Mass Defect, Binding Energy, and The Nuclear Force

To separate a nucleus to its individual components – the protons and neutrons, energy must be added to it. This energy is called binding energy, \(E_B\)  which is considered as the magnitude of the energy that holds the nucleons together. Since there is addition of energy, it follows that the rest energy, \(E_0\) of the separated nucleons is greater than the rest energy of the nucleus, therefore, the rest energy of the nucleus can be expressed mathematically as \(E_0\; – \;E_B\). The binding energy can be represented by the equation below:

\(E_B = (ZM_H + Nm_n -\; ^A_Z M) c^2\)     (1)

where

\(E_B\) = binding energy with \(Z\) protons and \(N\) neutrons

\(Z\) = atomic number

\(M_H\) = mass of hydrogen atom

\(N\) = number of neutrons

\(m_n\) = mass of neutron

\(^A_Z M\) = mass of neutral atom containing nucleus

\(c\) = speed of light in a vacuum which has a value of \(931.5\; MeV/u\)

 

\(ZM_H\) is the mass of \(Z\) protons and \(Z\) electrons combined as \(Z\) neutral \(^1_1H\) atoms. This is done to balance the \(Z\) electrons contained in \(^A_ZM\).

The mass of the nucleus is always less than the total mass of its nucleons by an amount of \(\Delta M = E_B/c^2\). This mass, \(\Delta M\) is called as the mass defect.

 


 

Example:

The neutral atomic mass of \(^{62}_{28}Ni\) is \(61.928345\; u\). Calculate:

  1. Mass defect
  2. Total binding energy
  3. Binding energy per nucleon

Given:

\(Z = 28\\ M_H = 1.007825\; u\\ N = A - Z = 62 - 28 = 34\\ m_n = 1.008665\; u\\ ^A_ZM = 61.928345\; u \)


Solution:

  1. The binding energy is just the mass defect multiplied by the square of the speed of light in vacuum. Thus, the mass defect, \(\Delta M\) is

\(\Delta M = ZM_H + Nm_n - ^A_Z M\\ \;\;\;\;\;\; = [(28)(1.007825\; u)] + [(34)(1.008665\; u)] – 61.928345\; u\\ \;\;\;\;\;\;= 0.585365\; u. \)

  1. Using the value of \(\Delta M\) in solution a and multiplying it by the square of the speed of light in vacuum, we can solve for the total binding energy.

\(E_B = (\Delta M)c^2\\ \;\;\;\; = (0.585365\; u)( 931.5\; MeV/u)\\ \;\;\;\; = 545.3\; MeV \)

  1. The binding energy per nucleon can be obtained by dividing the total binding energy by the mass number. Since \(A=62\), which is the mass number, the binding energy of each nucleon is,

\({E_B \over A} = {545.3\; MeV \over 62} = 8.795\; MeV. \)


The Nuclear Force

Despite the electrical repulsion of protons, the protons and neutrons can be held together by a force. This force, considering the nuclear structure, is called as the nuclear force. Nuclear force has the following characteristics:

  1. Nuclear force is independent on the charge. Both the proton and the neutron are bounded equally.
  2. It has a short range. Within its range, the nuclear force is much stronger than the electrical forces, which is the reason why nucleus can be stable.
  3. A particular nucleon cannot interact simultaneously with all the other nucleons in the nucleus, instead, it can interact immediately to the other nucleons within the nearest area to it. This is due to the nearly constant density of nuclear matter and the nearly constant binding energy per nucleon of the larger nuclides.
  4. The nuclear force favors the binding of pair of protons and pair of neutrons, in which each pair have opposite spins.