Energy Released in Nuclear Fusion Process

To find the energy equivalent of the mass difference $\Delta M c^2$ in the given nuclear fusion process, we compare the total binding energy of the initial nuclei with that of the final nucleus.

1. Initial Binding Energy ($BE_i$)

The initial state consists of:

• Two nuclei with mass number $A = 3$ and binding energy per nucleon of $5.6\text{ MeV}$: $BE_1 = 2 \times (3 \times 5.6\text{ MeV}) = 2 \times 16.8\text{ MeV} = 33.6\text{ MeV}$
• One nucleus with mass number $A = 4$ and binding energy per nucleon of $7.4\text{ MeV}$: $BE_2 = 1 \times (4 \times 7.4\text{ MeV}) = 29.6\text{ MeV}$

The total binding energy before the reaction is: $BE_i = BE_1 + BE_2 = 33.6\text{ MeV} + 29.6\text{ MeV} = 63.2\text{ MeV}$

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2. Final Binding Energy ($BE_f$)

The final state consists of one nucleus with mass number $A = 10$ and binding energy per nucleon of $6.1\text{ MeV}$: $BE_f = 10 \times 6.1\text{ MeV} = 61.0\text{ MeV}$

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3. Mass Difference / Energy Difference ($\Delta M c^2$)

The total rest mass energy $M c^2$ of a nucleus is related to its nucleons' mass by: $M c^2 = \sum m_{\text{nucleons}} c^2 - BE$

The change in energy in terms of mass defect magnitude is: $\Delta M c^2 = |BE_f - BE_i| = |61.0\text{ MeV} - 63.2\text{ MeV}| = 2.2\text{ MeV}$

Hence, the magnitude of $\Delta M c^2$ involved in the process is $2.2\text{ MeV}$.

Correct Option: C ($2.2$)