Mass defect and binding energy

Physics· Unit 27: Nuclear Physics, Section 1· 15 min read

1. Mass Defect: Definition and Origin★★☆☆☆⏱ 5 min

📘 Definition

Mass defect

\Delta m

The difference between the total mass of the individual, separate nucleons that make up a nucleus and the actual measured mass of the intact nucleus. Mass defect arises because some mass is converted to energy when nucleons bind together.

Example:

Calculated for any nuclide using the formula below

\Delta m = (Z m_p + (A-Z) m_n) - M_{\text{nucleus}}

Where $Z$ = proton number, $A$ = nucleon number, $m_{p}$ = proton mass, $m_{n}$ = neutron mass, $M_{nucleus}$ = mass of the intact nucleus. When using atomic data, electron masses cancel out automatically, so you can use atomic masses directly without correction.

📐 Worked Example

Calculate the mass defect of a nitrogen-14 nucleus, given: mass of nitrogen-14 atom = 14.003074 u, mass of ¹H atom = 1.007825 u, mass of neutron = 1.008665 u. Nitrogen-14 has $Z = 7$ , $A = 14$ .

1
Step 1: Calculate total mass of 7 hydrogen atoms (protons + electrons) and 7 neutrons:
2
$7 \times 1.007825 + 7 \times 1.008665 = 14.115430 \text{ u}$
3
Step 2: Subtract the atomic mass of nitrogen-14 to get mass defect (electron masses cancel):
4
$\Delta m = 14.115430 - 14.003074 = 0.112356 \text{ u}$

Exam tip:

Always double-check the number of protons and neutrons matches the nuclide's proton and nucleon numbers given in the question.

2. Binding Energy: Definition and Calculation★★☆☆☆⏱ 5 min

📘 Definition

Binding Energy

E_b

The minimum energy required to completely separate a nucleus into its individual free nucleons. It is equal to the energy released when the nucleus is formed from separate nucleons, and is calculated from mass defect using Einstein's mass-energy relation.

Example:

A higher binding energy means the nucleus is more tightly bound

For CIE exams, the standard conversion factor is $1 u = 931.5 MeV$ , so binding energy can be calculated directly from mass defect in u by simple multiplication.

📐 Worked Example

Calculate the total binding energy of the nitrogen-14 nucleus from the previous example, where $Δ m = 0.112356$ u.

1
Use the conversion factor 1 u = 931.5 MeV to convert mass defect to binding energy:
2
$E_b = \Delta m \times 931.5 = 0.112356 \times 931.5 \approx 104.6 \text{ MeV}$

✓ Quick check

Check your understanding:

What is the binding energy of a nuclide with mass defect 0.250 u?
- 233 MeV
- 3726 MeV
- 0.250 MeV
- 931.5 MeV
Reveal answer
233 MeV —
Correct: $0.250 \times 931.5 = 232.875 \approx 233$ MeV. Other options use incorrect conversion factors.

3. Binding Energy per Nucleon and Nuclear Stability★★★☆☆⏱ 5 min

📘 Definition

Binding Energy per Nucleon

\frac{E_b}{A}

The total binding energy of a nucleus divided by the number of nucleons ( $A$ ) in the nucleus. It is the standard measure of how tightly bound a nucleus is, and therefore how stable it is.

Example:

Higher binding energy per nucleon = more stable nuclide

A graph of binding energy per nucleon against nucleon number $A$ has a characteristic shape: it rises steeply for small $A$ , peaks at around $A = 56$ (iron), then gradually decreases for large $A$ . This shape explains why energy is released in both fission (large nuclei splitting) and fusion (small nuclei joining): both processes produce more tightly bound nuclides with higher binding energy per nucleon, converting mass to energy.

📐 Worked Example

Calculate the binding energy per nucleon for nitrogen-14, given total binding energy is 104.6 MeV and $A = 14$ .

1
Divide total binding energy by the number of nucleons $A$ :
2
$\frac{E_b}{A} = \frac{104.6}{14} \approx 7.47 \text{ MeV per nucleon}$

4. Common Pitfalls

Wrong move:

Subtracting electron mass from atomic masses to get nuclear mass

Why:

This introduces an unnecessary error, because electron masses cancel out when using atomic data

Correct move:

Use atomic masses directly, the electron masses will cancel automatically in mass defect calculations

Wrong move:

Calculating mass defect as $Δ m = M_{nucleus} - total nucleon mass$ , giving a negative value

Why:

This reverses the definition, leading to negative binding energy which is impossible

Correct move:

Mass defect is always positive: $Δ m = total mass of separate nucleons - nuclear mass$

Wrong move:

Using total binding energy instead of binding energy per nucleon to compare stability

Why:

Large nuclei always have higher total binding energy just because they have more nucleons, not because they are more stable

Correct move:

Always compare binding energy per nucleon when assessing relative nuclear stability

Wrong move:

Using 1 u = 931.5 J instead of 931.5 MeV for conversion

Why:

This gives a binding energy 10⁶ times too large, leading to wrong final answers

Correct move:

Always check units: CIE expects 1 u = 931.5 MeV for all binding energy calculations

5. Quick Reference Cheatsheet

Quantity	Formula/Rule	Key Note
Mass defect	$Δ m = Z m_{p} + (A - Z) m_{n} - M$	Electron masses cancel with atomic data
Total binding energy	$E_{b} = Δ m \times 931.5$ MeV	1 u = 931.5 MeV
Binding energy per nucleon	$\frac{E _{b}}{A} = \frac{E _{b}}{A}$	Higher value = more stable
Peak of binding energy curve	A ≈ 56 (Iron)	Most stable common nuclide

6. Frequently Asked

Why can we use atomic masses instead of nuclear masses for calculations?

The mass of the electrons cancels out on both sides of the equation when we use total atomic masses, so no explicit correction for electron mass is needed.

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.

2022 · 12
Calculate mass defect for helium nucleus
2023 · 22
Relate binding energy to nuclear stability
2021 · 11
Convert mass defect to energy in MeV

Going deeper

referenceCIE approved nuclear mass data tableUse for calculation practice

What's Next

Mass defect and binding energy are the foundation for understanding all nuclear energy processes, which are heavily tested in CIE A-Level Physics papers 1 and 2. The shape of the binding energy per nucleon curve directly explains why energy is released in nuclear fission (used in nuclear power) and nuclear fusion (the energy source of stars), the next core topics in nuclear physics. Mastery of mass defect calculations is also required for questions on radioactive decay energy, so a strong grasp of this sub-topic will help you with all subsequent nuclear physics questions.