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NEET PHYSICSNUCLEIMedium

Question

A certain mass of hydrogen is changed to helium by the process of fusion. The mass defect in the fusion reaction is 0.02866 u. The energy liberated per nucleon is: (given 1 u = 931 MeV)

A

2.67 MeV

B

26.7 MeV

C

6.675 MeV

D

13.35 MeV

Step-by-Step Solution

  1. Calculate Total Energy Released (E): The energy released is equivalent to the mass defect (Δm\Delta m) according to the relation E=Δmc2E = \Delta m c^2. Using the conversion factor provided (1 u = 931 MeV): E=0.02866 u×931 MeV/u26.68 MeVE = 0.02866 \text{ u} \times 931 \text{ MeV/u} \approx 26.68 \text{ MeV}
  2. Identify Number of Nucleons (A): The fusion of hydrogen into helium involves four hydrogen nuclei (protons) combining to form one helium nucleus (24He{}_{2}^{4}\mathrm{He}) . The mass number (AA) of the resulting helium nucleus is 4.
  3. Calculate Energy per Nucleon: Eper nucleon=Total EnergyMass NumberE_{\text{per nucleon}} = \frac{\text{Total Energy}}{\text{Mass Number}} Eper nucleon=26.68 MeV46.67 MeVE_{\text{per nucleon}} = \frac{26.68 \text{ MeV}}{4} \approx 6.67 \text{ MeV} This value is closest to option 6.675 MeV.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from NUCLEI. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSNUCLEIcertainhydrogenchangedheliumprocess

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