Sure! Here’s a step-by-step solution to finding the wavenumber of radiation (in cm⁻¹) required to induce a transition between the ground state and the first vibrational excited state of the CO molecule, modeled as a quantum harmonic oscillator. --- ### **Step 1: Calculate the Reduced Mass (\(\mu\))** The reduced mass is given by: \[ \mu = \frac{m_1 m_2}{m_1 + m_2} \] where \( m_1 = \) mass of \(^{12}\mathrm{C} \), \( m_2 = \) mass of \(^{16}\mathrm{O} \). - Atomic mass of \(^{12}\mathrm{C} = 12\,\mathrm{u} \) - Atomic mass of \(^{16}\mathrm{O} = 16\,\mathrm{u} \) - \( 1\,\mathrm{u} = 1.66054 \times 10^{-27}\,\mathrm{kg} \) So, \[ \mu = \frac{12 \times 16}{12 + 16} \times (1.66054 \times 10^{-27}\,\mathrm{kg}) = \frac{192}{28} \times (1.66054 \times 10^{-27}) = 6.8571 \times 1.66054 \times 10^{-27} = 1.139 \times 10^{-26}\,\mathrm{kg} \] **Explanation:** We calculate the reduced mass using the formula for two-body systems, converting atomic mass units to kg for SI consistency. --- ### **Step 2: Write the Vibrational Energy Difference** For a quantum harmonic oscillator, the vibrational energy levels are: \[ E_n = \left( n + \frac{1}{2} \right) h \nu \] where \( \nu \) is the vibrational frequency. The energy difference between the ground state (\( n=0 \)) and first excited state (\( n=1 \)) is: \[ \Delta E = E_1 - E_0 = [1.5 - 0.5]h\nu = h\nu \] The frequency \( \nu \) is related to the force constant \( k \) and reduced mass \( \mu \) by: \[ \nu = \frac{1}{2\pi} \sqrt{\frac{k}{\mu}} \] So, \[ \Delta E = h \cdot \frac{1}{2\pi} \sqrt{\frac{k}{\mu}} \] **Explanation:** We express the quantum energy difference in terms of the vibrational frequency, which depends on the force constant and reduced mass. --- ### **Step 3: Convert Energy Difference to Wavenumber (\(\tilde{\nu}\))** The wavenumber (\(\tilde{\nu}\)) in cm⁻¹ is: \[ \tilde{\nu} = \frac{\nu}{c} \] But since \(\nu = \frac{1}{2\pi} \sqrt{\frac{k}{\mu}}\): \[ \tilde{\nu} = \frac{1}{2\pi c} \sqrt{\frac{k}{\mu}} \] **Explanation:** We convert the frequency (Hz) to wavenumber (cm⁻¹), which is the standard spectroscopic unit. --- ## **Final Answer** **The wavenumber required to induce the transition is:** \[ \boxed{ \tilde{\nu} = \frac{1}{2\pi c} \sqrt{\frac{k}{\mu}} } \] where - \( k \) = force constant (in N/m) - \( \mu = \frac{12 \times 16}{12 + 16} \times 1.66054 \times 10^{-27}\,\mathrm{kg} \) - \( c \) = speed of light (\( 2.998 \times 10^{10}\,\mathrm{cm/s} \)) --- **Summary of Steps:** 1. **Find reduced mass \(\mu\) of CO molecule.** 2. **Write quantum vibrational energy difference \(\Delta E = h\nu\).** 3. **Express required wavenumber: \(\tilde{\nu} = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}}\).**

Let's complete the calculation step-by-step with numerical values. --- ### **Given Data:** - Force constant: \(k\) (leave in terms of \(k\)) - Atomic masses: - \(^{12}\mathrm{C}\): \(12\,\mathrm{u}\) - \(^{16}\mathrm{O}\): \(16\,\mathrm{u}\) - Conversion factor: \(1\,\mathrm{u} = 1.66054 \times 10^{-27}\,\mathrm{kg}\) - Speed of light: \(c = 2.998 \times 10^{10}\,\mathrm{cm/s}\) --- ### **Step 1: Reduced Mass \(\mu\)** \[ \mu = \frac{m_{C} \times m_{O}}{m_{C} + m_{O}} = \frac{12\,\mathrm{u} \times 16\,\mathrm{u}}{12\,\mathrm{u} + 16\,\mathrm{u}} = \frac{192\,\mathrm{u}^2}{28\,\mathrm{u}} = \frac{192}{28}\,\mathrm{u} \approx 6.8571\,\mathrm{u} \] Convert to kg: \[ \mu = 6.8571\,\mathrm{u} \times 1.66054 \times 10^{-27}\,\mathrm{kg/u} \approx 1.139 \times 10^{-26}\,\mathrm{kg} \] --- ### **Step 2: Vibrational Frequency \(\nu\)** \[ \nu = \frac{1}{2\pi} \sqrt{\frac{k}{\mu}} \] The energy difference between the ground and first excited state: \[ \Delta E = h \nu \] where Planck's constant: \[ h = 6.626 \times 10^{-34}\,\mathrm{Js} \] --- ### **Step 3: Wavenumber \(\tilde{\nu}\)** Wavenumber (cm\(^{-1}\)): \[ \tilde{\nu} = \frac{\nu}{c} = \frac{1}{2\pi c} \sqrt{\frac{k}{\mu}} \] Plugging in \(c = 2.998 \times 10^{10}\,\mathrm{cm/s}\): \[ \boxed{ \tilde{\nu} = \frac{1}{2\pi \times 2.998 \times 10^{10}} \sqrt{\frac{k}{1.139 \times 10^{-26}}} } \] Calculating the constants: \[ 2\pi \times 2.998 \times 10^{10} \approx 1.883 \times 10^{11} \] Therefore, \[ \boxed{ \tilde{\nu} \approx \frac{1}{1.883 \times 10^{11}} \times \sqrt{\frac{k}{1.139 \times 10^{-26}}} } \] or \[ \boxed{ \tilde{\nu} \approx 5.31 \times 10^{-12} \times \sqrt{\frac{k}{1.139 \times 10^{-26}}} } \] --- ### **Final Expression:** \[ \boxed{ \text{Wavenumber} \approx 5.31 \times 10^{-12} \times \sqrt{\frac{k}{1.139 \times 10^{-26}}}\ \text{cm}^{-1} } \] This is the general formula. To get a numerical answer, substitute the specific force constant \(k\) (in N/m). For example, if \(k = 1860\,\mathrm{N/m}\): \[ \sqrt{\frac{1860}{1.139 \times 10^{-26}}} = \sqrt{1.634 \times 10^{29}} \approx 1.28 \times 10^{14} \] Then, \[ \tilde{\nu} \approx 5.31 \times 10^{-12} \times 1.28 \times 10^{14} \approx 680\,\text{cm}^{-1} \] --- ### **Final numerical estimate for \(k = 1860\,\mathrm{N/m}\):** \[ \boxed{ \tilde{\nu} \approx 680\,\mathrm{cm}^{-1} } \] **In summary:** The wavenumber for the transition is approximately **680 cm\(^{-1}\)** given typical force constants for CO. Adjust accordingly based on your specific \(k\).

The solution utilizes the quantum harmonic oscillator model to determine the vibrational transition energy of the CO molecule. First, it calculates the reduced mass of the CO system from atomic masses converted to kilograms. Then, it relates the energy difference between vibrational levels to the vibrational frequency, which depends on the force constant and reduced mass. Finally, it converts the vibrational frequency into wavenumber (cm\(^{-1}\)) by dividing by the speed of light, providing the spectral position of the transition.