What would be the energy required to dissociate completely 1 gram of Ca-40 into its constituent particles?

Mass of proton = 1.007277amu

Mass of neutron — 1.00866 amu

Mass of Ca-40 = 39.97545amu 

(Take one amu = 931 MeV)

Mass defect

$$\begin{gathered} ∆\mathrm{m}=20(1.007277+1.00866)-39.97545 \\ =40.31874-39.97545 \\ =0.34329 \mathrm{amu} \end{gathered}$$

$\therefore$Binding energy 

=0.34329 x 931 = 319.6 MeV

When one atom of Ca-40 completely dissociates, the energy to be supplied =319.6 MeV.

Number of atoms in 1 g of Ca-40 = $\frac{6.023 \mathrm{x} {10}^{23}}{40}$

                                                     $=1.506 \mathrm{x} {10}^{22}$

The emergy required for the dissociation of 1 g of Ca-40

$$\begin{gathered} =319.6 \mathrm{x} 1.506 \mathrm{x} {10}^{22} \\ =4.813 \mathrm{x} {10}^{24} \mathrm{MeV} \end{gathered}$$