The radioactive decay of
84
214
𝑃
𝑜
(polonium-214) involves two alpha emissions and two beta emissions. What is the isotope resulting from this radioactive decay?

Question

The radioactive decay of 
84
214
𝑃
𝑜
 (polonium-214) involves two alpha emissions and two beta emissions. What is the isotope resulting from this radioactive decay?

BlackTom AI · Accepted Answer

First, restating the problem to ensure clarity: Po-214 undergoes two alpha decays and two beta decays. We need to determine the final isotope.

Option A: 70 206 Yb. This would require a substantial change in proton number from 84 to 70 (a drop of 14) and a mass number change from 214 to 206 (drop of 8). In alpha decay, each alpha emission reduces A by 4 and Z by 2. Two alphas would reduce A by 8 and Z by 4, not enough to reach Z 70. Additionally, Po (Z=84) cannot directly become Yb (Z=70) through the given sequence; the proton number would need a larger net decrease than alpha decays alone, and beta decays do not reduce Z.

Option B: 86 222 Rn. This would imply an increase in mass number by 8 and an increase in proton number by 2 from the starting Po-214. However, alpha decay always reduces both A and Z, and beta decay (beta-minus) increases Z by 1 without changing A. The net changes would not produce a heavier nucleus than the original Po-214 after the specified decays; therefore this option is inconsistent with two alphas followed by two betas.

Option C: 82 208 Pb. Starting from Po-214, two alphas would cut A by 8 and Z by 4, yielding Pb-206 (Z=80, A=206). A subsequent pair of beta decays would increase Z by 2 (two beta-minus decays) without altering A, giving Z=82 and A=206, which corresponds to Pb-206. The mass number here would still be 206, not 208, so this option would require an incorrect mass change after the sequence described.

Option D: 82 206 Pb. This matches the cumulative changes: two alphas reduce A by 8 and Z by 4 (Po-214 -> Pb-206), and two beta decays raise Z by 2 (Pb-206 -> Pb-206 with Z=82). The final nucleus would have Z=82 and A=206, which is lead-206, aligning with the stated decay sequence.

Across the options, the key steps are: each alpha emission lowers both A and Z by 4 and 2 respectively; each beta-minus emission raises Z by 1 with no mass change. Tracking these changes carefully shows how the starting Po-214 transforms through the specified decays to the final isotope, and comparing each option helps identify which one matches the net effects of two alphas followed by two betas. Note how options that propose inconsistent mass numbers or proton counts violate the established decay mechanics (alpha vs beta effects).

The radioactive decay of 84 214 𝑃 𝑜 (polonium-214) involves two alpha emissions and two beta emissions. What is the isotope resulting from this radioactive decay? 单项选择题

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