What is mass defect and how does it relate to nuclear stability

Question

Define mass defect. How does mass defect relate to nuclear binding energy and nuclear stability? Explain with reference to the binding energy per nucleon curve.

Solution — Step by Step

When protons and neutrons come together to form a nucleus, the actual mass of the nucleus is less than the sum of the masses of the individual nucleons. This difference is called the mass defect ( $\Delta m$ ).

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

where:

$Z$ = number of protons, $(A-Z)$ = number of neutrons
$m_p = 1.00728$ u (proton mass), $m_n = 1.00866$ u (neutron mass)
$M_{nucleus}$ = actual nuclear mass (from atomic mass data, subtract electron masses)

The mass “defect” hasn’t disappeared — it has been converted to energy.

By Einstein’s $E = mc^2$ , the mass defect converts to binding energy:

E_b = \Delta m \cdot c^2

Using atomic mass units: $1 \text{ u} = 931.5 \text{ MeV}/c^2$ , so:

E_b = \Delta m \times 931.5 \text{ MeV}

This energy was released when the nucleus formed. To break the nucleus back into free nucleons, you must supply this exact energy.

Binding energy is the energy required to completely separate all nucleons. More binding energy = more stable nucleus.

To compare stability across different nuclei, we use binding energy per nucleon (BE/A):

\frac{E_b}{A}

A higher BE/A means each nucleon is more tightly bound = more stable nucleus.

Key values on the BE/A curve:

Very light nuclei (H, He): BE/A is low (~1–7 MeV)
$^{56}$ Fe: Maximum BE/A ≈ 8.8 MeV — most stable nucleus
Heavy nuclei (U, Th): BE/A decreases to ~7.6 MeV — less stable

The BE/A curve peaks at iron-56 and slopes down on both sides. This shape explains two types of nuclear reactions:

Nuclear fusion (light nuclei): Combining two light nuclei (e.g., H + H → He) moves from low BE/A to higher BE/A — energy is released. The sun runs on this.

Nuclear fission (heavy nuclei): Splitting a heavy nucleus (e.g., U-235 → Ba + Kr) moves from lower BE/A to higher BE/A — energy is released. Nuclear reactors use this.

In both cases, products are closer to iron on the BE/A curve than the reactants — the system moves to a more stable state, releasing energy.

Why This Works

Mass defect is not mysterious — it is just the quantitative measure of how strongly the nuclear force binds nucleons together. The stronger the binding, the more energy was released when the nucleus formed, and the more mass was converted to energy.

The key insight: mass and energy are equivalent ( $E = mc^2$ ). The “missing” mass in a nucleus is not gone — it exists as binding energy, holding the nucleus together.

Stability is about how much energy it would take to undo that binding. Iron-56 requires the most energy per nucleon to disassemble — it is the end point of nucleosynthesis in stars.

Alternative Method

For a quick stability comparison: if only BE/A values are given for two nuclei, the one with higher BE/A is more stable. No calculation needed.

Common Mistake

Students confuse binding energy with the energy the nucleus “has stored.” Binding energy is the energy you must put in to break the nucleus apart — it is not energy the nucleus releases spontaneously. A stable nucleus doesn’t release its binding energy; an unstable one (radioactive) does — but through alpha, beta, gamma decay, not by breaking into free nucleons.

In JEE Main and NEET, binding energy per nucleon questions often come with a graph of BE/A vs mass number and ask: “Which nucleus is most stable?” (Answer: iron-56, peak of the curve). Or: “Why does fission of U-235 release energy?” (Because products have higher BE/A than U-235). Understand the graph, don’t just memorise numbers.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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