Question

If $200\,MeV$ energy is released in the fission of a single ${U^{235}}$ nucleus, the number of fissions required per second to produce $1\,KW$ power shall be: (Given $1\,eV = 1.6 \times {10^{ - 19}}\,J$ ).
(A) $3.125 \times {10^{13}}$
(B) $3.125 \times {10^{14}}$
(C) $3.125 \times {10^{15}}$
(D) $3.125 \times {10^{16}}$

Answer 1

Hint: To find the number of fissions required per second, by using the power formula of the nuclear fission reaction and then by using the given information of energy released and the power produced, then the number of fissions per sec can be determined.

Formula used:
The power produced in nuclear fission reaction,
$P = n\left( {\dfrac{E}{t}} \right)$
Where, $P$ is the power produced in nuclear fission reaction, $n$ is the number of fissions to produce the power, $E$ is the energy released in the nuclear fission reaction and $t$ is the time taken for the nuclear fission reaction.

Complete step by step answer:
Given that,
The energy released in the nuclear fission reaction, $E = 100\,MeV$.
The power produced in the nuclear fission reaction, $P = 1\,KW$.
The charge of the electron is, $e = 1.6 \times {10^{ - 19}}$
We have to find the number of fissions per second, so, $t = 1\,s$
The power produced in nuclear fission reaction,
$P = n\left( {\dfrac{E}{t}} \right)\,..................\left( 1 \right)$
By substituting the power produced during the nuclear fission reaction, the energy released from the nuclear fission reaction and the time taken for the nuclear fission reaction in the equation (1), then
$1\,KW = n\left( {\dfrac{{200\,MeV}}{1}} \right)$
Now substituting the values of kilo, mega and the charge of electron in the above equation, then the above equation is written as,
$1 \times {10^3} = n\left( {\dfrac{{200 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}}}{1}} \right)$
By multiplying the terms in numerator in RHS, then
$1 \times {10^3} = n \times 3.2 \times {10^{ - 11}}$
By keeping the term $n$ in one side and the other terms in other side, then the above equation is written as,
$\dfrac{{1 \times {{10}^3}}}{{3.2 \times {{10}^{ - 11}}}} = n$
On dividing the above equation, then the above equation is written as,
$n = 3.125 \times {10^{13}}$
Thus, the above equation shows the number of fissions per second.

Hence option (A) is the correct answer.

Note:
The final answer is the number of fissions, so it does not have the unit. The time is taken as $1\,s$ because in the question it is asked that the number of fissions per second, so time is taken as $1\,s$. In other words, to produce the $1\, KW$ power the total number of fissions required per second is $3.125 \times {10^{13}}$.