A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of `""_29^63Cu`atoms (of mass 62.92960 u).

#### Solution

Mass of a copper coin, *m*’ = 3 g

Atomic mass of `""_29Cu^63` atom, *m* = 62.92960 u

The total number `""_29Cu^63` of atoms in the coin, `N = (N_A xx m')/"Mass number"`

Where,

N_{A} = Avogadro’s number = 6.023 × 10^{23}_{ }atoms /g

Mass number = 63 g

`:. N = (6.023 xx 10^23 xx 3)/63 = 2.868 xx 10^22 "atoms/g"`

`""_29Cu^63` nucleus has 29 protons and (63 − 29) 34 neutrons

∴Mass defect of this nucleus, Δ*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

∴Δ*m*' = 29 × 1.007825 + 34 × 1.008665 − 62.9296

= 0.591935 u

Mass defect of all the atoms present in the coin, Δ*m* = 0.591935 × 2.868 × 10^{22}

= 1.69766958 × 10^{22} u

But 1 u = 931.5 MeV/*c*^{2}

∴Δ*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}= Δ*mc*^{2}

= 1.69766958 × 10^{22} × 931.5 `("MeV"/c^3)xx c^3`

= 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.