NEET Physics Nuclei Questions Solved

In the nuclear fusion reaction H12+H13He24+n given that the repulsive potential energy between the two nuclei is -7.7×10-14 J, the temperature at which the gases must be heated to initiate the reaction is nearly
[Boltzmann’s constant k=1.38×10-23 J/K)
(a)109 K            (b) 107 K
(c) 105 K           (d)  103 K

(a) Kinetic energy of the molecules of a gas at a temp. T is 32kT
 To initiate the reaction 32kT=7.7×10-14 J

32×1.38×10-23 T=7.7×10-14T=3.7×109 K

Difficulty Level:

  • 34%
  • 39%
  • 28%
  • 0%
Crack NEET with Online Course - Free Trial (Offer Valid Till August 23, 2019)