Question

The decay series for 238 92U is represented in Figure 19.1. Write the balanced nuclear equation for each of the following radioactive decays. a. Alpha-particle production by 226 88 Ra b. Beta-particle production by 214 82 Pb

Answer 1

Answer: The decay process of the radioisotopes are written below.

Explanation:

For the given options:

• Option a:

Alpha decay is defined as the process in which the nucleus of an atom disintegrates into two particles. The first one which is the alpha particle consists of two protons and two neutrons, also known as helium nucleus. The second particle is the daughter nuclei which is the original nucleus minus the alpha particle released.

$_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\alpha$

The chemical equation for the alpha decay process of Ra-226 isotope follows:

$_{88}^{226}\textrm{Ra}\rightarrow _{86}^{222}\textrm{Rn}+_2^4\alpha$

• Option b:

Beta decay is defined as the process in which the neutrons get converted into an electron and a proton. The released electron is known as the beta particle. In this process, the atomic number of the daughter nuclei gets increased by a factor of 1 but the mass number remains the same.

$_Z^A\textrm{X}\rightarrow _{Z+1}^{A}\textrm{Y}+_{-1}^0\beta$

The chemical equation for the alpha decay process of Pb-214 isotope follows:

$_{82}^{214}\textrm{Pb}\rightarrow _{83}^{214}\textrm{Bi}+_{-1}^0\beta$

Hence, the decay process of the radioisotopes are written above.