19.1 Mass Defect and Binding Energy

19.1 Mass Defect and Binding Energy

The results of experiments on the masses of different nuclei show that the mass of the nucleus is always less than the total mass of all the protons and neutrons making up the nucleus. The lost or missing mass is called mass defect $\Delta m$ given by the equation:

where $m_{p}$ is the mass of a proton, $m_{n}$ is the mass of a neutron and $m_{\mathrm{nucleus}}$ is the mass of the entire nucleus. The missing mass is converted to energy, used in the formation of the nucleus. This energy can be calculated from Einstein's mass-energy relation:

this energy is the potential energy called the Binding Energy (B.E) of the nucleus. It is defined as:

The work done on the nucleus to break it into constituents neutrons and protons. Binding energy is considered to have negative value similar to absolute gravitational potential value taken as zero on the surface of the Earth. A better measure is the building energy per nucleon. It can be considered as the average energy needed to separate a nucleus into its individual nucleons (neutrons and protons).

We usually use the masses of sub-atomic particles in atomic mass unit (amu or u) which is $1/12$th of the mass of an unbound natural atom of $_{6}^{12}\mathrm{C}$. $1\,\mathrm{u} = 1.66 \times 10^{-27}\,\mathrm{g}$ or $1\,\mathrm{u} = 1.66 \times 10^{-27}\,\mathrm{kg}$

Example 19.1

Find the mass defect and binding energy of the helium nucleus given that; Mass of proton $m_{p} = 1.672623 \times 10^{-27}\,\mathrm{kg}$ Mass of neutron $m_{n} = 1.674929 \times 10^{-27}\,\mathrm{kg}$ Mass of nucleus $= 6.646786 \times 10^{-27}\,\mathrm{kg}$

Solution

Using mass defect equation: $\Delta m = 2m_{p} + 2m_{n} - m_{\mathrm{nucleus}}$ $= 2(1.672623) \times 10^{-27}\,\mathrm{kg} + 2(1.674929) \times 10^{-27}\,\mathrm{kg} - 6.646786 \times 10^{-27}\,\mathrm{kg}$ $= 6.695104 \times 10^{-27}\,\mathrm{kg} - 6.646786 \times 10^{-27}\,\mathrm{kg}$ $\Delta m = 0.048318 \times 10^{-27}\,\mathrm{kg}$ Using Einstein's Equation: $B.E = \Delta m c^{2}$ $= 0.048318 \times 10^{-27}\,\mathrm{kg} \times (3.0 \times 10^{8}\,\mathrm{m\,s^{-1}})^{2}$ $B.E = 4.34862 \times 10^{-12}\,\mathrm{J}$ Changing into $\mathrm{eV}$ $B.E = (4.34862 \times 10^{-12}\,\mathrm{J}) / (1.6 \times 10^{-19}\,\mathrm{J\,(eV)^{-1}}) = 27\,\mathrm{MeV}$

For your Information

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Energy must be supplied to break the nucleus apart into its constituent protons and neutrons.

Some Atomic Masses