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Binding energy per nucleon for ⁿ_ZP and ²ⁿ_Z'Q are x and y respectively. How much energy would be released in the process ⁿ_ZP+ⁿ_ZP→²ⁿ_Z'Q ?

Asked in RS Academy GUJCET booklet · Energy released in fission and fusion

Answer: (3) 2ny-2nx

Step-by-step solution

Given: the binding energy per nucleon is x for ⁿ_ZP, which has n nucleons, and y for ²ⁿ_Z'Q, which has 2n.

Idea: the energy released is the total binding energy of the product minus the total binding energy of the reactants. A nucleus's total binding energy is its binding energy per nucleon times its own number of nucleons.

Before: two P nuclei, 2× nx=2nx.

After: one Q nucleus, 2n× y=2ny.

Energy released =2ny-2nx=2n(y-x).

So energy is given out only when y>x, that is when the heavier product is the more tightly bound nucleus — which is exactly the condition for fusion to be worth anything.

Why the other options are wrong

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