A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separated all the neutrons and protons form each other. for simplicity, assume that the coin is entirely made of `._29Cu^(63)` atoms (of mass 62.92960 u).

Explanation

Mass of a copper coin, m’ = 3 g
Atomic mass of `_(29)Cu^(63)` atom, m = 62.92960 u
The total number of `_(29)Cu^(63)` atoms in the coin, `N=(N_(A)xxm')/("mass number")`
Where,
`N_(A)="Avogadro’s number" = 6.023 × 10^(23) "atoms" //g`
Mass number = 63 g
`therefore N=(6.023xx10^(23)xx3)/(63)=2.868xx10^(22)" atoms "`
`_(29)Cu^(63)` nucleus has 29 protons and (63 − 29) 34 neutrons
`therefore "Mass defect of this nucleus", triangle m' = 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`
`therefore triangle m' = 29 × 1.007825 + 34 × 1.008665 − 62.9296`
`= 0.591935 " "u`
Mass defect of all the atoms present in the coin, `triangle m = 0.591935 × 2.868 × 1022`
`= 1.69766958 × 10^(22) u`
`But 1 u = 931.5 (MeV)/c^(2)`
`therefore triangle m = 1.69766958 × 10^(22) × 931.5 MeV//c^(2)`
Hence, the binding energy of the nuclei of the coin is given as:
`E_(b)= triangle mc^(2)`
`= 1.69766958 × 1022 × 931.5 ((MeV)/(c^(2)))xxc^(2)`
`= 1.581 × 10^(25) MeV`
`But 1 MeV = 1.6 × 10^(−13) J`
`E_(b) = 1.581 × 10^(25) × 1.6 × 10^(−13)`
`= 2.5296 × 10_(12) J`
This much energy is required to separate all the neutrons and protons from the given coin.