NEET2013PhysicsNuclear PhysicsActual
When a slow neutron is captured by a ₉₂²³⁵ U nucleus, a fission energy releasing 200 MeV . If power of nuclear reactor is 100 ~W then rate of nuclear fission is
Options
- A3.6 10^6 ~s ⁻¹
- B3.1 10¹² ~s ⁻¹
- C1.8 10^4 ~s ⁻¹
- D4.1 10^6 ~s ⁻¹
Correct answer
B. 3.1 10¹² ~s ⁻¹
Step-by-step solution
Number of fission per second = total power energy/fission . Here, total power =100 ~W energy/fission =200 MeV =200 10^6 1.6 10⁻¹⁹ ~J =3.2 10⁻¹¹ ~J fission rate = 100 3.2 10⁻¹¹ =3.1 10¹² ~s ⁻¹