Question

Four nuclei of an element undergo fusion to form a heavier nucleus, with a release of energy. Which of the two – the parent or the daughter nucleus – would have higher binding energy per nucleon?

Answer 1

We know that the binding energy per nucleon of a nucleus is directly proportional to the nucleus’s stability since the greater the value of mass defect, the greater the value of binding energy per nucleon. So, the more stable nucleus will have a higher binding energy per nucleon.

Complete answer:
When four nuclei of an atom fuse with each other to form a heavier nucleus, along with the emission of energy, the sum of the parent nuclei’s individual masses are greater than the mass of the daughter nucleus. This difference in the mass of the nuclei is referred to as the mass defect.
Now, we know, the binding energy per nucleon is directly proportional to the mass defect. The mass defect multiplied by the square of the speed of light is known as the binding energy of the daughter nucleus.
So, we must provide energy from outside, equal to the binding energy, to split the daughter nucleus into the parent nuclei.
Here in the given problem, the bigger nuclei are more stable than the four fused parent nuclei. Consequently, the binding energy per nucleon of the bigger daughter nucleus will be higher than the binding energy per nucleon of the four lighter parent nuclei.

Additional information:
When each nucleon acquires an average binding energy of $8MeV$ , the nucleus becomes most stable.

Note: The stability of the nucleus can be compared with a liquid drop. A large liquid drop tends to break up into smaller drops, whereas a large number of small drops tend to join to form a large drop. Similarly, a large nucleus (like ${U^{238}}$ ) and small nucleus (like ${H^2}$ ) both are unstable.