Let's tackle each question in **three clear steps with explanations**. --- ### **Part 1: Mn-56 Sample for Required Activity** **Given:** - Desired activity at end: \( A_f = 105 \) Bq (after 48 h) - Half-life of Mn-56: \( t_{1/2} = 2.578 \) h - Time interval: \( t = 48 \) h - Atomic mass of Mn-56: \( M = 55.939 \) u #### **Step 1: Find the Decay Constant (\(\lambda\))** \[ \lambda = \frac{\ln 2}{t_{1/2}} \] \[ \lambda = \frac{0.6931}{2.578\ \text{h}} \approx 0.2688\ \text{h}^{-1} \] **Explanation:** The decay constant tells us how quickly the Mn-56 nuclei decay per hour. --- #### **Step 2: Calculate the Required Number of Nuclei at the End (\(N_f\)), and at the Start (\(N_0\))** \[ A = \lambda N \implies N_f = \frac{A_f}{\lambda} \] \[ N_f = \frac{105}{0.2688} \approx 390.4 \] But this is per hour. Let's convert \(\lambda\) to per second for Bq: \[ \lambda = 0.2688\ \text{h}^{-1} \times \frac{1}{3600} = 7.466 \times 10^{-5}\ \text{s}^{-1} \] \[ N_f = \frac{105}{7.466 \times 10^{-5}} \approx 1.41 \times 10^6 \] Now, relate final and initial nuclei: \[ N_f = N_0 e^{-\lambda t} \implies N_0 = N_f e^{\lambda t} \] \[ \lambda t = 0.2688 \times 48 = 12.902 \] \[ N_0 = 1.41 \times 10^6 \times e^{12.902} \approx 1.41 \times 10^6 \times 4.01 \times 10^5 \approx 5.65 \times 10^{11} \] **Explanation:** We first determine how many Mn-56 atoms are needed at the end, then use the decay law to find how many are required at the start. --- #### **Step 3: Calculate the Mass of Mn-56 Needed** \[ \text{Number of nuclei} = \frac{\text{mass}}{\text{molar mass}} \times N_A \] \[ \text{mass} = \frac{N_0 \cdot M}{N_A} \] Where \(N_A = 6.022 \times 10^{23}\), \(M = 55.939\) g/mol. \[ \text{mass} = \frac{5.65 \times 10^{11} \times 55.939}{6.022 \times 10^{23}} \] \[ = \frac{3.165 \times 10^{13}}{6.022 \times 10^{23}} = 5.26 \times 10^{-11}\ \text{g} \] **Initial activity:** \[ A_0 = \lambda N_0 = 7.466 \times 10^{-5} \times 5.65 \times 10^{11} \approx 4.22 \times 10^7\ \text{Bq} \] **Explanation:** We convert the number of nuclei to mass using Avogadro’s number and find the initial activity using the decay constant. --- #### **Final Answers - Part 1** - **Mass of Mn-56 required at the beginning:** \(\boxed{5.26 \times 10^{-11}\ \text{g}}\) - **Initial activity:** \(\boxed{4.22 \times 10^{7}\ \text{Bq}}\) --- ### **Part 2: Cl-37 Reaction with Proton** **Given:** - Target: Cl-37 (\({}^{37}\)Cl) - Projectile: proton (\(p\)) - Neutron ejected (\(n\)), with kinetic energy = 1 MeV - Product nucleus mass: \(36.96678\) u - No gamma emission #### **Step 1: Write the Nuclear Reaction Equation** \[ {}^{37}_{17}\text{Cl} + {}^{1}_{1}\text{p} \rightarrow {}^{37}_{18}\text{Ar}^* \rightarrow {}^{36}_{18}\text{Ar} + {}^{1}_{0}\text{n} \] But the product mass is for \({}^{36}\)Ar. So, the reaction is: \[ {}^{37}_{17}\text{Cl} + {}^{1}_{1}\text{p} \rightarrow {}^{36}_{18}\text{Ar} + {}^{1}_{0}\text{n} \] **Explanation:** We balance mass and atomic numbers; the proton is absorbed, a neutron is ejected, forming Ar-36. --- #### **Step 2: Find the Q-value of the Reaction** Q-value (energy released): \[ Q = [m_\text{initial} - m_\text{final}]c^2 \] Where: - \(m_\text{initial} = m(\text{Cl-37}) + m(p)\) - \(m_\text{final} = m(\text{Ar-36}) + m(n)\) Atomic masses: - \(m(\text{Cl-37}) = 36.96590\) u - \(m(p) = 1.007825\) u - \(m(\text{Ar-36}) = 36.96678\) u (given) - \(m(n) = 1.008665\) u \[ Q = [36.96590 + 1.007825 - 36.96678 - 1.008665] \times 931.5\ \text{MeV} \] \[ = [37.973725 - 37.975445] \times 931.5 = (-0.001720) \times 931.5 = -1.60\ \text{MeV} \] **Explanation:** A negative Q-value means the reaction is endothermic; energy must be supplied by the kinetic energy of the proton. --- #### **Step 3: Find the Required Kinetic Energy of the Incident Proton** Since the neutron comes out with 1 MeV kinetic energy, the incident proton must supply both the Q-value deficit and the neutron’s kinetic energy. The minimum kinetic energy required is: \[ K_p = |Q| + KE_n \times \frac{m_n + m_{Ar-36}}{m_{Ar-36}} \] But more precisely, in the lab frame (assuming the target is at rest), and considering energy and momentum conservation, the minimum energy required (threshold energy) is: \[ K_{p,\text{min}} = -Q \cdot \left(1 + \frac{m_p}{m_{Ar-36}}\right) \] But here, we are told the neutron specifically has 1 MeV kinetic energy, so the proton’s kinetic energy must be at least: \[ K_p = |Q| + KE_n \] \[ K_p = 1.60\ \text{MeV} + 1.00\ \text{MeV} = 2.60\ \text{MeV} \] **Explanation:** The incident proton must provide enough energy to overcome the negative Q-value and impart the desired kinetic energy to the neutron. --- #### **Final Answers - Part 2** - **Reaction Equation:** \(\boxed{{}^{37}_{17}\text{Cl} + {}^{1}_{1}\text{p} \rightarrow {}^{36}_{18}\text{Ar} + {}^{1}_{0}\text{n}}\) - **Kinetic Energy of Incident Proton:** \(\boxed{2.6\ \text{MeV}}\) (assuming all kinetic energy goes to neutron; more precise answer requires two-body kinematics) --- ### **Summary Table** | Step | Explanation | Calculation | Final Answer | |------|-------------|-------------|--------------| | 1 | Decay constant, number needed | See above | \(\lambda = 7.466 \times 10^{-5}\ \text{s}^{-1}\), \(N_0 = 5.65 \times 10^{11}\) | | 2 | Mass and activity | See above | \(5.26 \times 10^{-11}\) g, \(4.22 \times 10^7\) Bq | | 3 | Cl-37 reaction | See above | \({}^{37}\text{Cl} + p \rightarrow {}^{36}\text{Ar} + n\), \(2.6\) MeV | --- **If you need more precise kinematic calculation for the proton energy, let me know!**

The concepts used involve radioactive decay and activity calculations to determine the initial amount of Mn-56 needed to achieve a specified activity after a given time, utilizing the decay law and decay constant. For the nuclear reaction, principles of conservation of mass and energy (Q-value calculation) are employed to find the reaction equation and the minimum incident proton energy required, considering the reaction's endothermic nature and the kinetic energy imparted to the emitted neutron.