Nuclear Mass, Binding Energy and Strong Nuclear Force

AP Physics 2· AP Physics 2 CED — Quantum, Atomic, and Nuclear Physics· 14 min read

1. Core Concepts Overview★★☆☆☆⏱ 3 min

This topic makes up a significant portion of the 20-25% exam weight assigned to AP Physics 2 Unit 7, with questions appearing in both multiple-choice and free-response sections. It connects mass-energy conversion to nuclear stability, and is required for all further nuclear physics topics on the exam.

📘 Definition

Key Nuclear Terms

Nuclear mass is the measured mass of a neutral atom or nucleus, typically reported in atomic mass units (u), where 1 u is defined as 1/12 the mass of a neutral carbon-12 atom. Binding energy is the energy needed to split a nucleus into free nucleons, and the strong nuclear force is the interaction that binds nucleons together.

2. Mass Defect and Mass-Energy Conversion★★★☆☆⏱ 4 min

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 ( $Δ m$ ). The difference arises because when nucleons bind, some mass is converted to binding energy that holds the nucleus together, per mass-energy equivalence.

📘 Definition

Mass Defect Formula

\Delta m

For AP calculations, we use neutral atomic masses because electron masses automatically cancel out in reactions. The formula is:

\Delta m = Z m_\text{H} + N m_n - m_\text{atom}

Where $Z$ is atomic number, $N = A - Z$ is neutron number, $m_{H}$ is the mass of a neutral hydrogen atom, $m_{n}$ is the mass of a free neutron, and $m_{atom}$ is the mass of the neutral atom. A convenient AP conversion is $1 u = 931.5 MeV / c^{2}$ , so binding energy is calculated directly as $E_{b} (MeV) = Δ m (u) \times 931.5$ , no extra $c^{2}$ term needed.

📐 Worked Example

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

1
1. Identify $Z$ and $N$ : Lithium has $Z = 3$ , so $N = 7 - 3 = 4$ .
2
1. Calculate the total mass of free constituents:
3
$3(1.00783) + 4(1.00866) = 3.02349 + 4.03464 = 7.05813\ \text{u}$
4
1. Calculate mass defect by subtracting the bound atomic mass:
5
$\Delta m = 7.05813 - 7.01600 = 0.04213\ \text{u}$
6
1. Convert mass defect to binding energy:
7
$E_b = 0.04213 \times 931.5 \approx 39.2\ \text{MeV}$

3. Binding Energy Per Nucleon and Nuclear Stability★★★☆☆⏱ 4 min

Total binding energy always increases with the number of nucleons, so it cannot be used to compare stability between different-sized nuclei. To compare stability, we use binding energy per nucleon, defined as $BE / A = E_{b} / A$ . Higher binding energy per nucleon means the nucleus is more tightly bound and more stable.

The binding energy per nucleon curve has a characteristic shape: it rises sharply for light nuclei ( $A < 20$ ), peaks at $A \approx 56$ (iron-56 is one of the most stable nuclei), then slowly decreases for heavier nuclei ( $A > 56$ ). This explains why energy is released in fusion of light nuclei and fission of heavy nuclei: both processes produce nuclei closer to the peak with higher average binding energy per nucleon.

📐 Worked Example

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

1
1. Energy released equals the increase in total binding energy of products compared to reactants.
2
1. Total binding energy of reactants is 4.46 MeV. A free neutron has 0 binding energy, so total binding energy of products is 7.72 MeV.
3
1. Calculate energy released:
4
$\Delta E = \text{BE}_\text{products} - \text{BE}_\text{reactants} = 7.72 - 4.46 = 3.26\ \text{MeV}$
5
This matches mass defect logic: products have less total mass than reactants, so the missing mass is converted to released energy.

4. Properties of the Strong Nuclear Force★★★☆☆⏱ 3 min

Protons repel each other via the long-range Coulomb force, so an attractive force is needed to hold the nucleus together: the strong nuclear force. AP Physics 2 requires you to remember four key properties:

It is very short-range: it only acts between adjacent nucleons, with a range of ~1-2 femtometers ( $1 fm = 1 0^{- 15} m$ ). Beyond 2 fm, it drops to nearly zero.
It is ~100 times stronger than Coulomb repulsion at short (1 fm) distances.
It is repulsive at distances less than ~0.5 fm, which prevents the nucleus from collapsing into a point.
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 immediate neighbors via the strong force, so adding more nucleons does not increase strong attraction per nucleon. Coulomb repulsion is long-range, so cumulative repulsion increases as the nucleus grows, lowering binding energy per nucleon for heavy nuclei. Heavy stable nuclei need 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
1. Recall: strong nuclear force is short-range, Coulomb repulsion between protons is long-range.
2
1. In a large nucleus with 50 protons, every proton experiences repulsive force from all 49 other protons, leading to a large net repulsive force that would break the nucleus apart.
3
1. Each nucleon only gets strong attraction from adjacent neighbors, so adding extra neutrons adds attractive strong force without adding extra Coulomb repulsion.
4
1. If tin-120 had equal numbers of protons and neutrons, cumulative Coulomb repulsion would destabilize the nucleus, so extra neutrons are required for stability.

5. Common Pitfalls

Wrong move:

Calculating $Δ m$ as $m_{atom} - (Z m_{p} + N m_{n})$ , resulting in a negative mass defect.

Why:

Students mix up which mass is larger; free unbound nucleons have more mass than the bound nucleus.

Correct move:

Always remember $Δ m = (sum of free masses) - (bound mass)$ , 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; uranium has higher total binding energy than iron but is much less stable.

Correct move:

Always use binding energy per nucleon ( $BE / A$ ) to compare nuclear stability between different-sized nuclei.

Wrong move:

Calculating energy released as $Δ E = (mass_{products} - mass_{reactants}) c^{2}$ , 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 = $(mass_{reactants} - mass_{products}) c^{2}$ , 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: $< 0.5 fm$ = repulsive, $0.5 - 2 fm$ = attractive, $> 2 fm$ = 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 $1 u = 931.5 MeV$ , so $E_{b} (MeV) = Δ m (u) \times 931.5$ with no extra $c^{2}$ term.

6. Quick Reference Cheatsheet

Category	Formula / Property	Notes
Mass Defect	$Δ m = Z m_{H} + N m_{n} - m_{atom}$	Uses atomic masses, electron masses cancel; $Δ m > 0$ always
Total Binding Energy	$E_{b} = Δ m c^{2}$	Total energy to split nucleus into free nucleons
u to MeV Conversion	$E_{b} (MeV) = Δ m (u) \times 931.5$	Convenient for AP, no extra $c^{2}$ needed
Binding Energy Per Nucleon	$BE / A = E_{b} / A$	Higher $BE / A$ = more stable; used for cross-size comparisons
Energy Released (Reaction)	$Δ E = (\sum m_{r} - \sum m_{p}) c^{2} = \sum BE_{p} - \sum BE_{r}$	Positive $Δ E$ = energy released
Strong Nuclear Force	$d < 0.5$ fm: repulsive; $0.5 < d < 2$ fm: attractive; $d > 2$ fm: ~0	Short-range, charge-independent, 100× stronger than Coulomb at 1 fm

7. Frequently Asked

Do I need to memorize atomic masses for the AP exam?

No, the AP exam always provides all required atomic masses and conversion factors for nuclear calculations.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Strong force stability explanation
2022 · FRQ
Mass defect calculation question

Going deeper

official syllabusAP Physics 2 Course and Exam DescriptionOfficial College Board curriculum reference

What's Next

This subtopic forms the foundation for all further nuclear physics concepts on the AP Physics 2 exam, including nuclear reactions, radioactive decay, fission, and fusion. Understanding binding energy per nucleon and the properties of the strong nuclear force is critical for justifying why energy is released in nuclear processes, a common free-response question topic. Mastery of mass defect calculations will also help you solve energy problems in nuclear decay and reaction questions that frequently appear on the multiple-choice section. Next, explore related topics to build complete mastery of Unit 7.