Answer
$5.21\times10^{-4}m$.
Work Step by Step
The value of the rotational inertia of CO is calculated in Example 29–2. Use equation 29–2 to find the energy of the photon.
$$\Delta E=\frac{\hbar ^2}{I}\ell$$
$$=\frac{(1.055\times10^{-34}J \cdot s)^2}{1.46\times10^{-46}kg \cdot m^2}(5)=3.812\times10^{-22}J$$
Now find the wavelength.
$$\lambda=\frac{c}{f}=\frac{hc}{hf}=\frac{hc}{\Delta E}$$
$$=\frac{(6.626\times10^{-34}J \cdot s)(3.00\times10^8m/s)}{ 3.812\times10^{-22}J }=5.21\times10^{-4}m$$
