A nucleus of 23290Th emits one α-particle and then two β− particles. What are the nucleon number and proton number of the final nucleus?
- α-decay: A = 232 − 4 = 228, Z = 90 − 2 = 88.
- First β− decay: A stays 228, Z = 88 + 1 = 89.
- Second β− decay: A stays 228, Z = 89 + 1 = 90.
- Z is 90 again, so the final nucleus is an isotope of thorium.
Answer: The final nucleus is 22890Th: nucleon number 228, proton number 90.