Question

Obtain the binding energy (in MeV) of a nitrogen nucleus ($_7{N^{14}}$).
Given mass = $_7{N^{14}}$ = 14.00307 amu

Answer 1

Hint: Binding energy, amount of energy required to separate a particle from a system of particles or to disperse all the particles of the system. Binding energy is especially applicable to subatomic particles in atomic nuclei, to electrons bound to nuclei in atoms, and to atoms and ions bound together in crystals. Here the following concept is used to solve the problem

Formula used:
\[\Delta M(931MeV) = \Delta E\]
Here, \[\Delta M\]= change in mass and \[\Delta E\]= change in energy.

Complete step by step answer:
The least amount of energy necessary to remove a particle from a system of particles or to deconstruct a system of particles into separate components is known as binding energy in physics and chemistry. The word is most commonly used in condensed matter physics, atomic physics, and chemistry in the former sense, whereas separation energy is used in nuclear physics in the latter.

A bound system's components are usually at a lower energy level than their unbound counterparts. A decrease in the total energy of a system is accompanied by a decrease in the total mass, according to relativity theory. The minimal energy necessary to deconstruct an atom's nucleus into its component protons and neutrons, known collectively as nucleons, is known as nuclear binding energy. Because the nucleus must gain energy in order for the nucleons to move apart, the binding energy is always positive.

The strong nuclear force attracts nucleons to one other. The total of the individual masses of the unbound component protons and neutrons is smaller than the mass of an atomic nucleus. The Einstein equation, \[E = m{c^2}\], may be used to determine the mass difference, where $E$ is the nuclear binding energy, $c$ is the speed of light, and $m$ is the mass difference.

You can use that to prove that a mass of 1 u is equivalent to an energy of 931.5 MeV.Hence the formula is modified as
\[\Delta M(931MeV) = \Delta E\]
Now given: Mass defect:\[\Delta m = 0.00307{\text{ }}u\]
Binding Energy is given by,
\[E = \Delta m \times 931{\text{ }}MeV\]
\[\Rightarrow E = 0.00307 \times 931\]
\[ \therefore E = 2.86{\text{ }}MeV.\]

Note: Remember to calculate the mass defect. It is the average atomic mass minus the Mass number. The phrase "nuclear binding energy" can also apply to the energy balance in procedures when the nucleus divides into fragments with many nucleons. When light nuclei fuse (nuclear fusion) or heavy nuclei split (nuclear fission), additional binding energy is accessible, and either process can result in the release of this binding energy. This energy may be made accessible as nuclear energy, which could be utilised to generate electricity or be used in a nuclear weapon. Excess energy is expelled as gamma rays and the kinetic energy of different ejected particles when a big nucleus breaks into fragments (nuclear fission products).