Showing posts with label Nuclear Fission reaction in nuclear physics. Show all posts
Showing posts with label Nuclear Fission reaction in nuclear physics. Show all posts

14 July 2020

Nuclear Fission

Very large nuclei having mass number greater than 230 tend to be unstable and can split into two or more parts. This is called Fission. Fission process is not a spontaneous one. It occurs when a neutron strikes an unstable nucleus.

In this case, the Uranium 235 will split into two nearly equal nuclei Barium-141 and Krypton-92 and release several free neutrons with huge amounts of energy.

92U235 + 0n1 → 56Ba141 + 36Kr92 + 3 0n1 + Energy
92U235 + 0n1 → 54Xe140 + 38Sr94 + 2 0n1 + Energy

These nuclei are isotopes of more stable elements. If left alone, they decay radioactively by emitting alpha or beta particles. The below figure clearly explain how the chain reaction takes place. Out come neutrons from fission process interact with fissile fuel present in the surrounding. Further this reaction produces totally nine neutrons. This cycle repeats to give a reaction.  Moderator like graphite, heavy water and beryllium are used to control the rate of reaction. If the chain reaction is not controlled, a nuclear explosion will occur.




CALCULATING  ENERGY RELEASED BY FISSION

Calculate the energy released in the following spontaneous fission reaction:
238U → 95Sr + 140Xe + 3n

given the atomic masses to be m(238U) = 238.050784 u, m(95Sr) = 94.919388 u, m(140Xe) = 139.921610 u, and m(n) =1.008665 u.

Solution

 The products have a total mass of
mproducts = 94.919388 u+139.921610 u+3(1.008665 u) = 237.866993 u

The mass lost is the mass of 238U minus mproducts, or

Δ= 238.050784 u − 237.8669933 u = 0.183791 u

so the energy released is E = (Δm) c2 = 171.2 MeV

Energy released by 1 Kg of uranium:

Number of atoms in 1 kg of uranium = (6.023 × 1026)/235

Energy released in one fission = 200 MeV

Energy produced by 1 kg of uranium during fission
E = (6.023 × 1026 × 200)/235 = 5.128 × 1026 MeV

E = (5.128 × 1026) × (1.6×10-13) J      [1 MeV = 1.6×10-13 J]

E = {(5.128 × 1026) × (1.6×10-13)}/{3.6× 106} k Wh

E = 2.26 × 107 k Wh                                    [1k Wh = 3.6×106 J]

Due to this reason, nuclear energy is being used for the generation of electricity. 

Most of the energy that is released during fission goes into the K.E of the fission fragments. The emitted neutrons, Beta and gamma rays and neutrinos carry off perhaps 20% of the total energy.

Read it carefully and try to solve simple problems Click Here

Problems in Electromagnetism

 PROBLEM NO. 1 (UG NEET 2024)