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Question:

The total number of α and β particles emitted in the nuclear reaction 23892U → 21482Pb is:

Solution:

Number of α particles emitted = (mass of U − mass of Pb) / mass of α particles = (238 − 214) / 4 = 6
Number of β particles emitted = 2 × no. of α particles emitted − (92 − 82) = 2
Reaction: 92U238 → 6α → 80X214 → 2β → 82Pb214
(6α, 2β), i.e., total 8 particles.