In a nuclear fusion reaction, two deuterium nuclei ({^{2}_{1}\text{H}}) combine to form a … — JAMB UTME

In a nuclear fusion reaction, two deuterium nuclei ( ${^{2}_{1}\text{H}}$ ) combine to form a helium-3 nucleus ( ${^{3}_{2}\text{He}}$ ) and a neutron ( ${^{1}_{0}\text{n}}$ ). The total mass of the two reactant deuterium nuclei is $4.0282\text{ u}$ , while the total mass of the resulting helium-3 and neutron products is $4.0247\text{ u}$ . Given that $1\text{ u} = 931.5\text{ MeV}$ , calculate the total energy released in this reaction in $\text{MeV}$ .

Answer: 3.26 MeV

Answer

The total energy released in the nuclear fusion reaction is approximately 3.26 MeV.

The energy released in a nuclear fusion reaction is determined by the mass defect, which is the difference between the total mass of reactants and the total mass of products. Subtracting

4.0247\text{ u}

from

4.0282\text{ u}

yields a mass defect of

0.0035\text{ u}

. Multiplying this mass defect by the mass-energy conversion factor of

931.5\text{ MeV/u}

gives an energy release of approximately

3.26\text{ MeV}

Step-by-Step Solution

Calculate the mass defect (Δm)

Δm = 4.0282 u - 4.0247 u = 0.0035 u

Mass defect is the difference between the initial mass of reactants and the final mass of products in a nuclear reaction.

Calculate the energy released (E) in MeV

E = 0.0035 u × 931.5 MeV/u = 3.26025 MeV

According to mass-energy equivalence, 1 unified atomic mass unit (u) liberates 931.5 MeV of energy.

Key Concept

Mass defect and energy release in nuclear fusion

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