Question

writing nuclear equations

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

Explanation:

Step1: Recall alpha - decay formula

In alpha - decay, the parent nucleus emits an alpha particle ($_{2}^{4}\text{He}$). The mass number of the parent nucleus decreases by 4 and the atomic number decreases by 2. Actinium has an atomic number of 89.
The nuclear equation for the alpha - decay of actinium - 227 ($_{89}^{227}\text{Ac}$) is $_{89}^{227}\text{Ac}
ightarrow_{87}^{223}\text{Fr}+_{2}^{4}\text{He}$.

Step2: Determine parent isotope for beryllium - 10

If beryllium - 10 ($_{4}^{10}\text{Be}$) is formed by alpha emission, we use the reverse of the alpha - decay logic. We add 4 to the mass number and 2 to the atomic number of beryllium. The atomic number of the parent isotope is $4 + 2=6$ (carbon), and the mass number is $10+4 = 14$. So the parent isotope is carbon - 14 ($_{6}^{14}\text{C}$).

Answer:

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

ightarrow_{87}^{223}\text{Fr}+_{2}^{4}\text{He}$

1. $_{6}^{14}\text{C}$