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NEET PHYSICSNUCLEIEasy

Question

The binding energy per nucleon of 37Li{}_{3}^{7}Li and 24He{}_{2}^{4}He nuclei are 5.60 MeV5.60 \text{ MeV} and 7.06 MeV7.06 \text{ MeV}, respectively. In the nuclear reaction 37Li+11H24He+24He+Q{}_{3}^{7}Li + {}_{1}^{1}H \rightarrow {}_{2}^{4}He + {}_{2}^{4}He + Q, the value of energy Q released is:

A

19.6 MeV

B

-2.4 MeV

C

8.4 MeV

D

17.3 MeV

Step-by-Step Solution

  1. Concept: The energy released (QQ) in a nuclear reaction is equal to the difference between the total binding energy of the products and the total binding energy of the reactants . Q=(BE)products(BE)reactantsQ = (BE)_{\text{products}} - (BE)_{\text{reactants}}
  2. Calculate Total Binding Energies:
  • Reactants:
  • Lithium (37Li{}_{3}^{7}Li): Has 7 nucleons. BELi=7×5.60 MeV=39.20 MeVBE_{Li} = 7 \times 5.60 \text{ MeV} = 39.20 \text{ MeV}.
  • Hydrogen (11H{}_{1}^{1}H): Is a single proton, so its binding energy is zero (BEH=0BE_{H} = 0).
  • Total BEreactants=39.20+0=39.20 MeVBE_{\text{reactants}} = 39.20 + 0 = 39.20 \text{ MeV}.
  • Products:
  • Helium (24He{}_{2}^{4}He): Has 4 nucleons. BEHe=4×7.06 MeV=28.24 MeVBE_{He} = 4 \times 7.06 \text{ MeV} = 28.24 \text{ MeV}.
  • There are two Helium nuclei. Total BEproducts=2×28.24 MeV=56.48 MeVBE_{\text{products}} = 2 \times 28.24 \text{ MeV} = 56.48 \text{ MeV}.
  1. Calculate Q-value: Q=56.48 MeV39.20 MeVQ = 56.48 \text{ MeV} - 39.20 \text{ MeV} Q=17.28 MeV17.3 MeVQ = 17.28 \text{ MeV} \approx 17.3 \text{ 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.

PHYSICSNUCLEIbindingenergynucleonnucleirespectively

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