$A$ 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 $_{29}^{63} Cu$ atoms (of mass $62.92960\; u$).

(N/A) Mass of a copper coin, $m' = 3\; g$.
Atomic mass of $_{29}^{63} Cu$ atom, $m = 62.92960\; u$.
The total number of $_{29}^{63} Cu$ atoms in the coin is $N = \frac{N_A \times m'}{\text{Mass number}}$.
Where $N_A = 6.023 \times 10^{23}\; \text{atoms/mol}$ and Mass number $= 63\; g/mol$.
$N = \frac{6.023 \times 10^{23} \times 3}{63} = 2.868 \times 10^{22}$ atoms.
$A$ $_{29}^{63} Cu$ nucleus has $29$ protons and $(63 - 29) = 34$ neutrons.
Mass defect of one nucleus, $\Delta m' = 29 \times m_H + 34 \times m_n - m$.
Using $m_H = 1.007825\; u$ and $m_n = 1.008665\; u$:
$\Delta m' = 29(1.007825) + 34(1.008665) - 62.9296 = 0.591935\; u$.
Total mass defect for all atoms, $\Delta m = 0.591935 \times 2.868 \times 10^{22} = 1.69767 \times 10^{22}\; u$.
Using $1\; u = 931.5\; MeV/c^2$, the binding energy $E_b = \Delta m \times 931.5\; MeV$.
$E_b = 1.69767 \times 10^{22} \times 931.5 = 1.581 \times 10^{25}\; MeV$.
Converting to Joules $(1\; MeV = 1.602 \times 10^{-13}\; J)$:
$E_b = 1.581 \times 10^{25} \times 1.602 \times 10^{-13} \approx 2.53 \times 10^{12}\; J$.