Question

Calculate the binding energy per nucleon for ${}_{10}^{20}Ne,{}_{26}^{56}Fe\text{ and }{}_{92}^{238}U.$ Given that the mass of the neutron is 1.008665amu, the mass of the proton is 1.007825 amu, the mass of ${}_{10}^{20}Ne$ is 19.9924 amu, mass of ${}_{26}^{56}Fe$ is 55.93492 amu, ${}_{92}^{238}U$ is 238.050783 amu.

Answer 1

Hint: The binding energy per nucleon is the ratio of the binding energy of a nucleus to the mass number. There is always a mass difference associated with the calculated mass and original mass of a nucleus; this difference in mass is converted into energy known as the binding energy. Calculated mass is the sum of the mass of the number of protons and mass of the number of neutrons present in a given nucleus. We multiply the mass difference with ${\text{c}}^{\text{2}}$ in order to get the binding energy.

Formula Used:
Calculated mass of a nucleus is,
$m_c=(A-Z)m_n+\left(Z\right)m_p$
Where,
(A-Z) is the number of neutrons in the nucleus.
$m_p$ is the mass of the proton.
$m_n$ is the mass of the neutron.
The mass difference is the difference between the calculated mass and the experimental mass.
$\mathrm{\Delta }m=m_c-m_e$ , where $m_e$ is the experimental mass.
Binding energy in terms of a.m.u is,
$B.E=\mathrm{\Delta }m{c}^2\times \left(931.5{\displaystyle \frac{MeV}{c^2}}\right)$
The binding energy per nucleon is given by,
$B.E\text{ per nucleon}={\displaystyle \frac{B.E}{A}}$

Complete step by step answer:
We know that the atomic nucleus consists of neutrons and protons. But if we take the mass of a proton and multiply it with the number of protons in the nucleus and also do the same thing with the neutron and add these two masses up, we are supposed to get the mass of that particular nucleus. But this is not the case; the true mass of the nucleus is always less than the calculated mass of it. This difference in mass is converted into energy, according to Einstein’s mass-energy equivalence.
We call this energy as the binding energy of the nucleus. The binding energy per nucleon is simply the ratio of the binding energy of a nucleus to its mass number. Atomic nuclei can be represented as,
${}_Z^{A}X$
Where,
A is the mass number, which is the sum of the number of neutrons and the number of protons present.
Z is the atomic number, which gives the number of protons in the nucleus.
So, we can calculate the mass of a nucleus using the formula,
$m_c=(A-Z)m_n+\left(Z\right)m_p$
Where,
(A-Z) is the number of neutrons in the nucleus.
$m_p$ is the mass of the proton.
$m_n$ is the mass of the neutron.
The mass difference is the difference between the calculated mass and the experimental mass.
$\mathrm{\Delta }m=m_c-m_e$ , where $m_e$ is the experimental mass.
This mass difference we have just got is multiplied with the square of the speed of light to attain the binding energy of the nucleus, so we can write,
$B.E=\mathrm{\Delta }m\times c^2$
The binding energy per nucleon is given by,
$B.E\text{ per nucleon}={\displaystyle \frac{B.E}{A}}$
The binding energy per a.m.u is given by, $931.5{\displaystyle \frac{MeV}{c^2}}$. So, if your mass difference is in a.m.u you can multiply the mass difference with the given factor to find the binding energy.
So, if we consider the case of Neon nuclei ${}_{10}^{20}Ne$, Z=10, (A-Z)=10, So the calculated mass of the nuclei is,
$m_c=(A-Z)m_n+\left(Z\right)m_p$
$$m_c=10\times 1.008665amu+10\times 1.007825amu$$
$\therefore m_c=20.165amu$
The experimental mass of Neon is 19.9924 amu. So, the mass difference is,
$\mathrm{\Delta }m=m_c-m_e$
$\mathrm{\Delta }m=20.165amu-19.9924amu$
$\mathrm{\Delta }m=0.1725amu$
So, the binding energy is given by,
$B.E=\mathrm{\Delta }m\times c^2$
Binding energy in terms of a.m.u is,
$B.E=\mathrm{\Delta }m{c}^2\times \left(931.5{\displaystyle \frac{MeV}{c^2}}\right)$
$B.E=160.68MeV$
Binding energy per nucleon is,
$B.E\text{ per nucleon}={\displaystyle \frac{B.E}{A}}={\displaystyle \frac{160.68MeV}{20}}$
$\therefore B.E\text{ per nucleon}=8.034MeV$
So, the binding per nucleon for Neon is 8.034MeV.
Next, we consider the case of Iron nuclei ${}_{92}^{238}U$, Z=92, (A-Z)=146. So the calculated mass of the nuclei is,
$m_c=(A-Z)m_n+\left(Z\right)m_p$
$$m_c=146\times 1.008665amu+92\times 1.007825amu$$
$\therefore m_c=239.985amu$
The experimental mass of Uranium is 239.985 amu. So, the mass difference is,
$\mathrm{\Delta }m=m_c-m_e$
$\mathrm{\Delta }m=239.985amu-238.050783$
$\mathrm{\Delta }m=1.934amu$
Binding energy in terms of a.m.u is,
$B.E=\mathrm{\Delta }m\times \left(931.5MeV\right)$
$B.E=1801.791MeV$
Binding energy per nucleon is,
$B.E\text{ per nucleon}={\displaystyle \frac{B.E}{A}}={\displaystyle \frac{1801.79MeV}{238}}$
$\therefore B.E\text{ per nucleon}=7.57MeV$
So, the binding per nucleon for Uranium is 7.57 MeV.
Next, we consider the case of Uranium nuclei ${}_{26}^{56}Fe$, Z=26, (A-Z)=30, So the calculated mass of the nuclei is,
$m_c=(A-Z)m_n+\left(Z\right)m_p$
$$m_c=30\times 1.008665amu+26\times 1.007825amu$$
$\therefore m_c=56.463amu$
The experimental mass of Iron is 56.463 amu. So, the mass difference is,
$\mathrm{\Delta }m=m_c-m_e$
$\mathrm{\Delta }m=56.463amu-55.93492amu$
$\mathrm{\Delta }m=0.5284amu$
Binding energy in terms of a.m.u is,
$B.E=\mathrm{\Delta }m\times \left(931.5MeV\right)$
$B.E=492.205MeV$
Binding energy per nucleon is,
$B.E\text{ per nucleon}={\displaystyle \frac{B.E}{A}}={\displaystyle \frac{492.205MeV}{56}}$
$\therefore B.E\text{ per nucleon}=8.79MeV$
So, the binding per nucleon for Iron is 8.79MeV.

Note: Binding energy can also be defined as the energy required to split an atomic nucleus into its constituent nucleons (neutrons and protons). The atomic mass unit (a.m.u) is the ${\displaystyle \frac{1}{12}}\text{th}$ of the mass of a carbon 12 atom. It is a unit used to express the mass of atoms and their constituents. Its binding energy is 931.5MeV. Iron shows the highest binding energy per nucleon among all the elements. So, it is considered to be the most stable nucleus in nature.