91影视

The effective spring constant is 1860 N/m

The bond length is0.113 nm

The mean separation between the nuclei of two bonded atoms in a molecule is known as bond length. The following factors affect the bond length between two bonded atoms- hybridization, the number of atoms present in a molecule, the number of bonds present and the size of the atom.

The energy of a photon that might be absorbed is given as-

$E=\mathrm{鈩�}\sqrt{\frac{魏}{渭}}卤I\frac{{\mathrm{鈩�}}^2}{渭a^2}路路路路路路路路路路路路路路路路路路路路路路路路路路路路路路路路路路路路路\left(1\right)$

Here,$\mathrm{鈩�}$is the modified Planck鈥檚 constant, k is the spring constant,$渭$is the effective mass of the atom andis the bond length.

The effective mass is given as-

$渭=\frac{mm\text{'}}{m+m\text{'}}$

Where and are the masses of the atoms participating in bonding.

The mass number of carbon and oxygen are respectively, so the effective mass of CO molecule is-

$$\begin{gathered} 渭=\frac{12.01脳16.00}{12.01+16.00}脳1.66脳{10}^{-27}\text{kg} \\ \text{= 1.14}脳{10}^{-26}\text{kg} \end{gathered}$$

For $魏=1860N/ma=0,113nm\text{and}渭=1.14脳{10}^{-26}\text{kg}$, the energy of photon is-

$$\begin{gathered} E=\left(1.05脳{10}^{-34}\text{Js}\right)\sqrt{\frac{1860N/m}{1.14脳{10}^{-26}\text{kg}}}卤I\frac{{\left(1.05脳{10}^{-34}\text{Js}\right)}^2}{\left(1.14脳{10}^{-26}\text{kg}\right)\left(0.113脳{10}^{-9}\text{m}\right)} \\ E=\left(0.265\text{eV}\right)卤I\left(0.0005\text{eV}\right) \\ =0.265\text{eV}\left(1卤I\frac{0.0005\text{eV}}{0.265\text{eV}}\right) \\ =0.265\text{eV}\left(1卤I0.002\right) \end{gathered}$$

$$\begin{gathered} E=\left(1.05脳{10}^{-34}\text{Js}\right)\sqrt{\frac{1860N/m}{1.14脳{10}^{-26}\text{kg}}}卤I\frac{{\left(1.05脳{10}^{-34}\text{Js}\right)}^2}{\left(1.14脳{10}^{-26}\text{kg}\right)\left(0.113脳{10}^{-9}\text{m}\right)} \\ E=\left(0.265\text{eV}\right)卤I\left(0.0005\text{eV}\right) \\ =0.265\text{eV}\left(1卤I\frac{0.0005\text{eV}}{0.265\text{eV}}\right) \\ =0.265\text{eV}\left(1卤I0.002\right) \end{gathered}$$

It is a common observation that, for the $螖n=1$ transition and the photon of energy $0.265\text{eV}$, there are many wavelengths present around the photon. These wavelengths are-

So, the required wavelengths are- $4.66渭m,4.67渭m,4.69渭m,4.70渭m$.

$$\begin{gathered} 位=\frac{hc}{E} \\ =\frac{\left(6.626脳{10}^{-34}\text{Js}\right)\left(6.24脳{10}^{18}\right)\left(3脳{10}^8\right)}{0.265\text{eV}\left(1卤I0.002\right)} \\ =\frac{4.68脳{10}^{-6}\text{m}}{\left(1卤I0.002\right)} \end{gathered}$$