Let's solve each part step by step: --- ### 3. (i) **Calculate the Q-value of the decay.** **Given:** - Kinetic energy of alpha particle, \( K_{\alpha} = 5.30 \) MeV **Theory:** In alpha decay: - The Q-value is the total energy released in the decay. - The alpha particle and the daughter nucleus share this energy. - By conservation of momentum, the kinetic energies are inversely proportional to their masses. Let: - Mass of alpha particle \( m_{\alpha} = 4 \) u - Mass of daughter nucleus \( m_D \approx 210 \) u **The Q-value formula:** \[ Q = K_{\alpha} + K_D \] where \( K_D \) is the recoil energy of the daughter nucleus. The ratio of kinetic energies (from conservation of momentum): \[ \frac{K_{\alpha}}{K_D} = \frac{m_D}{m_{\alpha}} \] \[ K_D = K_{\alpha} \cdot \frac{m_{\alpha}}{m_D} \] \[ Q = K_{\alpha} + K_D = K_{\alpha} + K_{\alpha} \cdot \frac{m_{\alpha}}{m_D} = K_{\alpha} \left(1 + \frac{m_{\alpha}}{m_D}\right) \] Plug in the values: \[ Q = 5.30 \left(1 + \frac{4}{210}\right) \] \[ = 5.30 \left(1 + 0.01905\right) \] \[ = 5.30 \times 1.01905 \] \[ = 5.39 \text{ MeV} \] **Final Answer:** \[ \boxed{Q = 5.39 \ \text{MeV}} \] --- ### 3. (ii) **Determine the recoil energy of the daughter nucleus (mass number ≈ 210).** From above, \[ K_D = K_{\alpha} \cdot \frac{m_{\alpha}}{m_D} \] \[ K_D = 5.30 \cdot \frac{4}{210} \] \[ = 5.30 \cdot 0.01905 \] \[ = 0.101 \text{ MeV} \] **Final Answer:** \[ \boxed{K_D = 0.101 \ \text{MeV}} \] --- ### 3. (iii) **Why is an electron treated as relativistic in high-energy beta decay, but a neutron is not?** **Explanation:** - An electron has a very small rest mass (\(0.511\) MeV/\(c^2\)). - In high-energy beta decay, the energy of the emitted electron is often comparable to or greater than its rest mass. - When a particle's kinetic energy is comparable to or exceeds its rest mass energy, relativistic effects become important (speed approaches a significant fraction of \(c\)). - A neutron, however, has a much larger rest mass (\( \approx 940\) MeV/\(c^2\)), so the same kinetic energy represents a tiny fraction of its rest mass, and relativistic effects are negligible. **Final Answer:** \[ \boxed{ \text{An electron is treated as a relativistic particle in high-energy beta decay because its rest mass is very small, so the emitted electron's kinetic energy is often comparable to or exceeds its rest mass energy, requiring relativistic treatment. A neutron's rest mass is much larger, so relativistic effects are negligible for the same kinetic energies.} } \] --- ### 4. **Describe how the Mossbauer Effect is used in industrial and chemical applications.** **Summary:** - The Mossbauer Effect involves recoil-free, resonant absorption and emission of gamma rays by nuclei in a solid. - It allows highly precise measurements of nuclear energy levels and their shifts due to interactions with the surrounding environment. **Applications:** - **Chemical analysis:** Used to determine oxidation states, electronic, and magnetic environments of atoms (especially iron compounds). - **Industrial uses:** Quality control in steel manufacturing, analysis of iron in ores, and study of corrosion processes. - **Physics research:** Used to study hyperfine interactions, isomer shifts, and magnetic ordering in materials. - **Environmental studies:** Tracing iron-containing pollutants. - **Biology:** Studying iron-containing proteins (e.g., hemoglobin). **Final Answer (sample 5-mark response):** \[ \boxed{ The Mossbauer Effect is used in industrial and chemical applications to provide highly sensitive and precise information about the atomic and magnetic environment of nuclei, especially iron. In industry, it is used for quality control of steel and detection of corrosion. Chemically, it helps determine the oxidation state and electronic structure of elements in compounds. The technique is valuable in studying iron-containing minerals, catalysts, and biological molecules, allowing detection of small changes in chemical bonding and local environment. } \] --- Let me know if you need any further explanation!

**Final Answers:** 3. (i) **Q-value of decay:** \[ \boxed{Q \approx 5.39 \text{ MeV}} \] 3. (ii) **Recoil energy of daughter nucleus:** \[ \boxed{K_D \approx 0.101 \text{ MeV}} \] 3. (iii) **Explanation for relativistic treatment:** \[ \boxed{ An electron is treated as a relativistic particle in high-energy beta decay because its rest mass is very small, so the emitted electron's kinetic energy is often comparable to or exceeds its rest mass energy, requiring relativistic treatment. A neutron's rest mass is much larger, so relativistic effects are negligible for the same kinetic energies. } \] 4. **Use of Mossbauer Effect in industrial and chemical applications:** \[ \boxed{ The Mossbauer Effect is used to analyze the atomic and magnetic environment of nuclei, especially iron, enabling precise detection of oxidation states, electronic, and magnetic properties. In industry, it aids in quality control of steels and corrosion studies; in chemistry, it helps determine chemical bonding and local environments of elements in compounds. } \]