Question

Obtain the binding energy of the nuclei ${}_{26}^{56}Fe\text{ and }{}_{83}^{209}Bi$ in units of MeV from the following data: (a) $\begin{align}
  & m\left( {}_{26}^{56}Fe \right)=55.934939u \\
 & m\left( {}_{83}^{209}Bi \right)=208.980388u \\
\end{align}$

Answer 1

Hint: The total binding energy is given by the defect in mass which is the difference between the calculated mass and actual mass. We can calculate the mass of the given atoms by finding out the number of protons and neutrons in them and then multiplying it by the mass of one proton and neutron respectively. Adding the two values will give us the calculated mass. To find it in MeV, we can multiply the obtained mass defect by 931.5.

Complete step by step answer:
We know that binding energy is the energy that acts inside the nucleus to keep all the subatomic particles together. In other words, we can say that it is the minimum energy required to separate a particle from its system.
Nuclear binding energy is the lost mass which is converted to energy released in formation of nucleus from individual nucleons.
We can find the binding energy by calculating the true mass defect. It is given by the difference between the calculated mass and actual mass.
Firstly, let us calculate the binding energy of the nuclei of iron.
We know that the atomic number of iron is 26 and its atomic mass is 56. Subtracting atomic number from the atomic mass will give us the number of neutrons present and number of protons is equal to the atomic number.
Therefore, the number of neutrons = (56-26) = 30.
Now we know that the mass of a proton is 1.007825u. Therefore, the mass of 26 protons = 26.20345u.
Similarly,
 We know that the mass of a neutron is 1.008665u. Therefore, mass of 30 neutrons = 30.25995u.
Therefore, the calculated mass of iron is (26.20345 + 30.25995) = 56.4634u. But the actual mass is given as 55.934939u.

Therefore, mass defect = calculated mass –actual mass = 56.4634u - 55.934939u = 0.528461u.
Therefore, total binding energy = $(\Delta m){{c}^{2}}=0.528461\times 931.5MeV=492.26MeV$
Similarly, let us calculate the binding energy of bismuth.
Number of neutrons = 209-83 = 126 neutrons.
Mass of 83 protons = 83.649475u
Mass of 126 neutrons = 127.09179u
Therefore, calculated mass = 83.649475u + 127.09179u = 210.74126u.
The mass of Bismuth is given to us as 208.980388u.

Therefore, mass defect = calculated mass – actual mass = 210.74126u - 208.980388u = 1.760872u
Therefore, total binding energy = $(\Delta m){{c}^{2}}=1.760872\times 931.5MeV=1640.26MeV$
Therefore, the total binding energy of ${}_{26}^{56}Fe\text{ and }{}_{83}^{209}Bi$ is $492.26MeV$ and $1640.26MeV$ respectively.

Note: The nucleus binding energy is always positive. This is due to the fact that the atomic mass is less than the constituent mass which means the constituent mass in terms of energy.
Like the binding energy, we can also calculate binding energy per nucleon. Binding energy per nucleon is given by dividing total binding energy by the total number of nucleons. Nucleons are the subatomic particles present inside the nucleus.