Let's solve the problem step-by-step, as requested. ### **Step 1: Calculate ΔE (energy difference) between \( n=4 \) and \( n=5 \) states** For a particle in a 1D box (infinite potential well) of length \( L \): \[ E_n = \frac{n^2 h^2}{8mL^2} \] where: - \( n \) = quantum number, - \( h \) = Planck's constant \((6.626 \times 10^{-34}\,\text{J·s})\), - \( m \) = mass of the particle, - \( L \) = length of the box. **For F\(_2\) molecule:** - Molar mass \( = 2 \times 19 = 38\,\text{g/mol} = 0.038\,\text{kg/mol} \) - Mass of one molecule: \( m = \frac{0.038}{6.022 \times 10^{23}} = 6.31 \times 10^{-26}\,\text{kg} \) **Length:** \( L = 3.0\,\text{cm} = 3.0 \times 10^{-2}\,\text{m} \) Calculate \( E_5 - E_4 \): \[ \Delta E = E_5 - E_4 = \frac{h^2}{8mL^2}(5^2 - 4^2) = \frac{h^2}{8mL^2}(25 - 16) = \frac{9h^2}{8mL^2} \] Plug in the values: \[ \Delta E = \frac{9 \times (6.626 \times 10^{-34})^2}{8 \times 6.31 \times 10^{-26} \times (3.0 \times 10^{-2})^2} \] Calculate denominator and numerator: - Numerator: \( 9 \times (6.626 \times 10^{-34})^2 = 9 \times 4.39 \times 10^{-67} = 3.95 \times 10^{-66} \) - Denominator: \( 8 \times 6.31 \times 10^{-26} \times 9 \times 10^{-4} = 8 \times 6.31 \times 9 \times 10^{-30} \) - \( (3 \times 10^{-2})^2 = 9 \times 10^{-4} \) - \( 8 \times 6.31 = 50.48 \) - \( 50.48 \times 9 = 454.32 \) - So denominator \( = 454.32 \times 10^{-30} = 4.5432 \times 10^{-28} \) Now: \[ \Delta E = \frac{3.95 \times 10^{-66}}{4.5432 \times 10^{-28}} = 8.7 \times 10^{-39}~\text{J} \] **Explanation:** This is the energy gap between the \( n=4 \) and \( n=5 \) states for an F\(_2\) molecule in a 3 cm box. --- ### **Step 2: Find \( n \) at which \( E_n = \frac{1}{4} k_B T \) at 450 K** Given: - \( k_B = 1.38 \times 10^{-23}~\text{J/K} \) - \( T = 450~\text{K} \) - \( E_n = \frac{n^2 h^2}{8mL^2} \) - Set \( E_n = \frac{1}{4}k_B T \): \[ \frac{n^2 h^2}{8mL^2} = \frac{1}{4} k_B T \implies n^2 = \frac{2 m L^2 k_B T}{h^2} \] Plugging in the values: - \( 2 m L^2 = 2 \times 6.31 \times 10^{-26} \times (3 \times 10^{-2})^2 = 2 \times 6.31 \times 9 \times 10^{-30} \) - \( 2 \times 6.31 = 12.62 \) - \( 12.62 \times 9 = 113.58 \) - So \( 113.58 \times 10^{-30} = 1.1358 \times 10^{-28} \) - \( k_B T = 1.38 \times 10^{-23} \times 450 = 6.21 \times 10^{-21} \) - Numerator: \( 1.1358 \times 10^{-28} \times 6.21 \times 10^{-21} = 7.056 \times 10^{-49} \) - Denominator: \( h^2 = (6.626 \times 10^{-34})^2 = 4.39 \times 10^{-67} \) So, \[ n^2 = \frac{7.056 \times 10^{-49}}{4.39 \times 10^{-67}} = 1.607 \times 10^{18} \implies n = \sqrt{1.607 \times 10^{18}} \approx 1.27 \times 10^9 \] **Explanation:** At \( n \approx 1.27 \times 10^9 \), the energy of the molecule equals \( \frac{1}{4}k_B T \) at 450 K. --- ### **Step 3: Find the energy separation between this level and the one above** Energy difference between \( n \) and \( n+1 \): \[ \Delta E = E_{n+1} - E_n = \frac{(n+1)^2 - n^2}{8mL^2}h^2 = \frac{(2n + 1) h^2}{8mL^2} \] Plug in \( n = 1.27 \times 10^9 \): - \( 2n + 1 \approx 2.54 \times 10^9 \) - We've already calculated \( \frac{h^2}{8 m L^2} \) in Step 1: - \( \frac{h^2}{8 m L^2} = \frac{4.39 \times 10^{-67}}{4.5432 \times 10^{-28}} = 9.66 \times 10^{-40}~\text{J} \) So, \[ \Delta E = 2.54 \times 10^9 \times 9.66 \times 10^{-40} = 2.45 \times 10^{-30}~\text{J} \] **Explanation:** At this high \( n \), the energy separation becomes extremely small (\( \sim 10^{-30} \) J), reflecting the near-continuous spectrum at high quantum numbers. --- ## **Final Answers** **1.** The energy difference between \( n=4 \) and \( n=5 \) for F\(_2\) in a 3.0 cm box is: **ΔE = \( 8.7 \times 10^{-39} \) J** **2.** The quantum number \( n \) where \( E_n = \frac{1}{4} k_B T \) at 450 K is: **\( n \approx 1.27 \times 10^9 \)** **3.** The energy separation between this level and the next is: **ΔE = \( 2.45 \times 10^{-30} \) J** --- ### **Summary Table** | Step | Quantity | Value | |------|----------|-------| | 1 | \( \Delta E_{4 \to 5} \) | \( 8.7 \times 10^{-39} \) J | | 2 | \( n_{1/4 k_B T} \) | \( 1.27 \times 10^9 \) | | 3 | \( \Delta E_{n \to n+1} \) | \( 2.45 \times 10^{-30} \) J | **Each step includes formulas, substitutions, and reasoning as requested.**

The problem involves applying quantum mechanical principles of a particle confined within a one-dimensional infinite potential well to determine energy levels and their differences. The core concept used is the quantization of energy, where a particle's energy depends on its quantum number \( n \) according to the formula \( E_n = \frac{n^2 h^2}{8 m L^2} \). By calculating the difference between specific energy levels (\( n=4 \) and \( n=5 \)), we employ the discrete energy spectrum characteristic of quantum confinement. To find the quantum number at which the energy matches a fraction of thermal energy (\( \frac{1}{4}k_B T \)), we set the energy expression equal to that thermal energy and solve for \( n \), illustrating the relationship between quantum states and thermal excitation. The energy separation between adjacent levels at high quantum numbers is derived from the incremental difference formula, which simplifies to a proportionality involving \( 2n+1 \). Overall, the approach combines quantum energy quantization with thermodynamic considerations to analyze how molecular confinement affects energy states and their thermal accessibility.