Consider the nuclear fusion reaction below: \text{^{2}_{1}H} + \text{^{3}_{1}H} \rightarro… — JAMB UTME

Consider the nuclear fusion reaction below:

\text{^{2}_{1}H} + \text{^{3}_{1}H} \rightarrow \text{^{4}_{2}He} + \text{^{1}_{0}n} + Q

Given the rest masses:
Mass of

\text{^{2}_{1}H} = 2.0141\text{ u}

Mass of

\text{^{3}_{1}H} = 3.0160\text{ u}

Mass of

\text{^{4}_{2}He} = 4.0015\text{ u}

Mass of

\text{^{1}_{0}n} = 1.0087\text{ u}

1\text{ u} = 931\text{ MeV}

, what is the total energy released (

Q

) in this fusion reaction?

18.53\text{ MeV}

The total initial mass of the reactants is

2.0141\text{ u} + 3.0160\text{ u} = 5.0301\text{ u}

. The total final mass of the products is

4.0015\text{ u} + 1.0087\text{ u} = 5.0102\text{ u}

. Subtracting the product mass from the reactant mass gives a mass defect of

\Delta m = 0.0199\text{ u}

. Multiplying this mass defect by the conversion factor

931\text{ MeV/u}

yields an energy release of

18.53\text{ MeV}

Calculate the total mass of the reactants

\text{Mass of reactants} = 2.0141\text{ u} + 3.0160\text{ u} = 5.0301\text{ u}

The total initial mass must be determined from the sum of the masses of deuterium and tritium.

Calculate the total mass of the products

\text{Mass of products} = 4.0015\text{ u} + 1.0087\text{ u} = 5.0102\text{ u}

The total final mass includes both the helium nucleus and the released neutron.

Determine the mass defect (

\Delta m

)

\Delta m = 5.0301\text{ u} - 5.0102\text{ u} = 0.0199\text{ u}

Mass defect is the difference between the total initial mass and the total final mass.

Convert the mass defect into energy released (

Q

)

Q = 0.0199 \times 931\text{ MeV} = 18.5269\text{ MeV} \approx 18.53\text{ MeV}

Using the mass-energy equivalence factor

1\text{ u} = 931\text{ MeV}

yields the energy in

\text{MeV}

Calculation of energy released in nuclear fusion reactions using mass defect

Estimated Time:2m 0s

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