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Nuclei · Question 18

JEE Main 2025 (Online) 2nd April Evening ShiftMCQEasy📈 Topic in 26% papers+4−1
1

Energy released when two deuterons (1H2 ) fuse to form a helium nucleus (2He4 ) is : (Given : Binding energy per nucleon of 1H2 = 1.1 MeV and binding energy per nucleon of 2He4 = 7.0 MeV)

  1. 26.8 MeV
  2. 8.1 MeV
  3. 23.6 MeV
  4. 5.9 MeV

Answer

(3) 23.6 MeV

Solution

1H2 + 1H2 2He4

BE of two deuterons = 2 (2 1.1) = 4.4 MeV

BE of helium = 4 7.0 = 28.0 MeV

Energy released = BE of products − BE of reactants

= 28.0 - 4.4 = 23.6 MeV

Answer: (3)

JEE Main 2025 (Online) 2nd April Evening Shift

Keep practising — Nuclei

2

Choose the correct nuclear process from the below options [p : proton, n : neutron, e- : electron, e+ : positron, : neutrino, : antineutrino]

  1. n p + e+ +
  2. n p + e- +
  3. n p + e+ +
  4. n p + e- +
JEE Main 2025 (Online) 28th January Morning Shift

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