NTA Abhyas JEE Main2020PhysicsNuclear PhysicsPractice
Li 7 fuses with a proton according to the nuclear reaction given below: p   +   7 L i   →   4 H e   +   4 H e Given that the atomic masses of 1 H , 4 He and 7 Li are 1 . 007825   u , 4 . 002603   u and 7 . 016004   u respectively, where u = 931 . 5   MeV / c 2 , then the Q - value of the reaction is
Options
- A17 . 35 MeV
- B18 . 06 MeV
- C177 . 35 MeV
- D170 . 35 MeV
Correct answer
A. 17 . 35 MeV
Step-by-step solution
The total mass of the initial particles m i   =   1 . 007825   + 7 . 016004 m i =   8 . 023829   u and the total mass of final particles m f   =   2 × 4 . 002603   =   8 . 005206   u Difference between the initial and final mass of particles =   m i   -   m f =   8 . 023829   -   8 . 005206 =   0 . 018623   u The Q value is given by Q   =   ∆ m c 2 =   0 . 018623 × 931 . 5   =   17 . 35