NEET2007PhysicsNuclear PhysicsActual
What is the amount of energy released by deuterium and tritium fusion? [Given: (M_ ₁^2 H =2.0141024 ~u , M_ ₁^3 H =3.0160504 ~u ), ( .M_ ₂^4 He =4.0026034 ~u , M_ ^1₀ n =1.0086654 ~u ] )
Options
- A(60.6 ~eV )
- B(12.6 ~eV )
- C(17.6 ~eV )
- D(28.3 ~eV )
Correct answer
C. (17.6 ~eV )
Step-by-step solution
( Deuterium ₁^2 H + Tritium ₁^3 H Helium ₂^4 He + neutron ₀^1 n +Q ) The energy released in the process is given by ( aligned & Q= [M_ ₁^2 H +M_ ₁^3 H -M_ ₂^4 He -M_ ₀^1 n ] c^2 & =[2.014102+3.016050-4.002603-1.008665] uc^2 & =(0.018884 u ) [931.5 MeV u ]=17.6 MeV . aligned )