Quantum Chemistry/Example 27

Calculate the frequency of a photon needed to excite CO from the ground vibrational state to the first vibrational state (k=1860 N/m).

$$v_0= \frac{1}{2\pi}\cdot\biggl(\frac{k}{\mu}\biggr)^\frac{1}{2} \qquad where \  \mu =  \frac{m_1\cdot m_2}{m_1 +m_2}

$$

C= 12.011u, O =15.999u

Plug the values into the reduced mass equation:

$$\mu = \frac{12.011u\cdot 15.999u}{12.011u +15.999u}

$$

$$\mu = 6.861u

$$

Convert the reduced mass from units of u to kg

$$1u = 1.66054\times 10^{-27}kg

$$

$$\mu = 1.139\times10^{-26}kg

$$

$$v_0= \frac{1}{2\pi}\cdot\biggl(\frac{1860N/m}{1.139\times10^{-26}kg}\biggr)^\frac{1}{2} \qquad

$$

$$\therefore v_0= 6.43\times10^{13}s^{-1}

$$