The atomic mass of a lithium nucleus ^{7}_{3}\text{Li} is 7.0160\text{ u}. Given that the … — JAMB UTME

The atomic mass of a lithium nucleus $^{7}_{3}\text{Li}$ is $7.0160\text{ u}$ . 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 lithium nucleus? ( $1\text{ u} = 931\text{ MeV}$ )

$37.89\text{ MeV}$ Answer
B
$5.41\text{ MeV}$
C
$36.59\text{ MeV}$
D
$0.041\text{ MeV}$

Answer

The total binding energy of the lithium nucleus is

37.89\text{ MeV}

The correct answer is derived by finding the mass defect

\Delta m = [3(1.0073) + 4(1.0087)] - 7.0160 = 0.0407\text{ u}

and multiplying by

931\text{ MeV/u}

to obtain

37.89\text{ MeV}

Step-by-Step Solution

Determine the number of protons and neutrons in

^{7}_{3}\text{Li}

Z = 3

protons,

N = A - Z = 7 - 3 = 4

neutrons

The mass number

A=7

and atomic number

Z=3

define the nuclear composition.

Calculate total mass of individual nucleons

Mass of nucleons

= 3(1.0073\text{ u}) + 4(1.0087\text{ u}) = 3.0219\text{ u} + 4.0348\text{ u} = 7.0567\text{ u}

Summing the masses of constituent protons and neutrons.

Calculate mass defect

(\Delta m)

\Delta m = 7.0567\text{ u} - 7.0160\text{ u} = 0.0407\text{ u}

Mass defect is the difference between total constituent mass and nuclear mass.

Convert mass defect to energy

E_b = 0.0407\text{ u} \times 931\text{ MeV/u} = 37.8917\text{ MeV} \approx 37.89\text{ MeV}

Using the equivalence

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

Key Concept

Mass Defect and Nuclear Binding Energy

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic