Question

Atomic mass of ${}_6^{13}C$ is $13.00335$ amu and its mass number is 13.0. If amu= 931 MeV, the binding energy of the neutrons present in the nucleus is
A. 0.24 MeV
B. 1.44 MeV
C. 1.68 MeV
D. 3.12 MeV

Answer 1

Hint: The mass of the nucleus formed by the combination of the nucleons like protons and neutrons is slightly lesser than the sum of individual masses of the nucleons. This difference in mass is called a mass defect. If one wonders why there is a decrease in mass, it is because when the nucleons are bound, they lose some energy to bind to each other. This lost energy corresponds to the decrease in mass accounted for, due to Einstein’s mass-energy equation.
The energy corresponding to the mass defect is called binding energy per nucleon and the value is equal to 931 MeV.

Formula used:
To calculate the Binding Energy, we need to make use of Einstein’s Energy-Mass equivalence formula, which is:
$E=m{c}^2$
where $c$ = speed of light in vacuum in $meter/sec$

Complete step by step solution:
Binding energy is the energy required to separate the nucleons from each other in the nucleus of the atom. The formula for binding energy is based on Einstein’s Energy-mass equivalence formula:
$E=m{c}^2$
Binding Energy per nucleon = 931 MeV
The nucleus consists of two types of nucleons called Protons and Neutrons. Atomic mass is the actual mass of the nucleus and Mass number is the number of protons and neutrons in the nucleus.
Hence, the mass of neutrons in amu,
$m_{ne}=M-A$
where,
M – atomic mass in amu
A – mass number
In ${}_6^{13}C$,
$$\begin{gathered} M=13.00335amu \\ A=13.0 \end{gathered}$$
Hence, mass of neutrons for the ${}_6^{13}C$ atom is –
$$\begin{gathered} m_{ne}=M-A \\ {m}_{ne}=13.00335-13.0 \\ {m}_{ne}=0.00335amu \end{gathered}$$
The binding energy of the neutrons is equal to the product of the mass of neutrons and the binding energy per neutron –
$$\begin{gathered} B=m_{ne}\times 931MeV \\ B=0.00335\times 931MeV \\ B=3.11885MeV \end{gathered}$$

$\therefore$ The binding energy of the neutrons present in the nucleus is 3.12 MeV. Hence option (D) is correct.

Note:
The energy units for microscopic units are expressed in terms of electron-volt ($eV$).
One electron-volt is equal to the energy gained by an electron when it is accelerated inside an electric field whose potential difference is equal to 1 volt.
The conversion from electron-volt to joule is as follows: $1eV=1.602\times {10}^{-19}J$.