Chapter 25, Problem 92A

Interpretation Introduction

Interpretation : The question is to find the isotope that remains after three beta particles and five alpha particles are lost from a thorium-234 isotope.

Concept Introduction: An unstable atomic nucleus loses energy by radiation during the process of radioactive decay. A substance is regarded as radioactive if it has unstable nuclei.

Answer to Problem 92A

Bismuth-214.

Explanation of Solution

Given information: Thorium-234 undergoes beta emission by emitting 3 beta particles and alpha emission by emitting 5 alpha particles.

From the given data, we get the below equation

  $Th\to 3\beta +5\alpha +X$

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.

Thus, by the law of conservation of mass,

  $\begin{array}{l}234=3(0)+5(4)+a\\ \Rightarrow a=214\\ 90+3(-1)+2(2)+b\\ \Rightarrow b=83\end{array}$

The radioactive isotope with mass number 214 and atomic number 83 is Bismuth and is represented as Bi.

Hence, the chemical equation becomes,

  $Th\to 3\beta +5\alpha +Bi$

Conclusion

The isotope that remains after three beta particles and five alpha particles are lost from a thorium-234 isotope is Bismuth-214.