write balanced nuclear equation for alpha emission for the isotope, $_{90}^{231}th$
write balanced nuclear equation for alpha emission for the isotope, $_{90}^{231}th$
Answer
Explanation:
Step1: Recall alpha - particle composition
An alpha - particle is represented as $_{2}^{4}\text{He}$.
Step2: Determine the daughter nucleus
For a radioactive nucleus $_{90}^{231}\text{Th}$ undergoing alpha - emission, the mass number of the daughter nucleus is $231 - 4=227$ and the atomic number is $90 - 2 = 88$. The element with atomic number 88 is radium (Ra).
Answer:
${90}^{231}\text{Th}\rightarrow{88}^{227}\text{Ra}+_{2}^{4}\text{He}$