A given coin has a mass of 3.0 g.

Solution:

Mass of a copper coin, m’ = 3 g

Atomic mass of ${ }_{29} \mathrm{Cu}^{63}$ atom, $m=62.92960 \mathrm{u}$

The total number of ${ }_{29} \mathrm{Cu}^{63}$ atoms in the coin, $N=\frac{N_{\mathrm{A}} \times m^{\prime}}{\text { Mass number }}$

Where,

NA = Avogadro’s number = 6.023 × 1023 atoms /g

 

Mass number = 63 g

$\therefore N=\frac{6.023 \times 10^{23} \times 3}{63}=2.868 \times 10^{22}$ atoms

${ }_{29} \mathrm{Cu}^{63}$ nucleus has 29 protons and $(63-29) 34$ neutrons

∴Mass defect of this nucleus, Δm' = 29 × mH + 34 × mn − m

Where,

Mass of a proton, mH = 1.007825 u

Mass of a neutron, mn = 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:

$E_{b}=\Delta m c^{2}$

$=1.69766958 \times 10^{22} \times 931.5\left(\frac{\mathrm{MeV}}{c^{2}}\right) \times c^{2}$

= 1.581 × 1025 MeV

But 1 MeV = 1.6 × 10−13 J

Eb = 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.