Question

Write a nuclear reaction for each step in the formation of 208₈₂Pb from 228₉₀Th, which proceeds by a series of decay reactions involving the step-wise emission of α, α, α, α, β, β, α particles, in that order.

Answer 1

The formation of 208₈₂Pb from 228₉₀Th involves a series of decay reactions including α and β particle emissions.

Step-by-step explanation:

The formation of 208₈₂Pb from 228₉₀Th proceeds through a series of decay reactions. The step-wise emission of α particles and β particles leads to the formation of the final isotope.

The nuclear reactions involved in the formation of 208₈₂Pb from 228₉₀Th are as follows:

1. α decay: 228₉₀Th ➞ 224₈₈Ra + 4₂He
2. α decay: 224₈₈Ra ➞ 220₈₆Rn + 4₂He
3. α decay: 220₈₆Rn ➞ 216₈₄Po + 4₂He
4. α decay: 216₈₄Po ➞ 212₈₂Pb + 4₂He
5. β decay: 212₈₂Pb ➞ 212₈₃Bi + 0₋₁e
6. β decay: 212₈₃Bi ➞ 212₈₄Po + 0₋₁e
7. α decay: 212₈₄Po ➞ 208₈₂Pb + 4₂He