Chapter 29, Problem 8P

(a)

To determine

To Calculate: The characteristic rotational energy for ${\text{O}}_{\text{2}}$ molecule.

Explanation of Solution

Given:

Bond length, $l=0.121\text{nm}$

Formula Used:

Characteristic rotational energy:

  $E=\frac{{\hslash }^2}{2I}$

Here, $\hslash$ is the reduced Planck’s constant and I is the moment of inertia,

  $I=2m{\left(\frac{l}{2}\right)}^2$

Here, m is the mass of atoms and l is the bond length.

Calculations:

Substitute the known values and solve.

  $\hslash =1.054\times {10}^{-34}\text{J}\cdot \text{s}$

Mass of oxygen atom, $m=16\text{u}$

  $\begin{array}{l}E=\frac{{\hslash }^2}{2\left[2m{\left(\frac{l}{2}\right)}^2\right]}\\ E=\frac{{\left(1.054\times {10}^{-34}\text{J}\cdot \text{s}\right)}^2}{2\left[2\left(16\text{u}\times \text{1}\text{.66}\times {\text{10}}^{-27}\text{kg/u}\right){\left(\frac{1.21\times {10}^{-10}\text{m}}{2}\right)}^2\right]}\\ E=2.86\times {10}^{-23}\text{J}\end{array}$

Conclusion:

Thus, the characteristic rotational energy for ${\text{O}}_{\text{2}}$ molecule is $2.86\times {10}^{-23}\text{J}$ .

(b)

To determine

To Calculate: The energy and wavelength emitted in $L=2$ to $L=1$ transition.

Explanation of Solution

Given:

The characteristic rotational energy for ${\text{O}}_{\text{2}}$ molecule is $2.86\times {10}^{-23}\text{J}$ .

  $L=2$ to $L=1$ transition

Formula Used:

Characteristic rotational energy:

  $E=\frac{{\hslash }^2}{2I}$

Here, $\hslash$ is the reduced Planck’s constant and I is the moment of inertia,

  $\Delta E=\frac{{\hslash }^2}{I}L$

Here, L is the upper energy state.

  $\Delta E=\frac{hc}{\lambda }$

Here, h is the Planck’s constant, c is the speed of light and $\lambda$ is the wavelength.

Calculations:

  $\begin{array}{l}\because E=\frac{{\hslash }^2}{2I}=2.86\times {10}^{-23}\text{J}\\ \therefore \frac{{\hslash }^2}{I}=2\times 2.86\times {10}^{-23}\text{J}\end{array}$

  $\begin{array}{l}\Delta E=\frac{{\hslash }^2}{I}L\\ \Delta E=2\times 2.86\times {10}^{-23}\text{J}\times 2\\ \Delta E=11.44\times {10}^{-23}\text{J}\end{array}$

  $\begin{array}{l}\frac{hc}{\lambda }=11.44\times {10}^{-23}\text{J}\\ \lambda \text{=}\frac{hc}{11.44\times {10}^{-23}\text{J}}\\ \lambda =\frac{\text{6}{\text{.63×10}}^{\text{-34}}\text{J}\cdot {\text{s×3×10}}^{\text{8}}\text{m/s}}{11.44\times {10}^{-23}\text{J}}\\ \lambda =1.74\times {10}^{-3}\text{m}\end{array}$

Conclusion:

Thus, energy emitted in $L=2$ to $L=1$ transition is $\Delta E=11.44\times {10}^{-23}\text{J}$ and the wavelength is $\lambda =1.74\times {10}^{-3}\text{m}$ .