Question

Assuming that $200MeV$ of energy is released per fission of $_{92}{U^{235}}$ find the number of fission per second, required to release $1kW$.
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: Uranium$ - 235$ is a uranium isotope that makes up $0.72$ percent of natural uranium. It is fissile, unlike the dominant isotope uranium$ - 238$, and can survive a fission chain reaction. It is the only fissile isotope found as a primordial nuclide in nature. The half-life of uranium$ - 235$ is $703.8$ million years.

Complete answer: Nuclear fission is the splitting of an atom's nucleus into two or more smaller nuclei, referred to as fission products. The fission of heavy elements is an exothermic reaction that releases enormous quantities of energy. For common fissile isotopes, the nuclei formed are usually of comparable but slightly different sizes, with a mass ratio of products of about $3:2$. The majority of fissions are binary, producing two charged fragments. Three positively charged fragments are formed approximately $2$ to $4$ times per $1000$ events, indicating ternary fission. In ternary processes, the smallest fragments vary from the size of a proton to the size of an argon nucleus.
Nuclear fission occurs when an unstable atom breaks into two or more smaller, more stable parts, releasing energy in the process. Extra neutrons are released during the fission process, which can then break more atoms, resulting in a chain reaction that releases a lot of energy. By soaking up neutrons, it is also possible to modulate the chain reaction.
According to the question –
Let us assume that the number of fissions released per second be n.
We already know that $1MeV = 1.6 \times {10^{ - 13}}J$
Then, the energy released per second $ = n \times 200 \times 1.6 \times {10^{ - 13}}J$
Energy required per second = $power \times time$
$ = 1KW \times 1s = 1000J$
$\therefore n \times 200 \times 1.6 \times {10^{ - 13}}J = 1000$
We can rewrite the above equation as –
$n = \dfrac{{1000}}{{3.2 \times {{10}^{ - 11}}}} = \dfrac{{10 \times {{10}^{13}}}}{{32}}$$ = 3.125 \times {10^{13}}$
The correct option is A.

Note:
When a neutron is bombarded with $U - 235$, the resulting $U - 236$ is unstable and fissions. The resulting elements have fewer nucleons than $U - 236$, with the remaining three neutrons released as high-energy particles capable of bombarding another $U - 235$ atom and continuing the chain reaction.