Question 368: Atoms & Nuclei

Question

In a nuclear fission process, a high mass nuclide $$(A \approx 236)$$ with binding energy $$7.6 \mathrm{~MeV} /$$ Nucleon dissociated into middle mass nuclides $$(\mathrm{A} \approx 118)$$, having binding energy of $$8.6 \mathrm{~MeV} / \mathrm{Nucleon}$$. The energy released in the process would be ______ $$\mathrm{MeV}$$.

Accepted Answer

To determine the energy released in a nuclear fission process, we use the difference in binding energy (BE) before and after the fission. The formula for energy released ($$Q$$ value) in the process is given by:
$$Q = (	ext{Total BE of products}) - (	ext{Total BE of reactants})$$
In this case, the reactant is a high mass nuclide with atomic mass $A \approx 236$ and a binding energy of $$7.6 \mathrm{MeV}/\mathrm{nucleon}$$. Each of the two middle mass nuclides formed as products has atomic mass $A \approx 118$ and a binding energy of $$8.6 \mathrm{MeV}/\mathrm{nucleon}$$.
Therefore, we calculate the total binding energy of reactant and products as follows:
For reactant:
$$	ext{BE}_{	ext{reactant}} = 236 	imes 7.6 \mathrm{MeV}$$
For products (since there are two identical products):
$$	ext{BE}_{	ext{products}} = 2 	imes (118 	imes 8.6) \mathrm{MeV}$$
Thus, the energy released ($$Q$$ value) is:
$$Q = 	ext{BE}_{	ext{products}} - 	ext{BE}_{	ext{reactant}}$$
$$Q = 2(118 	imes 8.6) - (236 	imes 7.6)$$
$$Q = 236 	imes (8.6 - 7.6)$$
$$Q = 236 	imes 1$$
$$Q = 236 \mathrm{MeV}$$
This calculation demonstrates how the difference in binding energy per nucleon before and after fission leads to the release of energy, consistent with the mass-energy equivalence principle.

Atoms & Nuclei

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