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NEET PHYSICSNUCLEIEasy

Question

The binding energy per nucleon of deuterium and helium atom is 1.1 MeV and 7.0 MeV. If two deuterium nuclei fuse to form a helium atom, the energy released is:

A

19.2 MeV

B

23.6 MeV

C

26.9 MeV

D

13.9 MeV

Step-by-Step Solution

  1. Identify Nucleons:
  • Deuterium (12H{}_{1}^{2}\text{H}) has a mass number A=2A=2 (1 proton, 1 neutron).
  • Helium (24He{}_{2}^{4}\text{He}) has a mass number A=4A=4 (2 protons, 2 neutrons).
  1. Calculate Initial Binding Energy (Reactants):
  • Binding Energy per nucleon for Deuterium = 1.1 MeV1.1 \text{ MeV}.
  • Total Binding Energy for one Deuterium nucleus = 2×1.1=2.2 MeV2 \times 1.1 = 2.2 \text{ MeV}.
  • Since two Deuterium nuclei fuse, total initial Binding Energy = 2×2.2=4.4 MeV2 \times 2.2 = 4.4 \text{ MeV}.
  1. Calculate Final Binding Energy (Products):
  • Binding Energy per nucleon for Helium = 7.0 MeV7.0 \text{ MeV}.
  • Total Binding Energy for the Helium nucleus = 4×7.0=28.0 MeV4 \times 7.0 = 28.0 \text{ MeV}.
  1. Calculate Energy Released (QQ):
  • The energy released in a nuclear reaction is the difference between the total binding energy of the products and the reactants .
  • Q=(BE)products(BE)reactantsQ = (BE)_{\text{products}} - (BE)_{\text{reactants}}
  • Q=28.0 MeV4.4 MeV=23.6 MeVQ = 28.0 \text{ MeV} - 4.4 \text{ MeV} = 23.6 \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.

PHYSICSNUCLEIbindingenergynucleondeuteriumhelium

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