A nucleus `._(Z)^(A)X` has mass represented by `m(A, Z)`. If `m_(p)` and `m_(n)` denote the mass of proton and neutron respectively and `BE` the blinding energy (in MeV), then

Options

A. `BE=[m(A,Z)-Zm_(p)-(A-Z)m_(n)]C^(2)`
B. `BE=[Zm_(p)+(A-Z)m_(n)-m(A,Z)C^(2)`
C. `BE=[Zm_(p)+Am_(n)-m(A,Z)]C^(2)`
D. `BE=m(A,Z)-Zm_(p)-(A-Z)m_(N)`

Answer

b

Explanation

In the case of formation of nucleus the evolution of energy equals to the binding energy of the nucl,eus takes place due to disappereacne of fraction of total mass .If the quantity of mass desappering is `trianglem`, then the binding energy is `BE=trianglemc^(2)`
From the above discussion, it is clear that the mass of the nulceus must be less than the sum of the masses of the constitutent neutrons and protons . We can then write
? `trianglem=Zx_(p)-Nm_(n)-m(A,Z)C^(2)`
`BE=[Zm_(p)+(A-Z)m_(n)-m(A,Z)c^(2)]`