The mass defect of a nitrogen nucleus ^{14}_{7}\text{N} is calculated to be 0.112\text{ u}… — JAMB UTME

The mass defect of a nitrogen nucleus $^{14}_{7}\text{N}$ is calculated to be $0.112\text{ u}$ . Taking $1\text{ u} = 931.5\text{ MeV}$ , what is the total binding energy of the nucleus in $\text{MeV}$ ?

Answer: 104.328 MeV

Answer

The total binding energy of the

^{14}_{7}\text{N}

nucleus is

104.328\text{ MeV}

The total binding energy of a nucleus is equal to the mass defect multiplied by the energy equivalent of one atomic mass unit (

931.5\text{ MeV/u}

). Calculating

0.112\text{ u} \times 931.5\text{ MeV/u}

gives

104.328\text{ MeV}

Step-by-Step Solution

Apply the binding energy formula

E_b = \Delta m \times 931.5\text{ MeV/u}

E_b = 0.112\text{ u} \times 931.5\text{ MeV/u} = 104.328\text{ MeV}

The binding energy is obtained by converting the mass defect directly into its energy equivalent.

Key Concept

Mass Defect and Binding Energy Conversion

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