Chapter 19, Problem 6STP

Interpretation Introduction

Interpretation: Balanced nuclear equations for ${}_{90}^{232}\text{Th}$ emitting an $\text{α}$ particle must be provided.

Concept introduction: In a nuclear reaction, unstable nuclide disintegrates into stable nuclide with emission of various radiations spontaneously. It is called radioactivity of nucleus. Emission of rays like alpha, beta and gamma rays occur in the process which changes the identity of nucleus.

During a nuclear change, an element undergoes change in its atomic mass or atomic number or both by emission of charged particles such as beta particle, alpha particle. Beta particle is one that has unit negative charge and alpha particle is one that has 2 unit positive charges and is equivalent to helium atom ${}_2^4\text{He}$ . Emission of proton, beta particle, alpha particle causes a change in atomic number of element.

Answer to Problem 6STP

Balanced nuclear equations for ${}_{90}^{232}\text{Th}$ emitting an $\text{α}$ particle is ${}_{90}^{232}\text{Th}{\to }_2^4\text{He}{+}_{88}^{228}\text{Rn}$ .

Explanation of Solution

Alpha particle is one that has 2 unit positive charges and is equivalent to helium atom ${}_2^4\text{He}$ . Therefore on release of $\text{α}$ particle from ${}_{90}^{232}\text{Th}$ , equation will be as follows:

  ${}_{90}^{232}\text{Th}{\to }_2^4\text{He}{+}_{88}^{228}\text{Rn}$

Conclusion

Balanced nuclear equations for ${}_{90}^{232}\text{Th}$ emitting an $\text{α}$ particle is ${}_{90}^{232}\text{Th}{\to }_2^4\text{He}{+}_{88}^{228}\text{Rn}$ .