A nitrogen nucleus ^{14}_{7}\text{N} has a nuclear mass of 13.9992\text{ u}. Given that th… — JAMB UTME

A nitrogen nucleus $^{14}_{7}\text{N}$ has a nuclear mass of $13.9992\text{ u}$ . Given that the mass of a proton is $1.0073\text{ u}$ and the mass of a neutron is $1.0087\text{ u}$ , what is the total binding energy of the nucleus in $\text{MeV}$ ? (Take $1\text{ u} = 931\text{ MeV}$ )

Answer: 105.02 MeV

Answer

The total binding energy of the

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

nucleus is

105.02\text{ MeV}

The total binding energy is computed by finding the total rest mass of 7 protons and 7 neutrons (14.1120 u), subtracting the actual nuclear mass of Nitrogen-14 (13.9992 u) to obtain a mass defect of 0.1128 u, and converting this mass defect to energy by multiplying by 931 MeV/u, yielding 105.02 MeV.

Step-by-Step Solution

Determine the number of constituent protons and neutrons

Z = 7

protons and

N = 14 - 7 = 7

neutrons

The atomic number is 7 and the mass number is 14.

Calculate the total mass of the constituent nucleons

M_{\text{nucleons}} = (7 \times 1.0073\text{ u}) + (7 \times 1.0087\text{ u}) = 7.0511\text{ u} + 7.0609\text{ u} = 14.1120\text{ u}

The sum of the individual rest masses of all isolated protons and neutrons.

Compute the mass defect

\Delta m = 14.1120\text{ u} - 13.9992\text{ u} = 0.1128\text{ u}

Mass defect is the difference between total mass of isolated nucleons and the bound nuclear mass.

Convert the mass defect into energy in MeV

E_b = 0.1128\text{ u} \times 931\text{ MeV/u} = 105.0168\text{ MeV} \approx 105.02\text{ MeV}

Applying the mass-energy equivalence factor

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

Key Concept

Mass Defect and Binding Energy

Estimated Time:2m 0s

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