Question

18．已知钍234的半衰期是24天，其衰变方程为$\left.\begin{array}{l}{234}\\{90}\end{array}\right.$Th→$\left.\begin{array}{l}{230}\\{88}\end{array}\right.$Ra+$\left.\begin{array}{l}{4}\\{2}\end{array}\right.$He（补全该衰变方程）．1g钍234经过72天后还剩$\frac{1}{8}$g．

Answer 1

分析 根据电荷数守恒、质量数守恒补全衰变方程．根据$m={m}_{0}（\frac{1}{2}）^{n}$求出衰变后还剩的质量．

解答 解：根据电荷数守恒、质量数守恒得，衰变方程为：$\left.\begin{array}{l}{234}\\{90}\end{array}\right.$Th→$\left.\begin{array}{l}{230}\\{88}\end{array}\right.$Ra+${\;}_{2}^{4}He$．
72天等于3个半衰期，则$m={m}_{0}（\frac{1}{2}）^{n}$=$1×（\frac{1}{2}）^{3}$=$\frac{1}{8}g$．
故答案为：$\left.\begin{array}{l}{4}\\{2}\end{array}\right.$He，$\frac{1}{8}$．

点评 解决本题的关键知道衰变的过程中电荷数守恒、质量数守恒，知道衰变中剩余质量和原质量之间的关系，难度不大．