Question

What are the possible results when uranium \[235\] undergoes nuclear fission reaction?
A) \[^{139}B{a_{56}}\]
B) \[^{95}K{r_{36}}\]
C) Both A and B
D) None of the above

Answer 1

Hint: We have to remember that the nuclear fission is a radioactive decay in which a molecule breaks itself into smaller fragments whereas in nuclear fusion, two small fragments combine to form a radioactive species. The particles in nuclear fission are unstably arranged inside a nucleus due to which when an external source is applied on a nucleus it disintegrates into certain chemical species.

Complete answer:
We have to remember that each atom of uranium-\[235\] contains \[92\] protons and \[143\] neutrons that come up with a total of \[235\] atoms. The particles are unstably arranged inside a nucleus due to which when an external source is applied on a nucleus it disintegrates into certain chemical species. Nuclear fission is the process in which a molecule breaks itself into small fragments or certain chemical species. When a nucleus splits it breaks down into two or more different neutrons. As the heavy element fuse during nuclear fission the reaction turns out to be exothermic and much more energy is released during this process.
We also remember that the neutrons that are released during the fission process, they can proceed a chain reaction of continuous fission. The forces that hold two or more nucleons together are nuclear forces. Uranium $235$ undergoes high energy and produces Uranium 236 which further fuses to give other nucleons and products.
Option A) this option can be correct as \[^{139}B{a_{56}}\] this is a product during nuclear fission of uranium \[235\].
Option B) this option can be correct as \[^{95}K{r_{36}}\] this is a product during nuclear fission of uranium \[235\].
Option C) this is a correct option as it has both A and B as correct answers.
Option D) this is an incorrect option as we got option C as the correct answer.

Note:
 We have to remember that the nuclear fission of uranium \[235\]can be described as follows:
\[^1{n_0}{ + ^{235}}{U_{92}} \to {[^{236}}{U_{92}}]{ \to ^{95}}K{r_{36}}{ + ^{139}}B{a_{56}}\] .Uranium $235$ undergoes high energy and produces Uranium $236$ which further fuses to give other nucleons and products.