Question

The number of neutrons emitted when $_{92}^{235}U$ undergoes controlled nuclear fission to form $_{54}^{142}Xe$ and $_{38}^{90}Sr$ is:

Answer 1

Figure out the number of neutrons present in the atoms on both LHS and RHS of the reaction by subtracting the sum of neutrons present in xenon and strontium from the neutrons present in uranium.

Complete step by step solution:
We can see here that the type of decay that the uranium nucleus undergoes is the nuclear fission type of decay. This type is most common in atoms that have excessively bulky nuclei. The nucleus divides into two or more than two parts as a result of this decay and forms new elements which may or may not further decay into other elements. The number of electrons or protons or neutrons should add up on both sides, if they do not, then those many electrons, protons, or neutrons are lost or released in the decay. So, we will see how many neutrons are lost in the decay by calculating the number of neutrons present in the atoms on both sides.
The unbalanced reaction is as follows:
$_{92}^{235}U\to _{54}^{142}Xe+_{38}^{90}Sr$
We see that uranium, xenon, and strontium have the atomic numbers 92, 54, and 38 respectively and the mass numbers 235, 142, and 90 respectively. From this, we can calculate the number of neutrons present by subtracting the atomic number from the atomic mass number.

& \text{Neutrons in }U=235-92=143 \\\ & \text{Neutrons in }Xe=142-54=88 \\\ & \text{Neutrons in }Sr=90-38=52 \\\ \end{aligned}$$ We will now add the number of neutrons present in xenon and strontium to see if they are equal to the number of neutrons in uranium. We know that $88+52=140$, so we can see that the RHS is short of 3 neutrons. Now let us check whether any of the neutrons have been converted to protons by the ${{\beta }^{-}}$- decay. We will add the atomic numbers of xenon and strontium and check whether they are equal to the number of protons in uranium. We know that $54+38=92$, so we can say that no neutrons have decayed into protons. At least 3 neutrons are emitted during this fission. The conditions for this decay dictate that the nucleus of uranium has to bombarded with neutrons for it to split into xenon and strontium, so 1 neutron will be required for the fission. So, the balanced reaction is: $$_{92}^{235}U+1\text{n}\to _{54}^{142}Xe+_{38}^{90}Sr+4\text{n}+energy$$ **Hence, we can see that 4 neutrons are emitted during this fission.** **Note:** Take note that if we consider the net reaction, only 3 neutrons are emitted, but this kind of decay is not possible without the bombardment of neutrons and that neutrons have to be considered as an emission too. Thus, the answer is 4 and not 3.