Exams › NEET › Physics

Applying principle of energy conservation, Energy of proton = total B.E. of 2α − energy of Li⁷ = 8 × 7.06 − 7 × 5.6 = 56.48 − 39.2 = 17.28 MeV

1. Moderator slows down the neutrons to thermal energies.
2. Fission rate total power = 5 energy 200 × 1.6 × 10⁻¹³ = 1.56 × 10¹⁴ s⁻¹
3. Fusion reaction.
4. 1 a.m.u = 931 MeV

Correct answer: Fusion reaction.

Solution

The question involves the calculation of energy released in a nuclear reaction, which is characteristic of a fusion process. Option C, 'Fusion reaction,' is the correct answer.

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 →