In a nuclear fusion reaction, two light nuclei fuse together to form a heavier nucleus. Th… — JAMB UTME

In a nuclear fusion reaction, two light nuclei fuse together to form a heavier nucleus. The total mass of the reactants before fusion is $4.028\text{ u}$ , and the total mass of the products after fusion is $4.003\text{ u}$ . Given that $1\text{ u} = 931.5\text{ MeV}$ , what is the total energy released in this reaction in $\text{MeV}$ ?

Answer: 23.2875 MeV

Answer

The total energy released in the reaction is 23.2875 MeV.

The energy released in nuclear fusion is calculated using the mass defect $\Delta m = m_{\text{reactants}} - m_{\text{products}}$. Subtracting

4.003\text{ u}

from

4.028\text{ u}

yields a mass defect of

0.025\text{ u}

. Multiplying this mass defect by

931.5\text{ MeV/u}

gives the total energy released as

23.2875\text{ MeV}

Step-by-Step Solution

Calculate the mass defect (\Delta m)

\Delta m = 4.028\text{ u} - 4.003\text{ u} = 0.025\text{ u}

Mass defect is the loss of mass during nuclear fusion that gets converted into energy.

Convert mass defect into energy released

E = 0.025\text{ u} \times 931.5\text{ MeV/u} = 23.2875\text{ MeV}

According to mass-energy equivalence, each atomic mass unit (u) of missing mass yields 931.5 MeV of energy.

Key Concept

Mass Defect and Energy Release in Nuclear Reactions

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