What would be the energy required to dissociate completely `1 g` of `Ca-40` into its constituent, particles? Given: Mass of proton `=1.00866 am u`,
Mass of neutron `=1.00866 am u`,
Mass of `Ca-40 =39.97454 am u`, (Take `1 am u =931 MeV`).

Options

A. ` 4.813 xx 10^(24)MeV`
B. ` 4.813 xx 10^(24)eV`
C. ` 4.813 xx 10^(23)MeV`
D. none of these

Explanation

To find the energy required to dissociate completely 1 g of Calcium-40 (Ca-40) into its constituent particles, we can follow these steps:

### Step 1: Determine the number of atoms in 1 g of Ca-40
The molar mass of Ca-40 is approximately 40 g/mol. Therefore, the number of moles in 1 g of Ca-40 is:

\[ \text{Number of moles} = \frac{1 \text{ g}}{40 \text{ g/mol}} = 0.025 \text{ mol} \]

Using Avogadro's number (\(6.022 \times 10^{23}\) atoms/mol), the number of atoms in 1 g of Ca-40 is:

\[ \text{Number of atoms} = 0.025 \text{ mol} \times 6.022 \times 10^{23} \text{ atoms/mol} \approx 1.51 \times 10^{22} \text{ atoms} \]

### Step 2: Calculate the mass defect (ΔM)
Calcium-40 has 20 protons and 20 neutrons. The mass of 20 protons and 20 neutrons is calculated as follows:

- Mass of 1 proton = 1.007277 amu
- Mass of 1 neutron = 1.00866 amu

So, the total mass of protons and neutrons is:

\[ \text{Total mass of protons} = 20 \times 1.007277 \text{ amu} = 20.14554 \text{ amu} \]
\[ \text{Total mass of neutrons} = 20 \times 1.00866 \text{ amu} = 20.1732 \text{ amu} \]

Adding these together gives:

\[ \text{Total mass} = 20.14554 + 20.1732 = 40.31874 \text{ amu} \]

Now, the mass defect (ΔM) is given by:

\[ \Delta M = \text{Total mass of nucleons} - \text{mass of Ca-40} \]
\[ \Delta M = 40.31874 \text{ amu} - 39.97454 \text{ amu} = 0.3442 \text{ amu} \]

### Step 3: Calculate the energy equivalent of the mass defect
Using the conversion factor \(1 \text{ amu} = 931 \text{ MeV}\):

\[ E = \Delta M \times c^2 = 0.3442 \text{ amu} \times 931 \text{ MeV/amu} \approx 320.0 \text{ MeV} \]

### Step 4: Calculate the total energy for 1 g of Ca-40
The energy released for one atom is approximately 320.0 MeV. For 1 g of Ca-40, the total energy (E_total) is:

\[ E_{\text{total}} = E \times \text{Number of atoms} \]
\[ E_{\text{total}} = 320.0 \text{ MeV} \times 1.51 \times 10^{22} \approx 4.83 \times 10^{24} \text{ MeV} \]

### Final Answer
The energy required to dissociate completely 1 g of Ca-40 into its constituent particles is approximately \(4.83 \times 10^{24} \text{ MeV}\).

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