What is the ratio of the binding energy per nucleon of nitrogen (14 7)N to helium (4 2)He Nuclei?
What is the ratio of the binding energy per nucleon of nitrogen (14 7)N to helium (4 2)He Nuclei?
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![### Understanding the Ratio of Binding Energy per Nucleon
**Question:**
What is the ratio of the binding energy per nucleon of nitrogen \((^{14}_{7}\text{N})\) to helium \((^{4}_{2}\text{He})\) nuclei?
**Explanation with Examples:**
##### Binding Energy and Its Importance:
Binding energy refers to the energy required to disassemble a nucleus into its constituent protons and neutrons. It is a crucial concept in nuclear physics, indicating the stability of a nucleus. Higher binding energy per nucleon usually suggests a more stable nucleus.
##### Calculating the Ratio:
To solve this problem, it is essential to know the binding energies per nucleon for both \(^{14}_{7}\text{N}\) and \(^{4}_{2}\text{He}\).
1. **Nitrogen-14 \((^{14}_{7}\text{N})\)**:
- **Mass Number (A):** 14
- **Atomic Number (Z):** 7
2. **Helium-4 \((^{4}_{2}\text{He})\)**:
- **Mass Number (A):** 4
- **Atomic Number (Z):** 2
##### Example Values (For Educational Purposes):
- **Binding Energy per Nucleon of \(^{14}_{7}\text{N}\)**: Let's assume it is approximately 7.5 MeV.
- **Binding Energy per Nucleon of \(^{4}_{2}\text{He}\)**: Let's assume it is approximately 7.1 MeV.
##### Calculation:
To find the ratio:
\[ \text{Ratio} = \frac{\text{Binding Energy per Nucleon of } ^{14}_{7}\text{N}}{\text{Binding Energy per Nucleon of } ^{4}_{2}\text{He}} \]
Using the example values:
\[ \text{Ratio} = \frac{7.5 \text{ MeV}}{7.1 \text{ MeV}} \approx 1.056 \]
Therefore, the ratio of the binding energy per nucleon of nitrogen-14 to helium-4 nuclei is approximately \(1.056\).
##### Conclusion:
Understanding the ratio of the binding energy per nucleon helps in comparing the stability of different nuclei. In this example, nitrogen-14 has a slightly higher binding energy](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6fcd7fab-9903-4b0b-a30c-af5d8b8fc75f%2F5011dbf5-3286-4c89-8c34-56a9a98bde73%2F8a73528_processed.png&w=3840&q=75)
Transcribed Image Text:### Understanding the Ratio of Binding Energy per Nucleon
**Question:**
What is the ratio of the binding energy per nucleon of nitrogen \((^{14}_{7}\text{N})\) to helium \((^{4}_{2}\text{He})\) nuclei?
**Explanation with Examples:**
##### Binding Energy and Its Importance:
Binding energy refers to the energy required to disassemble a nucleus into its constituent protons and neutrons. It is a crucial concept in nuclear physics, indicating the stability of a nucleus. Higher binding energy per nucleon usually suggests a more stable nucleus.
##### Calculating the Ratio:
To solve this problem, it is essential to know the binding energies per nucleon for both \(^{14}_{7}\text{N}\) and \(^{4}_{2}\text{He}\).
1. **Nitrogen-14 \((^{14}_{7}\text{N})\)**:
- **Mass Number (A):** 14
- **Atomic Number (Z):** 7
2. **Helium-4 \((^{4}_{2}\text{He})\)**:
- **Mass Number (A):** 4
- **Atomic Number (Z):** 2
##### Example Values (For Educational Purposes):
- **Binding Energy per Nucleon of \(^{14}_{7}\text{N}\)**: Let's assume it is approximately 7.5 MeV.
- **Binding Energy per Nucleon of \(^{4}_{2}\text{He}\)**: Let's assume it is approximately 7.1 MeV.
##### Calculation:
To find the ratio:
\[ \text{Ratio} = \frac{\text{Binding Energy per Nucleon of } ^{14}_{7}\text{N}}{\text{Binding Energy per Nucleon of } ^{4}_{2}\text{He}} \]
Using the example values:
\[ \text{Ratio} = \frac{7.5 \text{ MeV}}{7.1 \text{ MeV}} \approx 1.056 \]
Therefore, the ratio of the binding energy per nucleon of nitrogen-14 to helium-4 nuclei is approximately \(1.056\).
##### Conclusion:
Understanding the ratio of the binding energy per nucleon helps in comparing the stability of different nuclei. In this example, nitrogen-14 has a slightly higher binding energy
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