A nucleus has a mass defect of 2.0 × 10^-28\ kg. Find its binding energy (c = 3.00 × 10^8\ m s^-1). [2 marks]

SEAB Original-style practice: A helium-4 nucleus has a mass defect of 5.05 × 10^-29\ kg. Take c = 3.00 × 10^8\ m s^-1 and 1\ eV = 1.60 × 10^-19\ J. (a) Find the binding energy in joules. (b) Find the binding energy per nucleon in MeV.

(a) Binding energy: E = m\, c^2 = 5.05 × 10^-29 × (3.00 × 10^8)^2 = 5.05 × 10^-29 × 9.00 × 10^16 = 4.545 × 10^-12\ J. (b) Convert to eV: 4.545 × 10^-121.60 × 10^-19 = 2.84 × 10^7\ eV = 28.4\ MeV. Helium-4 has 4 nucleons, so binding energy per nucleon = 28.44 = 7.1\ MeV per nucleon. Markers reward E = m c^2 for the binding energy, conversion to MeV, and dividing by the nucleon number for binding energy per nucleon.

SingaporePhysicsSyllabus dot point

Why is the mass of a nucleus less than the sum of its parts, and how does this mass defect explain the energy released in fission and fusion?

Relate mass defect to binding energy through E = mc squared, interpret the binding-energy-per-nucleon curve, and explain energy release in fission and fusion

A focused answer to the H2 Physics learning outcome on nuclear binding energy. Mass defect and E = mc^2, binding energy per nucleon and its curve, and why fission of heavy nuclei and fusion of light nuclei both release energy.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to relate the mass defect of a nucleus to its binding energy through $E = mc^2$ , to interpret the binding-energy-per-nucleon curve, and to explain why both fission of heavy nuclei and fusion of light nuclei release energy. This is the energy source of stars and nuclear reactors.

The answer

Mass defect

The mass of a nucleus is always less than the total mass of its separate protons and neutrons. The difference is the mass defect $\Delta m$ :

\Delta m = (Z m_p + N m_n) - m_{\text{nucleus}}

where $Z$ is the number of protons and $N$ the number of neutrons. The "missing" mass has been converted to energy that binds the nucleus together.

Binding energy and E = mc squared

The binding energy is the energy equivalent of the mass defect, given by Einstein's mass-energy relation:

E = \Delta m\, c^2

It is the energy released when the nucleus is assembled from its separate nucleons, or equivalently the energy that must be supplied to pull it completely apart. Because $c^2$ is enormous, even a tiny mass defect corresponds to a large energy.

Binding energy per nucleon

To compare the stability of different nuclei, divide the binding energy by the number of nucleons:

\text{binding energy per nucleon} = \frac{\Delta m\, c^2}{A}

A higher binding energy per nucleon means a more tightly bound, more stable nucleus.

The binding-energy-per-nucleon curve

Plotting binding energy per nucleon against nucleon number $A$ gives a curve that rises steeply for light nuclei, peaks around $A = 56$ (the iron region, the most stable nuclei), then falls gradually for heavy nuclei. A nuclear reaction releases energy when it produces nuclei with a higher binding energy per nucleon, that is, when it moves toward the peak.

Fission and fusion

Fission: a heavy nucleus (such as uranium-235) splits into two medium-mass nuclei. These products lie nearer the peak with higher binding energy per nucleon, so energy is released.
Fusion: two light nuclei (such as isotopes of hydrogen) join to form a heavier nucleus. The product climbs the steep left side of the curve toward the peak, so energy is released, and far more per nucleon than fission.

Both processes release energy because they move nucleons to a more tightly bound state, with the lost mass appearing as energy via $E = mc^2$ .

Worked example

In a fusion reaction, two deuterium nuclei combine with a total mass defect of $6.0 \times 10^{-30}\ \text{kg}$ . Take $c = 3.00 \times 10^8\ \text{m s}^{-1}$ and $1\ \text{MeV} = 1.60 \times 10^{-13}\ \text{J}$ . Find the energy released (a) in joules and (b) in MeV.

Step 1: Apply the mass-energy relation

E = \Delta m\, c^2 = 6.0 \times 10^{-30} \times (3.00 \times 10^8)^2

Step 2: Evaluate in joules

E = 6.0 \times 10^{-30} \times 9.00 \times 10^{16} = 5.4 \times 10^{-13}\ \text{J}

Step 3: Convert to MeV

E = \frac{5.4 \times 10^{-13}}{1.60 \times 10^{-13}} = 3.4\ \text{MeV}

Step 4: Interpret

The reaction releases $5.4 \times 10^{-13}\ \text{J}$ , or about $3.4\ \text{MeV}$ . The energy comes entirely from the disappearance of $6.0 \times 10^{-30}\ \text{kg}$ of mass, illustrating that fusion converts a small mass defect into a large energy via $E = mc^2$ .

Common traps

Confusing mass defect with binding energy: The mass defect is a mass ( $\Delta m$ ); the binding energy is its energy equivalent, $\Delta m c^2$ .
Forgetting to square $c$: The relation is $E = \Delta m c^2$ ; $c^2 = 9.00 \times 10^{16}\ \text{m}^2\text{s}^{-2}$ .
Thinking heavier nuclei are always more stable: Stability peaks around iron-56; both very light and very heavy nuclei are less tightly bound per nucleon.

Saying fission and fusion release energy for the same reason in opposite directions wrongly. Both move toward the peak of the curve (higher binding energy per nucleon); fusion climbs from the light side, fission descends from the heavy side.

Using total binding energy instead of per nucleon for stability comparisons. Stability is judged by binding energy per nucleon, not the total.

Examples in context

Example 1. The Sun. The Sun fuses hydrogen into helium in its core. The helium product has a higher binding energy per nucleon than the hydrogen, so each reaction releases energy, with the tiny mass difference converted via $E = mc^2$ . This is the energy source that powers all stars.

Example 2. A nuclear reactor. A reactor splits uranium-235 in controlled fission. The medium-mass fragments are more tightly bound per nucleon than the original nucleus, so energy is released as the kinetic energy of the fragments, which heats water to drive turbines. The chain reaction is moderated to keep the energy release steady.

Try this

Q1. Define the mass defect of a nucleus. [2 marks]

Cue. The difference between the total mass of the separate nucleons and the mass of the assembled nucleus.

Q2. A nucleus has a mass defect of $2.0 \times 10^{-28}\ \text{kg}$ . Find its binding energy ( $c = 3.00 \times 10^8\ \text{m s}^{-1}$ ). [2 marks]

Cue. $E = \Delta m c^2 = 2.0 \times 10^{-28} \times 9.00 \times 10^{16} = 1.8 \times 10^{-11}\ \text{J}$ .

Q3. Explain, using the binding-energy-per-nucleon curve, why energy is released when light nuclei fuse. [3 marks]

Cue. Light nuclei have low binding energy per nucleon; fusing them forms a nucleus higher up the steep left side of the curve (more tightly bound), so the products are more stable and the difference is released as energy.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original5 marksA helium-4 nucleus has a mass defect of

5.05 \times 10^{-29}\ \text{kg}

. Take

c = 3.00 \times 10^8\ \text{m s}^{-1}

and

1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}

. (a) Find the binding energy in joules. (b) Find the binding energy per nucleon in MeV.

Show worked answer →

(a) Binding energy: $E = \Delta m\, c^2 = 5.05 \times 10^{-29} \times (3.00 \times 10^8)^2 = 5.05 \times 10^{-29} \times 9.00 \times 10^{16} = 4.545 \times 10^{-12}\ \text{J}$ .

(b) Convert to eV: $\dfrac{4.545 \times 10^{-12}}{1.60 \times 10^{-19}} = 2.84 \times 10^7\ \text{eV} = 28.4\ \text{MeV}$ .

Helium-4 has $4$ nucleons, so binding energy per nucleon $= \dfrac{28.4}{4} = 7.1\ \text{MeV per nucleon}$ .

Markers reward $E = \Delta m c^2$ for the binding energy, conversion to MeV, and dividing by the nucleon number for binding energy per nucleon.

Original5 marks(a) Explain what is meant by binding energy per nucleon and how the shape of its curve against nucleon number explains why both fission and fusion can release energy. (b) State which nucleus is the most stable.

Show worked answer →

(a) Binding energy per nucleon is the binding energy of a nucleus divided by its number of nucleons; it measures how tightly each nucleon is held and so how stable the nucleus is.

The curve rises steeply for light nuclei, peaks around mass number $56$ (iron region), then falls gradually for heavy nuclei. A reaction releases energy if it moves nucleons to a position of higher binding energy per nucleon (more stable). Fusing light nuclei climbs the steep left side toward the peak, releasing energy; splitting heavy nuclei moves down the gentle right side back toward the peak, also releasing energy.

(b) The most stable nucleus is around iron-56, at the peak of the curve.

Markers reward the definition (binding energy divided by nucleon number, a stability measure), the curve shape with the peak near iron, the energy-release argument for fusion (light) and fission (heavy), and iron as the most stable.