22.1 Mass Defect | Nuclear Physics | Physics 12

Mass-Energy Equivalence

Albert Einstein's special theory of relativity established that mass and energy are interconvertible. The relationship is given by:

$E = m c^{2}$

where $c = 3 \times 1 0^{8} m s^{- 1}$ is the speed of light in vacuum. This means that a small amount of mass corresponds to an enormous amount of energy. In nuclear physics, this equivalence is used to explain the energy released when nuclei are formed or when they undergo reactions.

$E = Δ m \cdot c^{2}$

where $Δ m$ is the change in mass (mass defect).

The Unified Atomic Mass Unit (u)

Because nuclear masses are extremely small, a convenient unit called the unified atomic mass unit (u) is used:

One unified atomic mass unit (u) is defined as exactly $\frac{1}{12}$ of the mass of a carbon-12 ( $_{6}^{12} C$ ) atom.

$1 u = 1.6605 \times 1 0^{- 27} kg$

Using $E = m c^{2}$ , the energy equivalent of 1 u is:

$E = (1.6605 \times 1 0^{- 27}) (3 \times 1 0^{8})^{2} \approx 931.5 MeV$

This conversion factor is extremely useful:

$1 u \equiv 931.5 MeV$

Mass Defect ( $Δ m$ )

When protons and neutrons (collectively called nucleons) combine to form a nucleus, the actual mass of the nucleus is always less than the sum of the masses of its individual nucleons.

This difference in mass is called the mass defect:

$Δ m = [Z m_{p} + (A - Z) m_{n}] - M_{nucleus}$

where:

$Z$ = atomic number (number of protons)
$A$ = mass number (total nucleons)
$m_{p} = 1.007276 u$ = mass of a proton
$m_{n} = 1.008665 u$ = mass of a neutron
$M_{nucleus}$ = measured mass of the nucleus

Why Does Mass Defect Occur?

When nucleons bind together, energy is released to the surroundings. By mass-energy equivalence, this released energy corresponds to a reduction in mass. The "missing" mass has been converted into the binding energy that holds the nucleus together.

$E_{b} = Δ m \cdot c^{2}$

Binding Energy

Binding energy ( $E_{b}$ ) is the energy required to completely separate a nucleus into its individual free nucleons. It is also the energy released when the nucleus is assembled from free nucleons.

$E_{b} = Δ m \cdot c^{2}$

Using the convenient conversion:

$E_{b} (in MeV) = Δ m (in u) \times 931.5 MeV/u$

Worked Example

Calculate the mass defect and binding energy of a Deuterium nucleus ( $_{1}^{2} H$ ).

Given:

$m_{p} = 1.007276 u$
$m_{n} = 1.008665 u$
$M_{nucleus} = 2.014102 u$
$Z = 1$ , $A = 2$

Step 1 — Sum of individual nucleon masses: $m_{p} + m_{n} = 1.007276 + 1.008665 = 2.015941 u$

Step 2 — Mass defect: $Δ m = 2.015941 - 2.014102 = 0.001839 u$

Step 3 — Binding energy: $E_{b} = 0.001839 \times 931.5 \approx 1.71 MeV$

Key Points

Quantity	Symbol	Formula
Mass defect	$Δ m$	$[Z m_{p} + (A - Z) m_{n}] - M_{nucleus}$
Binding energy	$E_{b}$	$Δ m \cdot c^{2}$
Energy of 1 u	—	$931.5 MeV$

The nucleus is always lighter than the sum of its free nucleons.
A larger mass defect means more binding energy and a more tightly bound nucleus.
Mass-energy equivalence ( $E = Δ m c^{2}$ ) is the fundamental principle connecting mass defect to binding energy.

Mass-Energy Equivalence

Albert Einstein's special theory of relativity established that mass and energy are interconvertible. The relationship is given by:

$E = m c^{2}$

$E = Δ m \cdot c^{2}$

where $Δ m$ is the change in mass (mass defect).

The Unified Atomic Mass Unit (u)

Because nuclear masses are extremely small, a convenient unit called the unified atomic mass unit (u) is used:

One unified atomic mass unit (u) is defined as exactly $\frac{1}{12}$ of the mass of a carbon-12 ( $_{6}^{12} C$ ) atom.

$1 u = 1.6605 \times 1 0^{- 27} kg$

Using $E = m c^{2}$ , the energy equivalent of 1 u is:

$E = (1.6605 \times 1 0^{- 27}) (3 \times 1 0^{8})^{2} \approx 931.5 MeV$

This conversion factor is extremely useful:

$1 u \equiv 931.5 MeV$

Mass Defect ( $Δ m$ )

When protons and neutrons (collectively called nucleons) combine to form a nucleus, the actual mass of the nucleus is always less than the sum of the masses of its individual nucleons.

This difference in mass is called the mass defect:

$Δ m = [Z m_{p} + (A - Z) m_{n}] - M_{nucleus}$

where:

$Z$ = atomic number (number of protons)
$A$ = mass number (total nucleons)
$m_{p} = 1.007276 u$ = mass of a proton
$m_{n} = 1.008665 u$ = mass of a neutron
$M_{nucleus}$ = measured mass of the nucleus

Why Does Mass Defect Occur?

$E_{b} = Δ m \cdot c^{2}$

Binding Energy

$E_{b} = Δ m \cdot c^{2}$

Using the convenient conversion:

$E_{b} (in MeV) = Δ m (in u) \times 931.5 MeV/u$

Worked Example

Calculate the mass defect and binding energy of a Deuterium nucleus ( $_{1}^{2} H$ ).

Given:

$m_{p} = 1.007276 u$
$m_{n} = 1.008665 u$
$M_{nucleus} = 2.014102 u$
$Z = 1$ , $A = 2$

Step 1 — Sum of individual nucleon masses: $m_{p} + m_{n} = 1.007276 + 1.008665 = 2.015941 u$

Step 2 — Mass defect: $Δ m = 2.015941 - 2.014102 = 0.001839 u$

Step 3 — Binding energy: $E_{b} = 0.001839 \times 931.5 \approx 1.71 MeV$

Key Points

Quantity	Symbol	Formula
Mass defect	$Δ m$	$[Z m_{p} + (A - Z) m_{n}] - M_{nucleus}$
Binding energy	$E_{b}$	$Δ m \cdot c^{2}$
Energy of 1 u	—	$931.5 MeV$

The nucleus is always lighter than the sum of its free nucleons.
A larger mass defect means more binding energy and a more tightly bound nucleus.
Mass-energy equivalence ( $E = Δ m c^{2}$ ) is the fundamental principle connecting mass defect to binding energy.