Question

writing nuclear equations

1. write the nuclear equation showing actinium - 227 undergoing a beta decay.
2. what parent isotope produced beryllium - 10 formed by a beta emission?

Explanation:

Step1: Recall beta - decay formula

In beta decay, a neutron in the nucleus is converted into a proton, an electron (beta - particle), and an antineutrino. The general equation for beta decay is $^{A}_{Z}X
ightarrow^{A}_{Z + 1}Y+^{\ \ 0}_{- 1}e+\bar{
u}$. Actinium has an atomic number $Z = 89$.
$^{227}_{89}Ac
ightarrow^{227}_{90}Th+^{\ \ 0}_{- 1}e+\bar{
u}$

Step2: Determine parent isotope for beryllium - 10

For beryllium ($Be$), $Z = 4$. In beta - emission, the daughter nucleus (beryllium - 10) has atomic number $Z = 4$. The parent nucleus has atomic number $Z=3$ (lithium, $Li$) since in beta - emission, the atomic number of the daughter is one more than that of the parent. The mass number remains the same in beta - emission. So the parent isotope is lithium - 10, $^{10}_{3}Li$.

Answer:

1. $^{227}_{89}Ac

ightarrow^{227}_{90}Th+^{\ \ 0}_{- 1}e+\bar{
u}$

1. $^{10}_{3}Li$