The total number of steps involved and number of beta particles emitted in the spontaneous decay of ${ }_{92}^{238} \mathrm{U} \rightarrow{ }_{82}^{206} \mathrm{Pb}$ respectively, are
(a)8 and 6
(b)14 and 6
(c)6 and 8
(d)14 and 8
Answer
Answer: B ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
²³⁸U → ²⁰⁶Pb: mass drops by 32 ⇒ 8 α; charge 92 − 16 = 76 must rise to 82 ⇒ 6 β⁻. Total 14 steps, 6 β particles.