hai laveti you come in dorm  and ask the question answer

thanks

orey laveti iam nagesh

Mass of a copper coin, m’ = 3 g
Atomic mass of atom, m = 62.92960 u
The total number of  atoms in the coin
Where,
N
A = Avogadro’s number = 6.023 × 1023 atoms /g
Mass number = 63 g
nucleus has 29 protons and (63 − 29) 34 neutrons
∴Mass defect of this nucleus, Δm'' = 29 × m
H + 34 × m
n − m
Where,
Mass of a proton, m
H = 1.007825 u
Mass of a neutron, m
n = 1.008665 u
∴Δm'' = 29 × 1.007825 + 34 × 1.008665 − 62.9296
= 0.591935 u
Mass defect of all the atoms present in the coin, Δm = 0.591935 × 2.868 × 1022

= 1.69766958 × 1022 u
But 1 u = 931.5 MeV/c2

∴Δm = 1.69766958 × 1022 × 931.5 MeV/c2
Hence, the binding energy of the nuclei of the coin is given as:
E
b= Δmc2

= 1.69766958 × 1022 × 931.5 
= 1.581 × 1025 MeV
But 1 MeV = 1.6 × 10−13 J
E
b = 1.581 × 1025 × 1.6 × 10−13 

= 2.5296 × 1012 J
This much energy is required to separate all the neutrons and protons from the given coin.