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Question: The energy released by fission of one uranium atom is 200 MeV. The number of fissions per second req...

The energy released by fission of one uranium atom is 200 MeV. The number of fissions per second required to produce 3.2 W power is:
(take 1 eV = 1.6×1019J1.6 \times {10^{ - 19}}J )
(A) 107{10^7}
(B) 1010{10^{10}}
(C) 1015{10^{15}}
(D) 1011{10^{11}}

Explanation

Solution

To answer this question, we should be first calculating the energy that has released in the joules form. Once done, we have to take into consideration the total amount of energy that is produced in one second. At the end we can find the number of fissions by dividing the above two mentioned quantities.

Complete step by step answer:
The energy that is released per fission is E = 200 MeV.
We can write 200 MeV as
200×106×1.6×1019J =3.2×1011J  200 \times {10^6} \times 1.6 \times {10^{ - 19}}J \\\ = 3.2 \times {10^{ - 11}}J \\\
Now the total energy that is produced in one second is Ep{E_p}.
We can write Ep{E_p} as 3.2 J, as it is given in the question.
So now the number of fission per second N is given by EpE\dfrac{{{E_p}}}{E}.
Put the values in the equation to get:
3.23.2×1011=1011\dfrac{{3.2}}{{3.2 \times {{10}^{ - 11}}}} = {10^{11}}
So we can say that the number of fissions per second that is required to produce 3.2 W power is 1011{10^{11}}.

Hence option D is correct.

Note: In the question, we have come across the term nuclear fission. For a better understanding of the answer we need to know the meaning of this process. Nuclear fission is defined as a process in which the nucleus of any specific atom suddenly splits or rather divides into two or even sometimes more number of small fission products. These fission products are known as nuclei. The process of nuclear fission also involves the formation of some by-products.
Nuclear fission is usually termed as elemental transformation.