Question

In the nuclear reaction:

${}_{92}{\text{U}}^{235}+\text{X}{\to }_{56}{\text{Ba}}^{141}{+}_{36}{\text{Kr}}^{92}+3\left({}_0{n}^1\right)+\text{Energy}$, $\text{X}$ stands for

• A. electron
• B. proton
• C. neutron
• D. none of these

Answer 1

The correct option is C

neutron

Step 1: Given Data and Assumption:

The following nuclear reaction is given:

${}_{92}{\text{U}}^{235}+\text{X}{\to }_{56}{\text{Ba}}^{141}{+}_{36}{\text{Kr}}^{92}+3\left({}_0{n}^1\right)+\text{Energy}$

Assume that, the mass number of $\text{X}$ is $a$ and charge is $b$.

Therefore, the symbol $\text{X}$ can be written as ${}_a{\text{X}}^b$.

Step 2: Conservation Laws:

Applying conservation of mass number for the nuclear reaction –

$$\begin{gathered} 235+b=141+92+3\times 1 \\ b=236-235 \\ b=1 \end{gathered}$$

Applying conservation of charge for the nuclear reaction –

$$\begin{gathered} 92+a=56+36+3\times 0 \\ a=92-92 \\ a=0 \end{gathered}$$

Therefore, $\text{X}{\equiv }_0{\text{X}}^1={}_0{n}^1$

Hence, the correct option is (C).