[math: Pa91234] Pa^{234}_{91} undergoes a series of decays to become [math: Rn86222] Rn^{222}_{86} How many types of what radioactive particles are emitted through this process?

Question

BlackTom AI · Accepted Answer

Consider the given nuclear decay: starting with Pa-234 (Z=91, A=234) ends at Rn-222 (Z=86, A=222). To track what particles are emitted, we analyze how each emission changes mass number A and atomic number Z.
Option A: 3 α and 1 β.
- Each alpha decay reduces A by 4 and Z by 2. Three alphas would reduce A by 12 (234 → 222) and Z by 6 (91 → 85).
- A single beta decay (β−) leaves A unchanged but increases Z by 1 (85 → 86).
- Net change: A goes from 234 to 222 and Z from 91 to 86, matching Pa-234 to Rn-222 exactly. This fits both mass and charge accounting.

Option B: 2 α and 2 β.
- Two alphas would drop A by 8 (234 → 226) and Z by 4 (91 → 87).
- Two β decays would raise Z by 2 (87 → 89) while leaving A at 226.
- Net result would be A=226 and Z=89, which does not match the target A=222 or Z=86.

Option C: 2 α and 1 β.
- Two alphas: A=226, Z=87 (as above).
- One β decay: Z increases by 1 to 88, A remains 226.
- This yields A=226 and Z=88, far from the required A=222 and Z=86.

Option D: 3 α and 3 β.
- Three alphas give A=222 and Z=85 (as in option A before β).
- Three β decays would increase Z by 3 (85 → 88) while A stays 222.
- Net would be A=222 but Z=88, not the target Z=86.

Summarizing, only the combination of 3 α emissions followed by 1 β emission achieves both the correct mass number and the correct atomic number to reach Rn-222 from Pa-234. The other options fail either the mass change, the charge change, or both.

[math: Pa91234] Pa^{234}_{91} undergoes a series of decays to become [math: Rn86222] Rn^{222}_{86} How many types of what radioactive particles are emitted through this process?单项选择题

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