Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
To determine which isotope could decay to produce [tex]${ }_{82}^{206}Pb$[/tex], you need to think about the characteristics of radioactive decay and the changes in atomic and mass numbers that occur during this process.
1. Radioactive decay: For an isotope to decay into [tex]${ }_{82}^{206}Pb$[/tex], it must start with a higher atomic number (number of protons) and/or mass number (sum of protons and neutrons) because during decay, both of these numbers can decrease. Keep in mind:
- Alpha Decay: The emission of an alpha particle (composed of 2 protons and 2 neutrons) reduces the atomic number by 2 and the mass number by 4.
- Beta Decay: An electron (or positron) emission changes the atomic number by 1 (either increasing or decreasing) without changing the mass number.
- Other decay processes: They typically tweak these numbers in less predictable ways.
Given options:
- 238 (no further context provided, so we will disregard this in lack of specific information).
- [tex]${ }_{87}^{222}Pb$[/tex]
- [tex]${ }_{72}^{178}Hf$[/tex]
- [tex]${ }_{77}^{192}Pb$[/tex]
Let's analyze each option:
### Option: [tex]${ }_{87}^{222}Pb$[/tex]
- Atomic Number: 87
- Mass Number: 222
The element with atomic number 87 is Francium (Fr), not lead (Pb). Assuming this is a hypothetical isotope notation error and continuing:
- 87 > 82: This would have to decay by losing 5 protons.
- 222 > 206: This however, could reasonably decay by losing 16 units in mass (e.g., 4 alpha decays (4 × 4)) or another combination of processes.
- Valid Characteristics: High mass and atomic number sufficient for decay to Pb-206.
### Option: [tex]${ }_{72}^{178}Hf$[/tex]
- Atomic Number: 72 (Hafnium/Hf)
- Mass Number: 178
- 72 < 82: This makes it impossible to decay into Pb-206 as you cannot increase atomic number through nuclear decay.
- Therefore, invalid.
### Option: [tex]${ }_{77}^{192}Pb$[/tex]
- Atomic Number: 77 (Iridium/Ir)
- Mass Number: 192
- 77 < 82: Iridium cannot decay to lead as it requires increasing its atomic number.
- 192 < 206: Again, this isotope doesn't provide enough mass to decay to achieve mass number 206.
- Therefore, invalid.
### Conclusion
The only option where both the atomic number and the mass number start higher and can be reduced through known decay processes to form [tex]${ }_{82}^{206}Pb$[/tex] is option [tex]${ }_{87}^{222}Pb$[/tex].
So, the correct atomic symbol representing an isotope that could undergo radioactive decay to produce [tex]${ }_{82}^{206}Pb$[/tex] is:
[tex]${ }_{87}^{222}Pb$[/tex]
1. Radioactive decay: For an isotope to decay into [tex]${ }_{82}^{206}Pb$[/tex], it must start with a higher atomic number (number of protons) and/or mass number (sum of protons and neutrons) because during decay, both of these numbers can decrease. Keep in mind:
- Alpha Decay: The emission of an alpha particle (composed of 2 protons and 2 neutrons) reduces the atomic number by 2 and the mass number by 4.
- Beta Decay: An electron (or positron) emission changes the atomic number by 1 (either increasing or decreasing) without changing the mass number.
- Other decay processes: They typically tweak these numbers in less predictable ways.
Given options:
- 238 (no further context provided, so we will disregard this in lack of specific information).
- [tex]${ }_{87}^{222}Pb$[/tex]
- [tex]${ }_{72}^{178}Hf$[/tex]
- [tex]${ }_{77}^{192}Pb$[/tex]
Let's analyze each option:
### Option: [tex]${ }_{87}^{222}Pb$[/tex]
- Atomic Number: 87
- Mass Number: 222
The element with atomic number 87 is Francium (Fr), not lead (Pb). Assuming this is a hypothetical isotope notation error and continuing:
- 87 > 82: This would have to decay by losing 5 protons.
- 222 > 206: This however, could reasonably decay by losing 16 units in mass (e.g., 4 alpha decays (4 × 4)) or another combination of processes.
- Valid Characteristics: High mass and atomic number sufficient for decay to Pb-206.
### Option: [tex]${ }_{72}^{178}Hf$[/tex]
- Atomic Number: 72 (Hafnium/Hf)
- Mass Number: 178
- 72 < 82: This makes it impossible to decay into Pb-206 as you cannot increase atomic number through nuclear decay.
- Therefore, invalid.
### Option: [tex]${ }_{77}^{192}Pb$[/tex]
- Atomic Number: 77 (Iridium/Ir)
- Mass Number: 192
- 77 < 82: Iridium cannot decay to lead as it requires increasing its atomic number.
- 192 < 206: Again, this isotope doesn't provide enough mass to decay to achieve mass number 206.
- Therefore, invalid.
### Conclusion
The only option where both the atomic number and the mass number start higher and can be reduced through known decay processes to form [tex]${ }_{82}^{206}Pb$[/tex] is option [tex]${ }_{87}^{222}Pb$[/tex].
So, the correct atomic symbol representing an isotope that could undergo radioactive decay to produce [tex]${ }_{82}^{206}Pb$[/tex] is:
[tex]${ }_{87}^{222}Pb$[/tex]
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.
