NEET Physics Nuclei Questions Solved

Watch Physics > Nuclei Videos
play button

If the binding energy per nucleon in Li7 and He4 nuclei are respectively 5.60 MeV and 7.06 MeV, then energy of reaction Li7+p2He42 is

(a) 19.6 MeV           (b) 2.4 MeV
(c) 8.4 MeV             (d) 17.3 MeV

(d) B.E. of Li7 = 39.20 MeV and He4=28.24 MeV
Hence binding energy of 2He4=56.48 MeV

Energy of reaction = 56.48-39.20 = 17.28 MeV .

Difficulty Level:

  • 21%
  • 17%
  • 13%
  • 50%
Crack NEET with Online Course - Free Trial (Offer Valid Till September 21, 2019)