large net repulsive force that would break the nucleus apart.
3. Each nucleon only experiences strong attraction from its immediate neighbors, so adding extra neutrons adds additional strong attractive force to hold the nucleus together without adding any extra Coulomb repulsion (since neutrons have no charge).
4. If tin-120 had an equal number of protons and neutrons ($Z=N=60$), the cumulative Coulomb repulsion would be large enough to destabilize the nucleus, so extra neutrons are required for stability.

Exam tip: AP exam questions almost always test the short-range property of the strong nuclear force — any question about why heavy nuclei are unstable will require you to connect short range to increasing Coulomb repulsion.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Calculating $\Delta m$ as $m_{\text{atom}}-(Z{m}_p+N{m}_n)$, resulting in a negative mass defect. Why: Students mix up which mass is larger, since free unbound nucleons have more mass than the bound nucleus. Correct move: Always remember $\Delta m=(\text{sum of free masses})-(\text{bound mass})$, so mass defect is always positive.
• Wrong move: Claiming that higher total binding energy means a more stable nucleus. Why: Students confuse total binding energy with binding energy per nucleon; a uranium nucleus has higher total binding energy than iron, but is much less stable. Correct move: Always use binding energy per nucleon ($\text{BE}/A$) to compare nuclear stability between different-sized nuclei.
• Wrong move: Calculating energy released as $\Delta E=({\text{mass}}_{\text{products}}-{\text{mass}}_{\text{reactants}})c^2$, resulting in negative energy released. Why: Students confuse the direction of mass conversion; more tightly bound products have less mass than reactants. Correct move: Energy released = $({\text{mass}}_{\text{reactants}}-{\text{mass}}_{\text{products}})c^2$, which is always positive for exothermic nuclear reactions.
• Wrong move: Claiming the strong nuclear force is long-range or attracts nucleons at all distances. Why: Students mix up strong force properties with Coulomb force properties. Correct move: Remember the distance rule: $<0.5\text{fm}$ = repulsive, $0.5-2\text{fm}$ = attractive, $>2\text{fm}$ = effectively zero; it is always short-range.
• Wrong move: Thinking the strong nuclear force holds electrons in orbit around the nucleus. Why: Students confuse the force holding the nucleus together with the force holding the atom together. Correct move: Strong force only acts between nucleons in the nucleus; electrostatic attraction holds electrons to the nucleus.
• Wrong move: Forgetting to convert atomic mass units correctly, getting an answer in joules when the question asks for MeV. Why: Students forget the convenient conversion factor for nuclear calculations. Correct move: Memorize that $1\text{u}=931.5\text{MeV}$, so $E_b(\text{MeV})=\Delta m(\text{u})\times 931.5$ with no extra $c^2$ term.

6. Practice Questions (AP Physics 2 Style)

Question 1 (Multiple Choice)

Which of the following correctly explains why heavy nuclei with $Z>80$ require more neutrons than protons to be stable? A: Neutrons experience strong nuclear repulsion that balances the attraction between protons. B: Extra neutrons add strong nuclear attraction without adding Coulomb repulsion to counteract the cumulative long-range repulsion between all protons. C: The strong nuclear force only acts between neutrons, not between protons, so extra neutrons are needed to bind the protons together. D: Neutrons decay into protons to balance the electrostatic force in heavy nuclei.

Worked Solution: Let's eliminate incorrect options: Option A is wrong, neutrons do not contribute extra repulsion in stable nuclei. Option C is wrong, the strong force acts between all nucleons, not just neutrons. Option D is wrong, neutron decay would add more protons and increase repulsion, making the nucleus less stable. Option B matches the properties of the strong and Coulomb forces: cumulative long-range Coulomb repulsion between protons requires extra neutrons that add strong attraction without repulsion. The correct answer is B.

Question 2 (Free Response)

(a) Calculate the mass defect and binding energy per nucleon for neon-20 (${}^{20}\text{Ne}$), given $m_{\text{atom}}{(}^{20}\text{Ne})=19.99244\text{u}$, $m_{\text{H}}=1.00783\text{u}$, $m_n=1.00866\text{u}$. (b) Explain why the binding energy per nucleon of neon-20 is lower than the binding energy per nucleon of iron-56. (c) Would energy be released if two neon-20 nuclei combine to form a single calcium-40 nucleus? Justify your answer, given $\text{BE}/A$ for neon-20 is ~8.0 MeV/nucleon and $\text{BE}/A$ for calcium-40 is ~8.5 MeV/nucleon.

Worked Solution: (a) For ${}^{20}\text{Ne}$, $Z=10$, $N=20-10=10$. Sum of masses: $10(1.00783)+10(1.00866)=20.1649\text{u}$. Mass defect: $\Delta m=20.1649-19.99244=0.17246\text{u}$. Total binding energy: $E_b=0.17246\times 931.5\approx 160.5\text{MeV}$. Binding energy per nucleon: $\text{BE}/A=160.5/20\approx 8.0\text{MeV/nucleon}$.

(b) The binding energy per nucleon curve peaks at $A\approx 56$, iron's mass number. The strong nuclear force is short-range, so each nucleon only interacts with adjacent neighbors, so binding energy per nucleon increases as we add nucleons up to the peak. Neon-20 has $A=20$, which is well before the peak, so it has lower binding energy per nucleon than iron-56.

(c) Yes, energy is released. Total binding energy of reactants: $2\times (20\text{nucleons}\times 8.0\text{MeV/nucleon})=320\text{MeV}$. Total binding energy of product: $40\text{nucleons}\times 8.5\text{MeV/nucleon}=340\text{MeV}$. The total binding energy of the product is higher than the reactants, so $\Delta E=340-320=20\text{MeV}$ of energy is released. This matches the rule that fusion of light nuclei with $A<56$ releases energy.

Question 3 (Application / Real-World Style)

A nuclear fission power plant produces 1 GW of electrical power ($1\times 10^9\text{W}$) with an efficiency of 33% (meaning only 1/3 of the total energy released by fission is converted to electricity). Each fission of a uranium-235 nucleus releases 200 MeV of energy. How many fission reactions occur per second in the reactor core of this power plant?

Worked Solution: First, calculate the total energy that must be produced by fission per second to get 1 GW of electrical power: total power = electrical power / efficiency = $1\times 10^9\text{W}/0.33\approx 3.03\times 10^9\text{J/s}$. Convert energy per fission from MeV to joules: $200\text{MeV}=200\times 10^6\text{eV}\times 1.602\times 10^{-19}\text{J/eV}\approx 3.20\times 10^{-11}\text{J per fission}$. The reaction rate $R$ is total power divided by energy per reaction: $R=3.03\times 10^9\text{J/s}/3.20\times 10^{-11}\text{J/fission}\approx 9.5\times 10^{19}\text{fissions per second}$. This means roughly $10^{20}$ uranium nuclei are split every second to produce enough electricity for a typical large power plant, matching real-world operating parameters.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundational prerequisite for all nuclear reaction topics that come next in Unit 7. You will immediately apply mass defect and binding energy calculations to analyze alpha decay, beta decay, and other spontaneous nuclear transmutations, as well as calculate energy released in radioactive decay. Without mastering the relationship between mass defect, binding energy, and the properties of the strong nuclear force, you will not be able to explain why large nuclei are unstable or calculate energy changes in any nuclear process — common FRQ topics on the AP exam. This topic also connects mass-energy equivalence from special relativity to real nuclear systems, tying together two core modern physics concepts.

• Nuclear Reactions and Radioactive Decay
• Fission and Fusion
• Fundamental Forces and Particle Physics