[math: Pa91234] undergoes a series of decays to become [math: Rn86222]How many types of what radioactive particles are emitted through this process?

Question

BlackTom AI · Accepted Answer

Alpha decay reduces the mass number (A) by 4 and the atomic number (Z) by 2; beta-minus (β−) decay increases Z by 1 while leaving A unchanged. These two decay rules are the key conservation relations used here.

1) Relevant knowledge points and concepts
- Alpha (α) emission: ΔA = −4, ΔZ = −2.
- Beta-minus (β−) emission: ΔA = 0, ΔZ = +1.
- Total A and Z must balance between parent and daughter after all emitted particles are accounted for.

2) Step-by-step reasoning
Let nα = number of α particles emitted, nβ = number of β− particles emitted.
Start with parent: A_initial = 234, Z_initial = 91.
Daughter: A_final = 222, Z_final = 86.

3) Calculations (showing intermediate steps)
Mass-number balance:
234 − 4·nα = 222
4·nα = 234 − 222 = 12
nα = 12 / 4 = 3

Charge-number balance:
91 − 2·nα + 1·nβ = 86
Substitute nα = 3:
91 − 2·3 + nβ = 86
91 − 6 + nβ = 86
85 + nβ = 86
nβ = 86 − 85 = 1

So nα = 3 and nβ = 1.

4) Why the correct answer is correct
These values satisfy both conservation equations simultaneously: three α emissions reduce A by 12 to 222 and reduce Z by 6 to 85; one β− emission raises Z by 1 to 86, matching the observed daughter nuclide. Thus the sequence requires 3 α and 1 β emission.

5) Why the other options are incorrect
- 2α and 1β: mass would be 234 − 8 = 226 (not 222).
- 2α and 2β: mass 226 (wrong) and charge balance would not match 86.
- 3α and 3β: mass would be correct (234 − 12 = 222) but charge would be 91 − 6 + 3 = 88 (not 86).

6) Conclusion
The correct emission sequence is 3α and 1β, which matches the provided answer "[math: 3α and 1β]".

[math: Pa91234] undergoes a series of decays to become [math: Rn86222]How many types of what radioactive particles are emitted through this process?单项选择题

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