Answer :
To determine which atomic symbol represents an isotope that undergoes radioactive decay to produce [tex]\(Pb-206\)[/tex], we need to consider the types of radioactive decay processes (such as alpha decay, beta decay, or gamma decay) and how they affect the atomic and mass numbers.
Here we will focus on alpha decay, a common decay process for heavy isotopes, in which an atom emits an alpha particle ([tex]\( ^4_2 He \)[/tex]), resulting in a new atom with its mass number decreased by 4 and its atomic number decreased by 2.
Let's analyze each option:
1. [tex]\( ^{238}_{92}U \)[/tex]:
- If [tex]\( ^{238}_{92}U \)[/tex] undergoes alpha decay, it will emit an alpha particle ([tex]\( ^4_2He \)[/tex]).
- This results in [tex]\(^{238-4}_{92-2}Pb \rightarrow ^{234}_{90}Th \)[/tex].
- This result is a thorium (Th), not lead (Pb-206). So this is not the correct choice.
2. [tex]\( ^{222}Pb \)[/tex]:
- Lead has an atomic number of 82, but here it is written without specifying the atomic number (presumably a typo or incorrect format).
- Even assuming it means [tex]\( ^{222}_{82} \text{Pb}\)[/tex], the isotope mass number must be 206.
- This isotope does not convert by alpha decay as [tex]\( ^{222-4}_{82-2}Pb \rightarrow ^{218}_{80}Hg \)[/tex], not [tex]\(Pb-206\)[/tex].
3. [tex]\( ^{178}_{72}Hf \)[/tex]:
- This species represents hafnium (Hf) with an atomic number of 72, not lead (Pb).
- Even assuming it decays, losing 66 mass units to become [tex]\(Pb-206\)[/tex] is unrealistic by a single decay process.
- So this is not the correct choice.
4. [tex]\( ^{192}Pb \)[/tex]:
- Lead has an atomic number 82 which matches.
- To transform this isotope of lead properly, it would require multi-step decays (beyond simple alpha or beta decays), making it forms unlikely here.
- Thus this might not fit our direct decay scheme.
In review:
Given [tex]\(Pb-206\)[/tex] directly, and understanding the decay more typically, we logically deduce the closer fit:
1st choice [tex]\( ^{238}_{92}U \)[/tex], completes feasible decay via Thorium into Lead-206 eventually:
Thus, the correct isotope that can undergo radioactive decay to produce [tex]\(Pb-206\)[/tex] is [tex]\( ^{238}_{92} U \)[/tex].
Here we will focus on alpha decay, a common decay process for heavy isotopes, in which an atom emits an alpha particle ([tex]\( ^4_2 He \)[/tex]), resulting in a new atom with its mass number decreased by 4 and its atomic number decreased by 2.
Let's analyze each option:
1. [tex]\( ^{238}_{92}U \)[/tex]:
- If [tex]\( ^{238}_{92}U \)[/tex] undergoes alpha decay, it will emit an alpha particle ([tex]\( ^4_2He \)[/tex]).
- This results in [tex]\(^{238-4}_{92-2}Pb \rightarrow ^{234}_{90}Th \)[/tex].
- This result is a thorium (Th), not lead (Pb-206). So this is not the correct choice.
2. [tex]\( ^{222}Pb \)[/tex]:
- Lead has an atomic number of 82, but here it is written without specifying the atomic number (presumably a typo or incorrect format).
- Even assuming it means [tex]\( ^{222}_{82} \text{Pb}\)[/tex], the isotope mass number must be 206.
- This isotope does not convert by alpha decay as [tex]\( ^{222-4}_{82-2}Pb \rightarrow ^{218}_{80}Hg \)[/tex], not [tex]\(Pb-206\)[/tex].
3. [tex]\( ^{178}_{72}Hf \)[/tex]:
- This species represents hafnium (Hf) with an atomic number of 72, not lead (Pb).
- Even assuming it decays, losing 66 mass units to become [tex]\(Pb-206\)[/tex] is unrealistic by a single decay process.
- So this is not the correct choice.
4. [tex]\( ^{192}Pb \)[/tex]:
- Lead has an atomic number 82 which matches.
- To transform this isotope of lead properly, it would require multi-step decays (beyond simple alpha or beta decays), making it forms unlikely here.
- Thus this might not fit our direct decay scheme.
In review:
Given [tex]\(Pb-206\)[/tex] directly, and understanding the decay more typically, we logically deduce the closer fit:
1st choice [tex]\( ^{238}_{92}U \)[/tex], completes feasible decay via Thorium into Lead-206 eventually:
Thus, the correct isotope that can undergo radioactive decay to produce [tex]\(Pb-206\)[/tex] is [tex]\( ^{238}_{92} U \)[/tex].
