Chapter 19, Problem 6A

Interpretation Introduction

Interpretation: Inference has to be drawn regarding $\text{α}$ particle on basis of reaction 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 6A

An alpha ( $\text{α}$ ) particle is ${}_2^4\text{He}$ . So option (a) is correct.

Explanation of Solution

From reaction: ${}_{92}^{238}\text{U}\to \text{α}\text{particle}+{}_{90}^{234}\text{Th}$ , It can be clearly understood that $\text{α}$ particle has a mass of 4 and atomic number of 2 as 4 unit of mass number and 2 unit of atomic number is lost by ${}_{92}^{238}\text{U}$ to form ${}_{90}^{234}\text{Th}$ .

Hence, alpha particle is one that has 2 unit positive charges and is equivalent to helium atom ${}_2^4\text{He}$ .

Conclusion

Since an alpha ( $\text{α}$ ) particle is ${}_2^4\text{He}$ . So option b, c, d and e incorrect.