The binding energy per nucleon for the parent nucleus is $E_1$ and that for the daughter nuclei is $E_2$. Then
A). $E_2 = 2E_1$
B). $E_1$ >$E_2$
C). $E_2$ > $E_1$
D). $E_1 = 2E_2$

Hint: The term binding energy, it is the amount of that energy required to separate the particle from the system of the particles or to scatter the all particles of that system.

Complete step by step solution:
The binding energy is mainly applied to that subatomic particle in the atomic nucleus, to that electrons bound to the nuclei of the atoms.
 In the Crystal form, the atoms and the ions bond together.
The binding energy of the nucleus is that required energy to the atomic nucleus to complete its protons and the neutrons or the energy that would be released by the joining of the single protons and the neutrons in a single nucleus.
After the decay, the daughter nuclei will be steadier.
 Hence, the Binding Energy per nucleon will be more than that of their parent nucleus.
So, the condition be $E_2$ > $E_1$
So, the option (C) is correct.

Note: With the help of a number of nucleons, the term binding energy is increased. This is because they are added together, and they must give a collection of the large amount.
The Nickel element has the highest binding energy.
Not all the nuclei are equally stable. The stability of the nuclei is depending on the value of mass defect.
The larger the value of the mass defect, there is a greater nuclear binding, and the nucleus is more stable.
The binding is negative because of the energy usage to split the nucleus into the single photon and the single neutrons.