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Sagot :
Let's identify the equation that represents alpha decay.
Alpha decay is a type of radioactive decay where an unstable nucleus ejects an alpha particle. An alpha particle consists of 2 protons and 2 neutrons, which corresponds to a helium-4 nucleus ([tex]\( ^4_2 \text{He} \)[/tex]).
Here are the given equations:
1. [tex]\( { }_{95}^{241} \text{Am} \rightarrow { }_{93}^{237} \text{Np} + { }_2^4 \text{He} \)[/tex]
2. [tex]\( { }_9^{18} \text{F} \rightarrow { }_8^{18} \text{O} + { }_1^0 \text{e} \)[/tex]
3. [tex]\( { }_6^{14} \text{C} \rightarrow { }_7^{14} \text{N} + { }_{-1}^0 \text{e} \)[/tex]
4. [tex]\( { }_{66}^{152} \text{Dy} \rightarrow { }_{66}^{152} \text{Dy} + \gamma \)[/tex]
Now let's go through each option:
1. [tex]\( { }_{95}^{241} \text{Am} \rightarrow { }_{93}^{237} \text{Np} + { }_2^4 \text{He} \)[/tex]
- In this equation, americium-241 ([tex]\( ^{241}_{95} \text{Am} \)[/tex]) decays into neptunium-237 ([tex]\( ^{237}_{93} \text{Np} \)[/tex]) by releasing an alpha particle ([tex]\( ^4_2 \text{He} \)[/tex]). This process is a clear example of alpha decay since the nucleus loses 2 protons and 2 neutrons.
2. [tex]\( { }_9^{18} \text{F} \rightarrow { }_8^{18} \text{O} + { }_1^0 \text{e} \)[/tex]
- This equation shows that fluorine-18 ([tex]\( ^{18}_{9} \text{F} \)[/tex]) decays into oxygen-18 ([tex]\( ^{18}_{8} \text{O} \)[/tex]) and emits a positron ([tex]\( ^0_1 \text{e} \)[/tex]). This represents positron emission, not alpha decay.
3. [tex]\( { }_6^{14} \text{C} \rightarrow { }_7^{14} \text{N} + { }_{-1}^0 \text{e} \)[/tex]
- This equation shows that carbon-14 ([tex]\( ^{14}_{6} \text{C} \)[/tex]) decays into nitrogen-14 ([tex]\( ^{14}_{7} \text{N} \)[/tex]) and emits a beta particle ([tex]\( ^0_{-1} \text{e} \)[/tex]). This is an example of beta decay, not alpha decay.
4. [tex]\( { }_{66}^{152} \text{Dy} \rightarrow { }_{66}^{152} \text{Dy} + \gamma \)[/tex]
- This equation involves dysprosium-152 ([tex]\( ^{152}_{66} \text{Dy} \)[/tex]) transitioning to the same element and emitting a gamma photon ([tex]\( \gamma \)[/tex]). This represents gamma decay, not alpha decay.
From our analysis, the correct equation that represents alpha decay is:
[tex]\( { }_{95}^{241} \text{Am} \rightarrow { }_{93}^{237} \text{Np} + { }_2^4 \text{He} \)[/tex]
Therefore, the correct answer is the first equation:
```
1
```
Alpha decay is a type of radioactive decay where an unstable nucleus ejects an alpha particle. An alpha particle consists of 2 protons and 2 neutrons, which corresponds to a helium-4 nucleus ([tex]\( ^4_2 \text{He} \)[/tex]).
Here are the given equations:
1. [tex]\( { }_{95}^{241} \text{Am} \rightarrow { }_{93}^{237} \text{Np} + { }_2^4 \text{He} \)[/tex]
2. [tex]\( { }_9^{18} \text{F} \rightarrow { }_8^{18} \text{O} + { }_1^0 \text{e} \)[/tex]
3. [tex]\( { }_6^{14} \text{C} \rightarrow { }_7^{14} \text{N} + { }_{-1}^0 \text{e} \)[/tex]
4. [tex]\( { }_{66}^{152} \text{Dy} \rightarrow { }_{66}^{152} \text{Dy} + \gamma \)[/tex]
Now let's go through each option:
1. [tex]\( { }_{95}^{241} \text{Am} \rightarrow { }_{93}^{237} \text{Np} + { }_2^4 \text{He} \)[/tex]
- In this equation, americium-241 ([tex]\( ^{241}_{95} \text{Am} \)[/tex]) decays into neptunium-237 ([tex]\( ^{237}_{93} \text{Np} \)[/tex]) by releasing an alpha particle ([tex]\( ^4_2 \text{He} \)[/tex]). This process is a clear example of alpha decay since the nucleus loses 2 protons and 2 neutrons.
2. [tex]\( { }_9^{18} \text{F} \rightarrow { }_8^{18} \text{O} + { }_1^0 \text{e} \)[/tex]
- This equation shows that fluorine-18 ([tex]\( ^{18}_{9} \text{F} \)[/tex]) decays into oxygen-18 ([tex]\( ^{18}_{8} \text{O} \)[/tex]) and emits a positron ([tex]\( ^0_1 \text{e} \)[/tex]). This represents positron emission, not alpha decay.
3. [tex]\( { }_6^{14} \text{C} \rightarrow { }_7^{14} \text{N} + { }_{-1}^0 \text{e} \)[/tex]
- This equation shows that carbon-14 ([tex]\( ^{14}_{6} \text{C} \)[/tex]) decays into nitrogen-14 ([tex]\( ^{14}_{7} \text{N} \)[/tex]) and emits a beta particle ([tex]\( ^0_{-1} \text{e} \)[/tex]). This is an example of beta decay, not alpha decay.
4. [tex]\( { }_{66}^{152} \text{Dy} \rightarrow { }_{66}^{152} \text{Dy} + \gamma \)[/tex]
- This equation involves dysprosium-152 ([tex]\( ^{152}_{66} \text{Dy} \)[/tex]) transitioning to the same element and emitting a gamma photon ([tex]\( \gamma \)[/tex]). This represents gamma decay, not alpha decay.
From our analysis, the correct equation that represents alpha decay is:
[tex]\( { }_{95}^{241} \text{Am} \rightarrow { }_{93}^{237} \text{Np} + { }_2^4 \text{He} \)[/tex]
Therefore, the correct answer is the first equation:
```
1
```
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