Question
226Ra88 decays by emitting three alpha particles and three beta particles. what's the resulting nuclei?
Is it 238Bi83?
(Number on left is mass # and on right is atomic #)
Is it 238Bi83?
(Number on left is mass # and on right is atomic #)
Answers
DrBob222
I don't think so.
88Ra226 ==>2He4 + 2He4 + 2He4 + -1e0 + -1e0 + -1e0 + X.
88-2-2-2+1+1+1 = 88-3=85At
(88 = 2+2+2-1-1-1 + X = 3+85)
226 -4-4-4 = 214
(226 = 4+4+4+X = 12 + 214
So I think the end product is 85At214. Check my thinking.
88Ra226 ==>2He4 + 2He4 + 2He4 + -1e0 + -1e0 + -1e0 + X.
88-2-2-2+1+1+1 = 88-3=85At
(88 = 2+2+2-1-1-1 + X = 3+85)
226 -4-4-4 = 214
(226 = 4+4+4+X = 12 + 214
So I think the end product is 85At214. Check my thinking.