Question

Obtain the binding energy of the nuclei ${}_{26}F{{e}^{56}}$ and${}_{83}B{{i}^{209}}$ in units of MeV from the following data: mass of hydrogen atom = 1.007825amu, mass of neutron = 1.08665amu, mass of ${}_{26}F{{e}^{56}}$ atom = 55.934939amu and mass of ${}_{83}B{{i}^{209}}$ atom = 208.980388amu. Also calculate binding energy per nucleon in the two cases.

Answer 1

Hint: From the given symbol of the element we will get the number of protons and number of neutrons present in them. Also the mass of hydrogen gives you the mass of a proton. Multiply the number of protons and neutrons with their masses and find their sum to get the total mass of the nucleons. Now subtract the given masses of the nucleus from the above will give you the mass defect and substitute this in the expression for binding energy to get the answer. Now divide this binding energy with the number of nucleons in each atom to get the binding energy per nucleon.

Formula used:
Binding energy,
$B.E=\Delta m{{c}^{2}}$

Complete step by step solution:
In the question we are given two nuclei ${}_{26}F{{e}^{56}}$ and${}_{83}B{{i}^{209}}$. We are also given mass of an atom of each nucleus, mass of neutron and mass of a hydrogen atom. With these given information, we are asked to find the binding energy of the nuclei and also the binding energy per nucleon.
We represent an element ‘X’ basically as,${}_{Z}{{X}^{A}}$, where, Z is the atomic number ( number of protons/ electrons) and A is the atomic mass (sum of the number of protons and neutrons).
We know that atomic hydrogen has just a proton in its nucleus to contribute towards its mass, so, mass of a hydrogen atom could be taken as mass of a proton. So,
Mass of hydrogen atom = mass of proton = 1.007825amu ………………………………… (1)
And we are given,
Mass of neutron = 1.08665amu ………………………………. (2)
For${}_{26}F{{e}^{56}}$, number of protons = 26
Number of neutrons = 56 - 26 = 30
Using (1), mass of 26 protons = $26\times 1.007825amu=26.20345amu$
Using (2), mass of 30 neutrons = $30\times 1.008665amu=30.25995amu$
So the total mass of nucleons =$26.20345amu+30.25995amu=56.4634amu$
But mass of a ${}_{26}F{{e}^{56}}$ nucleus is found to be = 55.934939amu
So, the mass defect can be given by,
$\Delta m=56.4634amu-55.934939amu$
$\Rightarrow \Delta m=0.528461amu$
Total binding energy is given by,
$B.E=\Delta m{{c}^{2}}$ ……………………………………….. (3)
Also, ${{c}^{2}}=931.5MeV$
Total binding energy here will be,
$B.{{E}_{Fe}}=0.528461\times 931.5MeV$
$B.{{E}_{Fe}}=492.26MeV$
Average Binding energy per nucleon will be binding energy divided by total number of nucleons.
So, average binding energy per nucleon here will be = $\dfrac{492.26}{56}=8.790MeV$
For${}_{83}B{{i}^{209}}$ , number of protons = 83
Number of neutrons = 209 - 83 = 126
Using (1), mass of 26 protons = $83\times 1.007825amu=83.649475amu$
Using (2), mass of 30 neutrons = $126\times 1.008665amu=127.09179amu$
So the total mass of nucleons =$83.649475amu+127.09179amu=210.741265amu$
But mass of a ${}_{83}B{{i}^{209}}$ nucleus is found to be = 208.980388amu
So, the mass defect can be given by,
$\Delta m=210.741265amu-208.980388amu$
$\Rightarrow \Delta m=1.760877amu$
 From (3), Total binding energy here will be,
$B.{{E}_{Bi}}=1.760877\times 931.5MeV$
$B.{{E}_{Fe}}=1640.26MeV$
Average Binding energy per nucleon will be binding energy divided by total number of nucleons.
So, average binding energy per nucleon here will be = $\dfrac{1640.26}{209}=7.848MeV$
Therefore, binding energies of ${}_{26}F{{e}^{56}}$ and ${}_{83}B{{i}^{209}}$ are 492.26MeV and 1640.26MeV respectively, and their average binding energy per nucleons are 8.790MeV and 7.848MeV respectively.

Note: We could define the binding energy as the smallest amount of energy that is required to remove a particle from a system of particles (nucleons in this case) or to disassemble a system of particles into their individual parts. This energy is based on the fact that a bound system is normally at a lower energy level than its unbound constituents. Also, this binding energy is always a positive number.