Mass-Energy and Nuclear Binding Energy

The Nucleus

It exists at the centre of an atom, containing entire positive charge and almost whole of mass. The electron revolve around the nucleus to form an atom. The nucleus consists of protons (+ve charge) and neutrons.

(i) A proton has positive charge equal in magnitude to that of an electron

(+1.6 x 10^–19 C) and a mass equal to 1840 times that of an electron.

(ii) A neutron has no charge and mass is approximately equal to that of proton.

(iii) The number of protons in a nucleus of an atom is called as the atomic number (Z) of that atom. The number of protons plus neutrons (called as Nucleons) in a nucleus of an atom is called as mass number (A) of that atom.

(iv) A particular set of nucleons forming an atom is called as nuclide. It is represented as _ZX^A.

(v) The nuclides having same number of protons (Z), but different number of nucleons (A) are called as isotopes.

(vi) The nuclide having same number of nucleons (A), but different number of protons (Z) are called as isobars.

(vii) The nuclide having same number of neutrons (A - Z) are called as isotones.

MASS DEFECT & BINDING ENERGY

The nucleons are bound together in a nucleus and the energy has to be supplied in order to break apart the constituents into free nucleons. The energy with which nucleons are bounded together in a nucleus is called as Binding Energy (B.E.). In order to free nucleons from a bounded nucleus this much of energy (= B.E.) is to be supplied.

It is observed that the mass of a nucleus is always less than the mass of constituent (free) nucleons. This difference in mass is called as mass defect and is denoted as $\Delta$m.

If m_n : mass of a neutron ; m_p : mass of a proton

M (Z, A) : mass of bounded nucleus

Then, $\Delta$m = Z . m_p + (A – Z). m_n – M (Z, A)

This mass-defect is in form of energy and is responsible for binding the nucleons together. From Einstein's law of inter-conversion of mass into energy:

E = mc² (c : speed of light; m : mass)

$\Rightarrow$ binding energy = $\Delta$m . c²

Generally, $\Delta$m is measured in amu units. So let us calculate the energy equivalent to 1 amu. It is calculated in eV (electron volts; 1 eV = 1.6 x 10^–19 J)

$E\left(\equiv 1amu=\frac{1.67\times {10}^{-27}{\left(3\times {10}^8\right)}^2}{1.6\times {10}^{-19}}\right)eV=931\times {10}^{6}eV=931MeV$

$\Rightarrow B.E.=\Delta m\left(931\right)MeV$

There is another quantity which is very useful in predicting the stability of a nucleus called as Binding energy per nucleons.

$B.E.\text{ per nucleons}=\frac{\Delta m\left(931\right)}{A}MeV$

From the plot of B.E./nucleons Vs mass number (A), we observe that:

(i) B.E./nucleons increases on an average and reaches a maximum of about 8.7 MeV for A º 50 – 80.

(ii) For more heavy nuclei, B.E./nucleons decreases slowly as A increases. For the heaviest natural element U²³⁸ it drops to about 7.5 MeV.

(iii) From above observation, it follows that nuclei in the region of atomic masses 50-80 are most stable.

Illustration 1: If mass of proton = 1.008 amu and mass of neutron = 1.009 amu, then the binding energy per nucleon for ₄Be⁹ (mass = 9.012 amu) will be –

(A) 0.0672 MeV (B) 0.672 MeV

(C) 6.72 MeV (D) 67.2 MeV

Solution: Mass defect

$\Delta m=(4\times 1.008+5\times 1.009)-9.012$

$=9.077-9.012=0.065amu$

$\therefore \frac{BE}{A}=\frac{0.065\times 931}{9}=6.72MeV$

Illustration 2: The energy released in the following $\beta$-decay process will be -

${}_0{n}^1\to {}_1{P}^1+{}_{-1}{e}^0+\bar{v}$

Given that \begin{align} {{m}_{n}}=1.6747\times {{10}^{-27}}\ kg \\   {{m}_{p}}=1.6725\times {{10}^{-27}}\ kg \\   {{m}_{e}}=0.00091\times {{10}^{-27}}\ kg \\\end{align}

(A) 0.931 MeV (B) 0.731 MeV

(C) 0.511 MeV (D) 0.271 MeV

Solution: Mass defect $\Delta m=(1.6747-1.6725-0.0091)\times {10}^{-27}$

$=0.0012\times {10}^{-27}kg$

$\therefore \Delta E=\frac{0.0012\times {10}^{-27}\times {(3\times {10}^8)}^2}{1.6\times {10}^{-13}}=0.731MeV$