Question

Suppose a nucleus with mass number $A = 240$ and B.E. per nucleon $= 7.6$ MeV, breaks into two nuclei, each of mass number $A = 120$ with B.E. per nucleon $= 8.5$ MeV. Calculate the energy released in the process.

Hint:

Energy released \(=\) total B.E. of the two products \(-\) B.E. of the parent \(= 2040 - 1824\) MeV.

Answer 1

Step 1: Understanding the Concept:
The binding energy of a nucleus is the energy needed to break it into free nucleons. In fission, the energy released equals the total binding energy of the products minus the binding energy of the parent.

Step 2: Binding Energy of the Parent:
\(A = 240\), B.E. per nucleon \(= 7.6\) MeV.
\[ BE_{parent} = 240 \times 7.6 = 1824\ \text{MeV} \]

Step 3: Binding Energy of the Two Products:
Each daughter has \(A = 120\) and B.E. per nucleon \(= 8.5\) MeV.
\[ BE_{products} = 2 \times (120 \times 8.5) = 2 \times 1020 = 2040\ \text{MeV} \]

Step 4: Energy Released:
\[ Q = BE_{products} - BE_{parent} = 2040 - 1824 = 216\ \text{MeV} \]
The products are more tightly bound, so the extra binding energy comes out as released energy.

Final Answer:
The energy released in the process is 216 MeV. \[ \boxed{216\ \text{MeV}} \]