Question

Obtain the binding energy of a nitrogen nucleus ( $_7{N^{14}}$ ) from the following data:
${m_H} = 1.00783\,a.m.u.$ , ${m_n} = 1.0087\,a.m.u.$ ; ${m_N} = 14.00307\,a.m.u.$

Answer 1

Binding energy is the minimum amount of energy required to remove a particle from a system of particles. Binding energy is calculated using the formula; $B.E. = \Delta m{c^2}$ . We know that $1\,a.m.u$ is equivalent to $931.5\,MeV$ of energy.

Complete step by step answer:
We need energy to remove a particle from an atom. This energy is to be given to the atom. An energy less than the required energy will not be able to remove the particle from the atom. This minimum required energy is known as binding energy.
Binding energy is calculated as follows:
$BE = \Delta m{c^2}$--equation $1$
Where $c$ is the speed of light.
And $\Delta m = Z \times {m_H} + N{m_n} - {m_N}$--equation $2$
Where $Z$ is the number of protons in Nitrogen and $Z = 7$

$N$ is the number of neutrons in the Nitrogen atom, having value $N = 7$. Also, the mass of the proton, neutron is given in a.m.u. we will have to convert it into MeV.
$1\,a.m.u$ is equivalent to $\dfrac{{931.5}}{{{c^2}}}MeV$
Using the above relation and equation $2$ , we have
$\Delta m = 7 \times 1.00783 + 7 \times 1.0087 - 14.00307$
$\Rightarrow \Delta m = 0.11236\,a.m.u$
Now the binding energy will be given as:
$BE = 108.6633\,MeV$
$\therefore BE = 108.6633\,MeV$

Hence, the binding energy of a Nitrogen nucleus is $108.6633\,MeV$.

Additional information:
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 and to atoms and ions bound together in crystals. Nickel has the highest binding energy.

Note: $1\,a.m.u$ is equivalent to $\dfrac{{931.5}}{{{c^2}}}MeV$.Be careful not to confuse between the atomic masses of Nitrogen and mass of proton and neutron. The atomic number of protons is $Z = 7$. Convert the final answer into MeV.