Consider the nuclear fission, $$Ne^{20} \rightarrow 2He^4 + C^{12}$$. Given that the binding energy/nucleon of $$Ne^{20}$$, $$He^4$$ and $$C^{12}$$ are 8.03 MeV, 7.86 MeV, respectively. Identify the correct statement:

We have to examine the reaction $$^{20}\!Ne \;\longrightarrow\; 2\,^{4}\!He \;+\; ^{12}\!C.$$

First recall the idea of binding energy. The binding energy of a nucleus is the energy released when its constituent nucleons come together. A larger binding energy therefore means a more stable (lower-mass) nucleus. In any nuclear reaction, the net energy change is obtained by comparing the total binding energy of the products with that of the reactants:

$$\Delta E \;=\; \bigl(\text{B.E. of products}\bigr) \;-\; \bigl(\text{B.E. of reactants}\bigr).$$

If $$\Delta E \lt 0,$$ energy has to be supplied; if $$\Delta E \gt 0,$$ energy is released.

The binding energies per nucleon which we shall use are stated in the question:

$$^{20}\!Ne:\; 8.03\;\text{MeV per nucleon},\qquad ^{4}\!He:\; 7.07\;\text{MeV per nucleon},\qquad ^{12}\!C:\; 7.86\;\text{MeV per nucleon}.$$

Now we convert these per-nucleon figures into total binding energies for each nucleus, simply by multiplying by the number of nucleons present (the mass number A).

For the reactant $$^{20}\!Ne$$:

$$\text{B.E.}_{^{20}\!Ne}= 20 \times 8.03\;\text{MeV} = 160.60\;\text{MeV}.$$

For the products:

There are two α-particles $$\bigl(^{4}\!He\bigr)$$, each containing 4 nucleons.

$$\text{B.E. of one }^{4}\!He = 4 \times 7.07\;\text{MeV} = 28.28\;\text{MeV}.$$

$$\text{B.E. of two }^{4}\!He = 2 \times 28.28\;\text{MeV} = 56.56\;\text{MeV}.$$

For $$^{12}\!C$$ we have 12 nucleons.

$$\text{B.E.}_{^{12}\!C} = 12 \times 7.86\;\text{MeV} = 94.32\;\text{MeV}.$$

So the total binding energy of the products is

$$\text{B.E.}_{\text{products}} = 56.56\;\text{MeV} + 94.32\;\text{MeV} = 150.88\;\text{MeV}.$$

With all totals in hand, we evaluate the energy change:

$$\Delta E = \text{B.E.}_{\text{products}} - \text{B.E.}_{\text{reactant}} = 150.88\;\text{MeV} - 160.60\;\text{MeV} = -9.72\;\text{MeV}.$$

The negative sign tells us that the products possess less binding energy than the initial nucleus. Because binding energy and mass are inversely related, the products are effectively heavier than the reactant. To create these heavier, less-bound products we must supply the energy deficit:

$$9.72\;\text{MeV must be supplied}.$$

Hence, the correct answer is Option B.