Two deuterium nuclei ({^{2}_{1}\text{H}}) undergo nuclear fusion to form a helium-3 nucleu… — JAMB UTME

Two deuterium nuclei (

{^{2}_{1}\text{H}}

) undergo nuclear fusion to form a helium-3 nucleus (

{^{3}_{2}\text{He}}

) and a neutron (

{^{1}_{0}\text{n}}

) according to the nuclear equation:

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

Given the atomic mass values:
Mass of

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

Mass of

{^{3}_{2}\text{He}} = 3.0160\text{ u}

Mass of

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

Taking

1\text{ u} = 931.5\text{ MeV}

, what is the energy released in this fusion reaction?

The energy released in the fusion reaction is

3.26\text{ MeV}

The total mass of the reactants is

2 \times 2.0141\text{ u} = 4.0282\text{ u}

, while the total mass of the products is

3.0160\text{ u} + 1.0087\text{ u} = 4.0247\text{ u}

. Subtracting product mass from reactant mass gives a mass defect

\Delta m = 0.0035\text{ u}

. Multiplying this mass defect by the energy equivalent factor of

931.5\text{ MeV/u}

yields

3.26\text{ MeV}

Calculate the total mass of the reactants

Total reactant mass

= 2 \times 2.0141\text{ u} = 4.0282\text{ u}

Two deuterium nuclei are fusing on the left side of the reaction equation.

Calculate the total mass of the products

Total product mass

= 3.0160\text{ u} + 1.0087\text{ u} = 4.0247\text{ u}

The reaction produces one helium-3 nucleus and one neutron.

Determine the mass defect (\Delta m)

\Delta m = 4.0282\text{ u} - 4.0247\text{ u} = 0.0035\text{ u}

Mass defect is the difference between total reactant mass and total product mass.

Calculate the energy released in MeV

E = 0.0035\text{ u} \times 931.5\text{ MeV/u} = 3.26025\text{ MeV} \approx 3.26\text{ MeV}

Multiplying the mass defect in atomic mass units (u) by

931.5\text{ MeV/u}

gives the total energy released.

Mass Defect and Energy Release in Nuclear Fusion

Estimated Time:1m 30s

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