Answer :
Sure! Let's complete each of the given nuclear fusion equations step-by-step using the information provided.
### First Equation
[tex]\[ { }_2^2 H +{ }_1^3 H \rightarrow{ }_8^4 He \][/tex]
1. Identify the nuclei involved:
- \({ }_2^2 H\): This represents Deuterium, which has 2 protons and 2 nucleons (protons + neutrons).
- \({ }_1^3 H\): This represents Tritium, which has 1 proton and 3 nucleons.
2. Determine the product nucleus (Helium-4):
- The formula \({ }_8^4 He\) in the provided equation seems to be a typographical error. In reality, \({ }_2^4 He\) represents a Helium-4 nucleus, which has 2 protons and 4 nucleons.
3. Balance the equation by adding missing particles:
- Total protons before reaction: \(2 (from \ Deuterium) + 1 (from \ Tritium) = 3\)
- Total nucleons before reaction: \(2 (from \ Deuterium) + 3 (from \ Tritium) = 5\)
- Helium-4 (\({ }_2^4 He\)) has 2 protons and 4 nucleons, missing one nucleon to balance the equation.
- A neutron, which is represented as \({ }_0^1 n\) (0 protons and 1 nucleon), balances the nucleons and can be added to complete the equation.
Thus, the completed first equation is:
[tex]\[ { }_2^2 H +{ }_1^3 H \rightarrow{ }_2^4 He +{ }_0^1 n \][/tex]
So, the answers for A and B are:
- A: \({ }_2^4 He\)
- B: \({ }_0^1 n\)
### Second Equation
[tex]\[ { }_7^{14} N +{ }_1^1 H \rightarrow{ }_0^c E \][/tex]
1. Identify the nuclei involved:
- \({ }_7^{14} N\): This represents Nitrogen-14, which has 7 protons and 14 nucleons.
- \({ }_1^1 H\): This represents a proton with 1 proton and 1 nucleon.
2. Determine the product nuclei:
- \({ }_0^c E\) indicates a missing element with 0 protons. For a reaction with Nitrogen-14 and proton, possible products that make sense are a stable nucleus plus another nucleus.
3. Balancing the equation:
- Total protons before reaction: \(7 (from \ Nitrogen-14) + 1 (from \ Proton) = 8\)
- Total nucleons before reaction: \(14 (from \ Nitrogen-14) + 1 (from \ Proton) = 15\)
- A typical reaction produces Carbon-12 (\({ }_6^{12} C\)) and another nucleus, which in this case could be Helium-4 (\({ }_2^4 He\)) to balance the total nucleons and protons.
Thus, the completed second equation is:
[tex]\[ { }_7^{14} N +{ }_1^1 H \rightarrow{ }_6^{12} C +{ }_2^4 He \][/tex]
So, the answers for C and D are:
- C: \({ }_6^{12} C\)
- D: \({ }_2^4 He\)
The fully balanced equations are:
1. \({ }_2^2 H +{ }_1^3 H \rightarrow{ }_2^4 He +{ }_0^1 n\)
2. \({ }_7^{14} N +{ }_1^1 H \rightarrow{ }_6^{12} C +{ }_2^4 He\)
Here are the completed parts:
- A: \({ }_2^4 He\)
- B: \({ }_0^1 n\)
- C: \({ }_6^{12} C\)
- D: \({ }_2^4 He\)
- \(E\): (Not required as we completed D)
Thus, those are the solutions for the problem.
### First Equation
[tex]\[ { }_2^2 H +{ }_1^3 H \rightarrow{ }_8^4 He \][/tex]
1. Identify the nuclei involved:
- \({ }_2^2 H\): This represents Deuterium, which has 2 protons and 2 nucleons (protons + neutrons).
- \({ }_1^3 H\): This represents Tritium, which has 1 proton and 3 nucleons.
2. Determine the product nucleus (Helium-4):
- The formula \({ }_8^4 He\) in the provided equation seems to be a typographical error. In reality, \({ }_2^4 He\) represents a Helium-4 nucleus, which has 2 protons and 4 nucleons.
3. Balance the equation by adding missing particles:
- Total protons before reaction: \(2 (from \ Deuterium) + 1 (from \ Tritium) = 3\)
- Total nucleons before reaction: \(2 (from \ Deuterium) + 3 (from \ Tritium) = 5\)
- Helium-4 (\({ }_2^4 He\)) has 2 protons and 4 nucleons, missing one nucleon to balance the equation.
- A neutron, which is represented as \({ }_0^1 n\) (0 protons and 1 nucleon), balances the nucleons and can be added to complete the equation.
Thus, the completed first equation is:
[tex]\[ { }_2^2 H +{ }_1^3 H \rightarrow{ }_2^4 He +{ }_0^1 n \][/tex]
So, the answers for A and B are:
- A: \({ }_2^4 He\)
- B: \({ }_0^1 n\)
### Second Equation
[tex]\[ { }_7^{14} N +{ }_1^1 H \rightarrow{ }_0^c E \][/tex]
1. Identify the nuclei involved:
- \({ }_7^{14} N\): This represents Nitrogen-14, which has 7 protons and 14 nucleons.
- \({ }_1^1 H\): This represents a proton with 1 proton and 1 nucleon.
2. Determine the product nuclei:
- \({ }_0^c E\) indicates a missing element with 0 protons. For a reaction with Nitrogen-14 and proton, possible products that make sense are a stable nucleus plus another nucleus.
3. Balancing the equation:
- Total protons before reaction: \(7 (from \ Nitrogen-14) + 1 (from \ Proton) = 8\)
- Total nucleons before reaction: \(14 (from \ Nitrogen-14) + 1 (from \ Proton) = 15\)
- A typical reaction produces Carbon-12 (\({ }_6^{12} C\)) and another nucleus, which in this case could be Helium-4 (\({ }_2^4 He\)) to balance the total nucleons and protons.
Thus, the completed second equation is:
[tex]\[ { }_7^{14} N +{ }_1^1 H \rightarrow{ }_6^{12} C +{ }_2^4 He \][/tex]
So, the answers for C and D are:
- C: \({ }_6^{12} C\)
- D: \({ }_2^4 He\)
The fully balanced equations are:
1. \({ }_2^2 H +{ }_1^3 H \rightarrow{ }_2^4 He +{ }_0^1 n\)
2. \({ }_7^{14} N +{ }_1^1 H \rightarrow{ }_6^{12} C +{ }_2^4 He\)
Here are the completed parts:
- A: \({ }_2^4 He\)
- B: \({ }_0^1 n\)
- C: \({ }_6^{12} C\)
- D: \({ }_2^4 He\)
- \(E\): (Not required as we completed D)
Thus, those are the solutions for the problem.
