Let's analyze the problem step-by-step. --- ### Given Data: - **Mass of nitrogen molecule (N₂):** \( m_{N_2} = 28 \times 1.661 \times 10^{-27} \text{ kg} \) Simplify: \( m_{N_2} = 28 \times 1.661 \times 10^{-27} = 46.508 \times 10^{-27} = 4.6508 \times 10^{-26} \text{ kg} \) - **Box length:** \( L = 1 \text{ cm} = 1 \times 10^{-2} \text{ m} \) - **Quantum levels:** \( n = 1 \) and \( n= 2 \) --- ### Part (i): Calculate the energy separation between the \( n=1 \) and \( n=2 \) levels --- #### Step 1: Recall the energy levels in a 1D infinite potential well: \[ E_n = \frac{n^2 h^2}{8 m L^2} \] where: - \( h = 6.626 \times 10^{-34} \text{ Js} \) - \( m \) is the mass of the particle (the entire molecule here, as per the problem statement) - \( L \) is the length of the box --- #### Step 2: Compute \( E_1 \) and \( E_2 \): \[ E_1 = \frac{1^2 \times h^2}{8 m L^2} \] \[ E_2 = \frac{2^2 \times h^2}{8 m L^2} = \frac{4 h^2}{8 m L^2} \] --- #### Step 3: Find the energy difference: \[ \Delta E = E_2 - E_1 = \frac{4 h^2}{8 m L^2} - \frac{h^2}{8 m L^2} = \frac{(4 - 1) h^2}{8 m L^2} = \frac{3 h^2}{8 m L^2} \] --- #### Step 4: Plug in the values: \[ \Delta E = \frac{3 \times (6.626 \times 10^{-34})^2}{8 \times 4.6508 \times 10^{-26} \times (1 \times 10^{-2})^2} \] Calculate numerator: \[ (6.626 \times 10^{-34})^2 = 43.9 \times 10^{-68} = 4.39 \times 10^{-67} \] Numerator: \[ 3 \times 4.39 \times 10^{-67} = 13.17 \times 10^{-67} \] Denominator: \[ 8 \times 4.6508 \times 10^{-26} \times 1 \times 10^{-4} = 8 \times 4.6508 \times 10^{-30} = 37.2064 \times 10^{-30} \] Now, compute: \[ \Delta E = \frac{13.17 \times 10^{-67}}{37.2064 \times 10^{-30}} = \frac{13.17}{37.2064} \times 10^{-67 + 30} = 0.3539 \times 10^{-37} = 3.539 \times 10^{-38} \text{ Joules} \] --- ### **Final answer for Part (i):** \[ \boxed{ \Delta E \approx 3.54 \times 10^{-38} \text{ Joules} } \] --- ### Part (ii): Calculate the frequency and wavelength of radiation for excitation from \( n=1 \) to \( n=2 \) --- #### Step 1: Find the frequency \( \nu \) The energy of the photon corresponds to the energy difference: \[ E_{photon} = \Delta E = h \nu \] Thus: \[ \nu = \frac{\Delta E}{h} = \frac{3.54 \times 10^{-38}}{6.626 \times 10^{-34}} \approx 5.34 \times 10^{-5} \text{ Hz} \] --- #### Step 2: Find the wavelength \( \lambda \) Using the relation: \[ c = \lambda \nu \Rightarrow \lambda = \frac{c}{\nu} \] where: - \( c = 3.00 \times 10^{8} \text{ m/s} \) Calculate: \[ \lambda = \frac{3.00 \times 10^{8}}{5.34 \times 10^{-5}} \approx 5.61 \times 10^{12} \text{ m} \] Expressed in nanometers: \[ 1 \text{ nm} = 10^{-9} \text{ m} \] \[ \lambda \approx 5.61 \times 10^{12} \text{ m} = 5.61 \times 10^{21} \text{ nm} \] --- ### **Final answers for Part (ii):** | Quantity | Value | |------------|--------| | **Frequency \( \nu \)** | \(\boxed{5.34 \times 10^{-5} \text{ Hz}}\) | | **Wavelength \( \lambda \)** | \(\boxed{5.61 \times 10^{12} \text{ nm}}\) | --- ### **Summary:** - **(i) Energy separation between \( n=1 \) and \( n=2 \):** approximately \( 3.54 \times 10^{-38} \text{ Joules} \) - **(ii) Corresponding radiation:** - Frequency: approximately \( 5.34 \times 10^{-5} \text{ Hz} \) - Wavelength: approximately \( 5.61 \times 10^{12} \text{ nm} \) *Note:* The extremely small energy difference and correspondingly tiny frequency indicate that the energy levels are very close, reflecting the large size of the box and the molecule's mass.