22.2 Binding Energy

Mass-Energy Equivalence

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

$E=m{c}^2$

where $c=3\times 10^8\text{ m/s}$ is the speed of light. A small amount of mass corresponds to an enormous amount of energy.

Energy equivalent of 1 atomic mass unit (u):

$E=1\text{ u}\times c^2=(1.66054\times 10^{-27}\text{ kg})(3\times 10^8\text{ m/s}{)}^2\approx 931.5\text{ MeV}$

This conversion factor ($1\text{ u}=931.5\text{ MeV}$) is used extensively in nuclear calculations.

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Mass Defect ($\Delta m$)

When protons and neutrons combine to form a nucleus, the measured mass of the nucleus is always less than the sum of the masses of its individual nucleons.

This difference is called the mass defect:

$\boxed{\Delta m=\left[Z{m}_p+(A-Z)m_n\right]-M_{\text{nucleus}}}$

where:

• $Z$ = number of protons, $A$ = mass number
• $m_p=1.007276\text{ u}$, $m_n=1.008665\text{ u}$
• $M_{\text{nucleus}}$ = measured mass of the nucleus

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Binding Energy

The mass defect is converted into binding energy — the energy released when the nucleus is assembled from its constituent nucleons. Equivalently, it is the energy that must be supplied to completely disassemble the nucleus:

$\boxed{E_b=\Delta m\cdot c^2}$

In practical units:

$E_b\text{ (in MeV)}=\Delta m\text{ (in u)}\times 931.5\text{ MeV/u}$

Worked Example: Calculate the binding energy of Deuterium (${}_1^2\text{H}$), given $M_{\text{nucleus}}=2.01355\text{ u}$.

$\Delta m=(1\times 1.007276+1\times 1.008665)-2.01355=0.002391\text{ u}$

$E_b=0.002391\times 931.5\approx 2.23\text{ MeV}$

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Binding Energy Per Nucleon

To compare the stability of different nuclei, we use binding energy per nucleon:

$\frac{B.E.}{A}=\frac{E_b}{A}$

A higher value means the nucleus is more tightly bound and more stable.

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The B.E./A vs. Nucleon Number Graph

The graph of binding energy per nucleon ($B.E./A$) against mass number ($A$) has the following key features:

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Relevance to Fission and Fusion

Nuclear Fusion

Light nuclei (low $A$, low B.E./A) combine to form a heavier nucleus closer to the peak:

${}_1^2\text{H}{+}_1^3\text{H}{\to }_2^4\text{He}{+}_0^{1}n+17.6\text{ MeV}$

The product has higher B.E./A → energy is released.

Nuclear Fission

Heavy nuclei (high $A$, lower B.E./A) split into medium-mass fragments closer to the peak:

${}_{92}^{235}\text{U}{+}_0^{1}n{\to }_{56}^{141}\text{Ba}{+}_{36}^{92}\text{Kr}+3_0^{1}n+\sim 200\text{ MeV}$

The fragments have higher B.E./A than uranium → energy is released.

In both cases, energy is released because the total binding energy of the products is greater than that of the reactants. The difference in binding energy appears as kinetic energy of the products.