In a nuclear fusion reaction, a lithium-6 nucleus ({^{6}_{3}\text{Li}}) fuses with a deute… — JAMB UTME

In a nuclear fusion reaction, a lithium-6 nucleus ( ${^{6}_{3}\text{Li}}$ ) fuses with a deuterium nucleus ( ${^{2}_{1}\text{H}}$ ) to form two alpha particles ( ${^{4}_{2}\text{He}}$ ). Given the atomic masses of ${^{6}_{3}\text{Li}} = 6.0151\text{ u}$ , ${^{2}_{1}\text{H}} = 2.0141\text{ u}$ , and ${^{4}_{2}\text{He}} = 4.0026\text{ u}$ , and taking $1\text{ u} = 931.5\text{ MeV}$ , what is the total energy released in this reaction in $\text{MeV}$ ?

Answer: 22.356 MeV

Answer

The total energy released in the fusion reaction is 22.356 MeV.

The total mass before fusion (

{^{6}_{3}\text{Li}} + {^{2}_{1}\text{H}}

) is

8.0292\text{ u}

, while the total mass after fusion (

2 \times {^{4}_{2}\text{He}}

) is

8.0052\text{ u}

. The resulting mass defect is

\Delta m = 0.0240\text{ u}

. Multiplying this mass loss by

931.5\text{ MeV/u}

gives an energy release of

22.356\text{ MeV}

Step-by-Step Solution

Calculate the total initial mass of the reactants

m_{\text{initial}} = 6.0151\text{ u} + 2.0141\text{ u} = 8.0292\text{ u}

Add the rest masses of the lithium-6 nucleus and the deuterium nucleus together.

Calculate the total final mass of the products

m_{\text{final}} = 2 \times 4.0026\text{ u} = 8.0052\text{ u}

The reaction produces two helium-4 nuclei (alpha particles), so multiply the mass of one alpha particle by 2.

Determine the mass defect

\Delta m = m_{\text{initial}} - m_{\text{final}} = 8.0292\text{ u} - 8.0052\text{ u} = 0.0240\text{ u}

Subtract the final mass of products from the initial mass of reactants to find the mass lost.

Convert the mass defect into energy released

E = \Delta m \times 931.5\text{ MeV/u} = 0.0240 \times 931.5 = 22.356\text{ MeV}

Apply Einstein's mass-energy equivalence principle using the standard conversion factor

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

Key Concept

Mass Defect and Energy Release in Nuclear Fusion Reactions

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