Exams › NEET › Physics

Binding energy per nucleon for fission products is higher relative to Binding energy per nucleon for parent nucleus, i.e., more masses are lost and are obtained as kinetic energy of fission products. So, the given ratio < 1.

1. B.E. per nucleon is smaller for lighter as well as heavier nucleus. But fusion reaction occurs for small mass number nuclei and fission reaction occurs for larger mass number nuclei to attain reaction binding energy per nucleon.
2. m₃ < (m₁ + m₂) (∵ m₁ + m₂ = m₃ + E) E = [m₁ + m₂ - m₃] c²
3. Mass defect = B.E / c²
4. Mass of nucleus = Mass of proton + mass of neutron - mass defect

Correct answer: B.E. per nucleon is smaller for lighter as well as heavier nucleus. But fusion reaction occurs for small mass number nuclei and fission reaction occurs for larger mass number nuclei to attain reaction binding energy per nucleon.

Solution

Option A correctly explains that fusion occurs for lighter nuclei and fission for heavier nuclei to achieve higher binding energy per nucleon, which is consistent with the energy release in these processes.

Related NEET Physics questions

• Two nuclei have their mass numbers in the ratio of 1 : 3. The ratio of their nuclear densities would be:
• If radius of the 27/13 Al nucleus is taken to be R_Al, then the radius of 125/53 Te nucleus is nearly:
• If the nuclear radius of 27/13 Al is 3.6 Fermi, the approximate nuclear radius of 64/29 Cu in Fermi is:
• The radius of germanium (Ge) nuclide is measured to be twice the radius of 9/4 Be. The number of nucleons in Ge are:
• If the nucleus 27/13 Al has nuclear radius of about 3.6 fm, then 125/32 Te would have its radius approximately as:
• The nuclei of which one of the following pairs of nuclei are isobars?

⚔️ Practice NEET Physics free + battle 1v1 →