Question
Question: On bombarding \[{{U}^{235}}\] by a slow neutron, 200 MeV energy is released. If the power output of ...
On bombarding U235 by a slow neutron, 200 MeV energy is released. If the power output of atomic reactor is 1.6 MW, then the rate of fission will be:
A. 5×1022/s
B. 5×1016/s
C. 8×1016/s
D. 20×1016/s
Solution
In this question we are given the output power and energy released per fission. We have been asked to calculate the rate of fission. Now, we know that the rate of fission is given by the output power over the energy released per fission. U235 is an isotope of uranium, which is a radioactive element that is used in many nuclear reactions.
Formula used:
R=EP
Where,
R is the rate of reaction
P is the power output in MW
E is the energy in joules J
Complete step by step answer:
Fission is a reaction that occurs when a neutron strikes in a larger atom, this forces the atom to excite and split into two smaller atoms. These are also known as fission products. When an atom is split a large amount of energy is released. To calculate the rate of reaction, we will take the power output over the energy released per fission.
Therefore,
R=EP
Now, we have been given the energy in MeV, we shall therefore convert this into joules
So, we know
1 eV =1.6×10−19
Therefore,
200 MeV = 200×1.6×10−19 MJ
Now, after substituting the given values in our equation
We get,
R=200×1.6×10−191.6
On solving,
We get,
R=5×1016/s
Therefore, the correct answer is option B.
Note:
Fission is a process where the neutron strikes a larger atom and splits in two smaller atoms. However, fusion is a process where two atoms fuse together to form a heavier atom. Fusion is beneficial over fission as it produces four times energy as fission. An example of fusion is the sun, where the hydrogen atoms are fused together to form helium atoms releasing a large amount of energy.