Physics - Modern Physics Question with Solution | TestHub

PhysicsModern PhysicsNuclear physicsMedium2 minPYQ_2013

PhysicsMediumsingle choice

When a slow neutron is captured by a $_{92}^{235} U$ nucleus, a fission energy releasing $200 MeV$ . If power of nuclear reactor is $100 W$ then rate of nuclear fission is

Options:

Answer:

Solution:

Number of fission per second $= \frac{total power}{energy/fission}$ . Here, total power $= 100 W$ energy/fission $= 200 MeV = 200 \times 1 0^{6} \times 1.6 \times 1 0^{- 19} J$ $= 3.2 \times 1 0^{- 11} J$ $∴$ fission rate $= \frac{100}{3.2 \times 1 0 ^{- 11}} = 3.1 \times 1 0^{12} s^{- 1}$

Stream:AIIMSSubject:PhysicsTopic:Modern PhysicsSubtopic:Nuclear physics

⏱ 2mℹ️ Source: PYQ_2013

Doubts & Discussion

Loading discussions...