Asked in NUMS 2019 (cancelled sitting) 2019Moderate

The binding per nucleon of deuteron (21H) and helium nucleus (42He) are 1.1 MeV and 7 MeV respectively. If two deuterons merge to form a single helium nucleus, the energy released will be:

Correct answer: C. 23.6 MeV

  • A. 04.8 MeV
  • B. 13.6 MeV
  • C. 23.6 MeV
  • D. 25.8 MeV
  • E. 26.2 MeV

Explanation

The energy released when two deuterons fuse to form a helium nucleus can be calculated using the formula:1H² + H² →2 He* + Qwhere Q is the energy released in the reaction. The total binding energy of the helium nucleus (42He) is 4 times the binding energy per nucleon, which is given as 7 MeV. Therefore, the total binding energy of helium nucleus is 4 x 7 = 28 MeV.The binding energy per nucleon of a deuteron (2H) is given as 1.1 MeV. Since a deuteron is composed of 2 nucleons, the total binding energy of each deuteron is 2 x 1.1 = 2.2 MeV.When two deuterons fuse to form a helium nucleus, the total binding energy of the products will be:2 x 2.2 MeV + Q = 28 MeVSolving for Q, we get:Q = 28 MeV - 4.4 MeV = 23.6 MeVTherefore, the energy released when two deuterons fuse to form a helium nucleus is 23.6 MeV.

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About Composition of Atomic Nuclei

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