Question

The binding energy of nucleus is given by $$E_b=\mathrm{\Delta }m{c}^2=\left[Z{m}_p+\left(A-Z\right)m_n-m_N\right]c^2=\left[Z{m}_p+\left(A-Z\right)m_n-m_N\right]\times 931.49MeV/u$$

Answer 1

Hint Here we understand the binding energy equation and then solve the equation.so Once the mass defect is determined, by applying the formula, nuclear binding energy can be determined by converting mass to energy. You can scale this energy into per-mole amounts and per-nucleon when it is measured from joules for a nucleus.
Useful formula
Express binding energy as
 $E_b=\mathrm{\Delta }m{c}^2$
Where,
$\mathrm{\Delta }m{c}^2$ is mass defect,

Complete step by step procedure
Given by,
Binding energy,
$$E_b=\mathrm{\Delta }m{c}^2=\left[Z{m}_p+\left(A-Z\right)m_n-m_N\right]c^2=\left[Z{m}_p+\left(A-Z\right)m_n-m_N\right]\times 931.49MeV/u$$
Now,
Energy is released when nucleus is assembled from constituent nucleons; this is binding energy of nucleus
Release comes from work done by strong nuclear force
According to the binding energy formula,
 $E_b=\mathrm{\Delta }m{c}^2$
 ∆m is difference in combined masses of constituent nucleons $m_j$ , and mass of nucleus $m_{nuc}$
$\mathrm{\Delta }m=X_j{m}_j-m_{nuc}$
Classifying nuclei
$Z$ is number of protons either atomic number
$N$ is number of neutrons
$A$ is number of nucleons
$$\left(protons+neutrons;massnumber\right)$$
$$A=Z+N$$
We know the mass of constituents
Here,
${\sum }_j{m}_j=Z{m}_p+N{m}_n=Z{m}_p\left(A-Z\right)m_n$
 where $m_p$ is proton mass, $m_n$ is neutron mass
So, binding energy
Therefore,
$E_b=\left[Z{m}_p+\left(A-Z\right)m_n-m_N\right]c^2$
Now,
We combine to solve the mass effect,
Here,
We take
$Z{m}_p+\left(A-Z\right)m_n=m\left(A,Z\right)$
Then, we know the mass effect,
Depending on the sum of the masses of the isolated elements, nuclear binding energy poses a major difference between the real mass of the nucleus and its predicted mass.
Mass effect $={\displaystyle \frac{B.E}{e^2}}$
Hence,

Thus, the binding energy is $m\left(A,Z\right)=Z{m}_n-\left(A-Z\right)m_n-{\displaystyle \frac{B.E}{e^2}}$

Note The binding energy is so strong that it maintains a large amount of mass. In any case, the real mass is less than the sum of the individual masses of the constituent neutrons and protons, because when the nucleus is formed, energy is expelled. In cases where the nucleus breaks into fragments that consist of more than one nucleon, the binding energies are often related to the nuclear binding energy.