A nucleus contains 6 protons and 6 neutrons. Given that the mass of a proton is 1.0078\tex… — JAMB UTME

A nucleus contains $6$ protons and $6$ neutrons. Given that the mass of a proton is $1.0078\text{ u}$ , the mass of a neutron is $1.0087\text{ u}$ , and the total binding energy of the nucleus is $93.10\text{ MeV}$ , what is the mass of the nucleus in atomic mass units ( $\text{u}$ )? (Take $1\text{ u} = 931\text{ MeV}$ )

Answer: 11.999 u

Answer

The mass of the nucleus is

11.9990\text{ u}

To find the mass of the nucleus, first calculate the combined mass of the 6 free protons and 6 free neutrons:

6(1.0078\text{ u}) + 6(1.0087\text{ u}) = 12.0990\text{ u}

. Next, determine the mass defect from the binding energy:

\Delta m = \frac{93.10\text{ MeV}}{931\text{ MeV/u}} = 0.1000\text{ u}

. Finally, subtract the mass defect from the total mass of individual nucleons to get the nuclear mass:

12.0990\text{ u} - 0.1000\text{ u} = 11.9990\text{ u}

Step-by-Step Solution

Calculate the total mass of the individual free protons and neutrons.

Total mass of nucleons =

6(1.0078\text{ u}) + 6(1.0087\text{ u}) = 12.0990\text{ u}

The mass of the constituent particles before binding is the sum of their individual rest masses.

Calculate the mass defect from the given binding energy.

Mass defect

\Delta m = \frac{93.10\text{ MeV}}{931\text{ MeV/u}} = 0.1000\text{ u}

Binding energy and mass defect are related by

E_b = \Delta m \times 931\text{ MeV/u}

Subtract the mass defect from the total nucleon mass to obtain the nuclear mass.

Mass of nucleus =

12.0990\text{ u} - 0.1000\text{ u} = 11.9990\text{ u}

Mass defect represents the mass lost when nucleons bind together into a nucleus.

Key Concept

Mass Defect and Binding Energy Relationship

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