Let's solve the two parts step by step: --- ## (a) Laboratory Bombarding Energy \( K) for Resonance ### **Step 1: Identify the Resonant State** - The resonance is at an excitation energy of **9.63 MeV** in \( {}^{16}\text{O}^* \). - The reaction is \( {}^{12}\text{C} + \alpha \to {}^{16}\text{O}^* \). - The **Q-value** for the reaction is the binding energy difference for \( {}^{16}\text{O} \to {}^{12}\text{C} + \alpha \), which is **7.2 MeV** (as shown in the diagram). ### **Step 2: Calculate the Required Bombarding Energy** The relation between excitation energy (\( E_x \)), Q-value (\( Q \)), and bombarding energy (\( K \)) for the reaction in the **lab frame** (with \( \alpha \) particle as projectile and \( {}^{12}\text{C} \) as target): \[ E_x = Q + \frac{M_T + m_p}{M_T} K \] where \( M_T \) is the target mass (\( {}^{12}\text{C} \)), \( m_p \) is the projectile mass (\( \alpha \)), and \( K \) is the laboratory bombarding energy. But for a simple estimate (since \( M_T \gg m_p \)), we often use: \[ E_x = Q + K_{\text{cm}} \] But \( K_{\text{cm}} = \frac{m_p}{M_T + m_p} K_{\text{lab}} \) So, \[ K_{\text{lab}} = \frac{(E_x - Q)(M_T + m_p)}{m_p} \] For \( E_x = 9.63 \) MeV, \( Q = 7.2 \) MeV, \( m_p = 4 \), \( M_T = 12 \): \[ K_{\text{lab}} = (9.63 - 7.2) \times \frac{12 + 4}{4} = 2.43 \times 4 = 9.72 \text{ MeV} \] **Explanation:** The resonance will be observed when the kinetic energy of the alpha particle in the lab is about **9.7 MeV**. --- ## (b) Approximate Lifetime of the Excited State ### **Step 1: Use the Uncertainty Principle** The width \( \Gamma \) and lifetime \( \tau \) are related by: \[ \tau = \frac{\hbar}{\Gamma} \] Given: - \( \Gamma = 510 \text{ keV} = .510 \text{ MeV} \) - \( \hbar = 6.582 \times 10^{-22} \text{ MeV} \cdot \text{s} \) ### **Step 2: Calculate Lifetime** \[ \tau = \frac{6.582 \times 10^{-22} \text{ MeV} \cdot \text{s}}{.510 \text{ MeV}} = 1.29 \times 10^{-21} \text{ s} \] **Explanation:** The lifetime of the excited state is approximately \( 1.3 \times 10^{-21} \) seconds. --- ## **Final Answers** **(a)** The laboratory bombarding energy \( K \) at which resonance will be observed is: \[ \boxed{9.7 \text{ MeV}} \] **(b)** The approximate lifetime of the excited state is: \[ \boxed{1.3 \times 10^{-21} \text{ s}} \]

Let's solve the two parts step by step: --- ## (a) Laboratory Bombarding Energy \( K \) for Resonance ### **Step 1: Identify the Resonant State** - The resonance occurs at an excitation energy of **9.63 MeV** in \( {}^{16}\text{O}^* \). - The reaction is \( {}^{12}\text{C} + \alpha \to {}^{16}\text{O}^* \). - The **Q-value** for the reaction is **7.2 MeV**. ### **Step 2: Calculate the Required Bombarding Energy** Using the relationship between excitation energy (\( E_x \)), Q-value (\( Q \)), and bombarding energy (\( K \)): \[ E_x = Q + K_{\text{cm}} \] Where \( K_{\text{cm}} \) is the kinetic energy in the center of mass frame. For a simple estimate: \[ K_{\text{lab}} = \frac{(E_x - Q)(M_T + m_p)}{m_p} \] Substituting values: - \( E_x = 9.63 \) MeV - \( Q = 7.2 \) MeV - Mass of \( \alpha \) particle \( m_p = 4 \) - Mass of \( {}^{12}\text{C} \) \( M_T = 12 \) Calculating: \[ K_{\text{lab}} = (9.63 - 7.2) \times \frac{12 + 4}{4} = 2.43 \times 4 = 9.72 \text{ MeV} \] **Explanation:** The resonance will be observed when the kinetic energy of the alpha particle in the lab is approximately **9.7 MeV**. --- ## (b) Approximate Lifetime of the Excited State ### **Step 1: Use the Uncertainty Principle** The width \( \Gamma \) and lifetime \( \tau \) are related by: \[ \tau = \frac{\hbar}{\Gamma} \] Given: - \( \Gamma = 510 \text{ keV} = 0.510 \text{ MeV} \) - \( \hbar = 6.582 \times 10^{-22} \text{ MeV} \cdot \text{s} \) ### **Step 2: Calculate Lifetime** \[ \tau = \frac{6.582 \times 10^{-22} \text{ MeV} \cdot \text{s}}{0.510 \text{ MeV}} = 1.29 \times 10^{-21} \text{ s} \] **Explanation:** The lifetime of the excited state is approximately \( 1.3 \times 10^{-21} \) seconds. --- ## **Final Answers** **(a)** The laboratory bombarding energy \( K \) at which resonance will be observed is: \[ \boxed{9.7 \text{ MeV}} \] **(b)** The approximate lifetime of the excited state is: \[ \boxed{1.3 \times 10^{-21} \text{ s}} \]