A lithium-6 nucleus ^{6}_{3}\text{Li} has a measured nuclear mass of 6.0151\text{ u}. Give… — JAMB UTME

A lithium-6 nucleus $^{6}_{3}\text{Li}$ has a measured nuclear mass of $6.0151\text{ u}$ . Given that the mass of a proton is $1.0078\text{ u}$ and the mass of a neutron is $1.0087\text{ u}$ , calculate the binding energy per nucleon of the nucleus in $\text{MeV}$ . (Take $1\text{ u} = 931\text{ MeV}$ )

Answer: 5.34 MeV

Answer

The binding energy per nucleon of the

^{6}_{3}\text{Li}

nucleus is approximately

5.34\text{ MeV}

(or

5.34\text{ MeV/nucleon}

The total mass of 3 free protons and 3 free neutrons is

6.0495\text{ u}

. Subtracting the actual nuclear mass (

6.0151\text{ u}

) yields a mass defect of

0.0344\text{ u}

. Multiplying by

931\text{ MeV/u}

gives a total binding energy of

32.0264\text{ MeV}

. Dividing by the

6

nucleons in lithium-6 yields

5.34\text{ MeV}

per nucleon.

Step-by-Step Solution

Determine the number of protons and neutrons and compute the total constituent mass.

Protons

Z = 3

, Neutrons

N = 3

. Total nucleon mass

m_{\text{nucleons}} = 3(1.0078\text{ u}) + 3(1.0087\text{ u}) = 6.0495\text{ u}

Free nucleons have a combined mass greater than the bound nucleus.

Calculate the mass defect (

\Delta m

\Delta m = 6.0495\text{ u} - 6.0151\text{ u} = 0.0344\text{ u}

The difference between total constituent mass and measured nuclear mass represents the lost mass converted into binding energy.

Convert mass defect to total binding energy in

\text{MeV}

E_b = 0.0344\text{ u} \times 931\text{ MeV/u} = 32.0264\text{ MeV}

Using the equivalence

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

Calculate binding energy per nucleon by dividing by mass number

A = 6

\frac{32.0264\text{ MeV}}{6} = 5.3377\text{ MeV} \approx 5.34\text{ MeV}.

Binding energy per nucleon measures the stability per particle in the nucleus.

Key Concept

Mass Defect and Binding Energy per Nucleon

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