Answer :
To address this problem step-by-step, let's break it into parts:
### 5(D). Distinguish Between Fission and Fusion Reactions
Fission:
1. Process: The nucleus of a heavy atom (such as Uranium-235) splits into two or more smaller nuclei, along with a few neutrons and a large amount of energy.
2. Example: [tex]\( ^{235}U + n \rightarrow ^{139}Ba + ^{94}Kr + 3n + \text{energy} \)[/tex]
3. Characteristics:
- Primarily used in nuclear reactors and atomic bombs.
- Produces radioactive waste.
- Requires a critical mass to sustain the reaction.
Fusion:
1. Process: Two light nuclei (such as hydrogen isotopes) combine to form a heavier nucleus, accompanied by the release of energy.
2. Example: [tex]\( ^{2}H + ^{2}H \rightarrow ^{3}He + n + \text{energy} \)[/tex]
3. Characteristics:
- Powers the sun and other stars.
- Produces less radioactive waste compared to fission.
- Requires extremely high temperatures and pressures (achieved in stars).
### 11. Line Fusion of Two Deuterons
The given reaction:
[tex]\[ ^{2}H + ^{2}H \rightarrow ^{3}He + n + Q \][/tex]
Given atomic masses:
- [tex]\( ^{2}H \)[/tex] (Deuterium) = 2.0154 u
- [tex]\( ^{3}He \)[/tex] = 3.0174 u
- [tex]\( n \)[/tex] (Neutron) = 1.009 u
- 1 atomic mass unit (u) = 931 MeV
Calculation of the energy Q released in the reaction:
1. Initial Mass:
- Mass of two deuterons: [tex]\( 2 \times 2.0154 \text{u} = 4.0308 \text{u} \)[/tex]
2. Final Mass:
- Mass of [tex]\( ^{3}He + n \)[/tex]: [tex]\( 3.0174 \text{u} + 1.009 \text{u} = 4.0264 \text{u} \)[/tex]
3. Mass Defect:
- [tex]\( \Delta m = \text{Initial Mass} - \text{Final Mass} \)[/tex]
- [tex]\( \Delta m = 4.0308 \text{u} - 4.0264 \text{u} = 0.0044 \text{u} \)[/tex]
4. Energy Released:
- [tex]\( Q = \Delta m \times 931 \text{MeV/u} \)[/tex]
- [tex]\( Q = 0.0044 \text{u} \times 931 \text{MeV/u} \)[/tex]
- [tex]\( Q \approx 4.095 \text{MeV} \)[/tex]
Thus, the energy [tex]\( Q \)[/tex] released in the reaction is approximately [tex]\( 4.095 \text{MeV} \)[/tex].
### (B) Answer the Following Questions
i. Isotopes:
- Isotopes are atoms of the same element that have the same atomic number but different mass numbers due to different numbers of neutrons.
- Examples:
- Hydrogen Isotopes: Protium ([tex]\(^1H\)[/tex]), Deuterium ([tex]\(^2H\)[/tex]), Tritium ([tex]\(^3H\)[/tex])
- Carbon Isotopes: Carbon-12 ([tex]\(^{12}C\)[/tex]), Carbon-13 ([tex]\(^{13}C\)[/tex]), Carbon-14 ([tex]\(^{14}C\)[/tex])
ii. Radioactivity:
- Radioactivity is the spontaneous emission of particles or electromagnetic radiation from the unstable nucleus of an atom as it transforms into a more stable configuration.
iii. Distinguish Between Natural and Artificial Radioactivity:
- Natural Radioactivity:
- Occurs naturally in elements such as uranium, thorium, and radon.
- Example: [tex]\( ^{238}U \rightarrow ^{234}Th + \alpha \)[/tex]
- Artificial Radioactivity:
- Induced by human actions, such as neutron bombardment or particle acceleration.
- Example: [tex]\( ^{14}N + \alpha \rightarrow ^{17}O + p \)[/tex]
### (C) Decay Equation and Isotope Information
Given decay equation:
[tex]\[ X \rightarrow ^{4}_{2}He + ... \][/tex]
Calculate:
i. Values of [tex]\(M\)[/tex] and [tex]\(Z\)[/tex]:
- For the reaction: [tex]\( X \rightarrow ^4_2He + Y \)[/tex]
- Suppose [tex]\( X \)[/tex] has mass number [tex]\( A \)[/tex] and atomic number [tex]\( Z \)[/tex]
- After emitting an alpha particle ([tex]\( ^4_2He \)[/tex]), the new element [tex]\( Y \)[/tex] would have:
- Mass number [tex]\( M = A - 4 \)[/tex]
- Atomic number [tex]\( Z = Z - 2 \)[/tex]
ii. Binding Energy [tex]\( BE \)[/tex] of [tex]\(X\)[/tex]:
Given:
- Mass of [tex]\(X = 7 \text{u}\)[/tex]
- Mass of proton = [tex]\(1.00784 \text{u}\)[/tex]
- Mass of neutron = [tex]\(1.00574 \text{u}\)[/tex]
- 1 [tex]\(u\)[/tex] = 931 MeV
- Calculate binding energy per nucleon:
1. Total mass of nucleons before binding:
- Suppose [tex]\(X\)[/tex] has [tex]\(Z\)[/tex] protons and [tex]\(N = A-Z\)[/tex] neutrons.
- Total mass of protons = [tex]\(Z \times 1.00784 \text{u}\)[/tex]
- Total mass of neutrons = [tex]\((A-Z) \times 1.00574 \text{u}\)[/tex]
2. Mass defect:
- Mass defect ([tex]\( \Delta m \)[/tex]) = Total mass of nucleons - Mass of [tex]\(X\)[/tex]
- [tex]\( \Delta m = [Z \times 1.00784 + (A - Z) \times 1.00574] - A \)[/tex]
3. Binding Energy:
- [tex]\( BE = \Delta m \times 931 \text{ MeV} \)[/tex]
- [tex]\( BE \text{ per nucleon} = \frac{BE}{A} \)[/tex]
By plugging in the appropriate numbers for the protons and neutrons along with the atomic mass of [tex]\(X\)[/tex], you can calculate the binding energy.
This concludes the detailed explanation and the answers for the given parts of the question.
### 5(D). Distinguish Between Fission and Fusion Reactions
Fission:
1. Process: The nucleus of a heavy atom (such as Uranium-235) splits into two or more smaller nuclei, along with a few neutrons and a large amount of energy.
2. Example: [tex]\( ^{235}U + n \rightarrow ^{139}Ba + ^{94}Kr + 3n + \text{energy} \)[/tex]
3. Characteristics:
- Primarily used in nuclear reactors and atomic bombs.
- Produces radioactive waste.
- Requires a critical mass to sustain the reaction.
Fusion:
1. Process: Two light nuclei (such as hydrogen isotopes) combine to form a heavier nucleus, accompanied by the release of energy.
2. Example: [tex]\( ^{2}H + ^{2}H \rightarrow ^{3}He + n + \text{energy} \)[/tex]
3. Characteristics:
- Powers the sun and other stars.
- Produces less radioactive waste compared to fission.
- Requires extremely high temperatures and pressures (achieved in stars).
### 11. Line Fusion of Two Deuterons
The given reaction:
[tex]\[ ^{2}H + ^{2}H \rightarrow ^{3}He + n + Q \][/tex]
Given atomic masses:
- [tex]\( ^{2}H \)[/tex] (Deuterium) = 2.0154 u
- [tex]\( ^{3}He \)[/tex] = 3.0174 u
- [tex]\( n \)[/tex] (Neutron) = 1.009 u
- 1 atomic mass unit (u) = 931 MeV
Calculation of the energy Q released in the reaction:
1. Initial Mass:
- Mass of two deuterons: [tex]\( 2 \times 2.0154 \text{u} = 4.0308 \text{u} \)[/tex]
2. Final Mass:
- Mass of [tex]\( ^{3}He + n \)[/tex]: [tex]\( 3.0174 \text{u} + 1.009 \text{u} = 4.0264 \text{u} \)[/tex]
3. Mass Defect:
- [tex]\( \Delta m = \text{Initial Mass} - \text{Final Mass} \)[/tex]
- [tex]\( \Delta m = 4.0308 \text{u} - 4.0264 \text{u} = 0.0044 \text{u} \)[/tex]
4. Energy Released:
- [tex]\( Q = \Delta m \times 931 \text{MeV/u} \)[/tex]
- [tex]\( Q = 0.0044 \text{u} \times 931 \text{MeV/u} \)[/tex]
- [tex]\( Q \approx 4.095 \text{MeV} \)[/tex]
Thus, the energy [tex]\( Q \)[/tex] released in the reaction is approximately [tex]\( 4.095 \text{MeV} \)[/tex].
### (B) Answer the Following Questions
i. Isotopes:
- Isotopes are atoms of the same element that have the same atomic number but different mass numbers due to different numbers of neutrons.
- Examples:
- Hydrogen Isotopes: Protium ([tex]\(^1H\)[/tex]), Deuterium ([tex]\(^2H\)[/tex]), Tritium ([tex]\(^3H\)[/tex])
- Carbon Isotopes: Carbon-12 ([tex]\(^{12}C\)[/tex]), Carbon-13 ([tex]\(^{13}C\)[/tex]), Carbon-14 ([tex]\(^{14}C\)[/tex])
ii. Radioactivity:
- Radioactivity is the spontaneous emission of particles or electromagnetic radiation from the unstable nucleus of an atom as it transforms into a more stable configuration.
iii. Distinguish Between Natural and Artificial Radioactivity:
- Natural Radioactivity:
- Occurs naturally in elements such as uranium, thorium, and radon.
- Example: [tex]\( ^{238}U \rightarrow ^{234}Th + \alpha \)[/tex]
- Artificial Radioactivity:
- Induced by human actions, such as neutron bombardment or particle acceleration.
- Example: [tex]\( ^{14}N + \alpha \rightarrow ^{17}O + p \)[/tex]
### (C) Decay Equation and Isotope Information
Given decay equation:
[tex]\[ X \rightarrow ^{4}_{2}He + ... \][/tex]
Calculate:
i. Values of [tex]\(M\)[/tex] and [tex]\(Z\)[/tex]:
- For the reaction: [tex]\( X \rightarrow ^4_2He + Y \)[/tex]
- Suppose [tex]\( X \)[/tex] has mass number [tex]\( A \)[/tex] and atomic number [tex]\( Z \)[/tex]
- After emitting an alpha particle ([tex]\( ^4_2He \)[/tex]), the new element [tex]\( Y \)[/tex] would have:
- Mass number [tex]\( M = A - 4 \)[/tex]
- Atomic number [tex]\( Z = Z - 2 \)[/tex]
ii. Binding Energy [tex]\( BE \)[/tex] of [tex]\(X\)[/tex]:
Given:
- Mass of [tex]\(X = 7 \text{u}\)[/tex]
- Mass of proton = [tex]\(1.00784 \text{u}\)[/tex]
- Mass of neutron = [tex]\(1.00574 \text{u}\)[/tex]
- 1 [tex]\(u\)[/tex] = 931 MeV
- Calculate binding energy per nucleon:
1. Total mass of nucleons before binding:
- Suppose [tex]\(X\)[/tex] has [tex]\(Z\)[/tex] protons and [tex]\(N = A-Z\)[/tex] neutrons.
- Total mass of protons = [tex]\(Z \times 1.00784 \text{u}\)[/tex]
- Total mass of neutrons = [tex]\((A-Z) \times 1.00574 \text{u}\)[/tex]
2. Mass defect:
- Mass defect ([tex]\( \Delta m \)[/tex]) = Total mass of nucleons - Mass of [tex]\(X\)[/tex]
- [tex]\( \Delta m = [Z \times 1.00784 + (A - Z) \times 1.00574] - A \)[/tex]
3. Binding Energy:
- [tex]\( BE = \Delta m \times 931 \text{ MeV} \)[/tex]
- [tex]\( BE \text{ per nucleon} = \frac{BE}{A} \)[/tex]
By plugging in the appropriate numbers for the protons and neutrons along with the atomic mass of [tex]\(X\)[/tex], you can calculate the binding energy.
This concludes the detailed explanation and the answers for the given parts of the question.
