Chapter 25, Problem 87A

Interpretation Introduction

Interpretation: The question is to find the type of particle that struck the target.

  ${}_{98}^{249}\text{Cf}+?\to 4_0^1\text{n}{+}_{106}^{263}\text{Sg}$

Concept Introduction: The metal Californium is radioactive. Californium emits neutrons with great intensity. It is used in portable metal detectors, to distinguish between water and oil layers in oil wells, to identify gold and silver ore, and to detect metal fatigue and stress in aircraft.

Answer to Problem 87A

The type of particle that struck the target is $O$ .

Explanation of Solution

Given information: The given incomplete chemical reaction is,

  ${}_{98}^{249}\text{Cf}+?\to 4_0^1\text{n}{+}_{106}^{263}\text{Sg}$

According to the law of conservation of mass, mass is neither created nor destroyed during a chemical reaction.

The atomic number and the mass number should be equal on both sides.

Let the mass number and the atomic number of the unknown particle be a and b respectively and the unknown particle be X.

  ${}_{98}^{249}\text{Cf}+X\to 4_0^1\text{n}{+}_{106}^{263}\text{Sg}$

Thus, the mass number and the atomic number of the unknown particle is

  $\begin{array}{l}249+a=4(1)+263\\ 249+a=267\\ a=267-249\\ \Rightarrow a=18\\ 98+b=0+106\\ 98+b=106\\ b=106-98\\ \Rightarrow b=8\end{array}$

The chemical particle with mass number 18 and atomic number 8 is Oxygen and is represented by the letter O.

Hence, the chemical equation becomes,

  ${}_{98}^{249}\text{Cf}+O\to 4_0^1\text{n}{+}_{106}^{263}\text{Sg}$

Conclusion

The type of particle that struck the target is $O$ .