Chapter 29, Problem 9P

(a)

To determine

To Calculate: The energy and wavelength emitted in $L=1$ to $L=0$ transition.

Explanation of Solution

Given:

The equilibrium separation between H atoms in hydrogen molecule is $l=0.074\text{nm}$

  $L=1$ to $L=0$ transition.

Formula Used:

Energy emitted during the transition:

  $\Delta E=\frac{{\hslash }^2}{I}L$

Here, I is the moment of inertia, $\hslash$ is the reduced Planck’s constant and L is the upper energy state.

  $I=2m{\left(\frac{l}{2}\right)}^2$

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

  $\Delta E=\frac{hc}{\lambda }$

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

Calculations:

  $\begin{array}{l}I=2m{\left(\frac{l}{2}\right)}^2\\ I=2\left(\text{1u×1}{\text{.66×10}}^{\text{-27}}\text{kg/u}\right){\left(\frac{0.074\times {10}^{-9}\text{m}}{2}\right)}^2\\ I=4.54\times {10}^{-48}\text{kg}\cdot {\text{m}}^{\text{2}}\end{array}$

  $\begin{array}{l}\Delta E=\frac{{\hslash }^2}{I}L\\ \Delta E=\frac{{\left(1.054\times {10}^{-34}\text{J}\cdot \text{s}\right)}^2}{4.54\times {10}^{-48}\text{kg}\cdot {\text{m}}^{\text{2}}}\left(1\right)\\ \Delta E=2.45\times {10}^{-21}\text{J}\end{array}$

  $\begin{array}{l}\frac{hc}{\lambda }=2.45\times {10}^{-21}\text{J}\\ \lambda \text{=}\frac{hc}{2.45\times {10}^{-21}\text{J}}\\ \lambda =\frac{\text{6}{\text{.63×10}}^{\text{-34}}\text{J}\cdot {\text{s×3×10}}^{\text{8}}\text{m/s}}{2.45\times {10}^{-21}\text{J}}\\ \lambda =8.13\times {10}^{-5}\text{m}\end{array}$

Conclusion:

Thus, energy emitted in $L=1$ to $L=0$ transition is $2.45\times {10}^{-21}\text{J}$ and the wavelength is $8.13\times {10}^{-5}\text{m}$ .

(b)

To determine

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

Explanation of Solution

Given:

The equilibrium separation between H atoms in hydrogen molecule is $l=0.074\text{nm}$

  $L=2$ to $L=1$ transition.

Formula Used:

Energy emitted during the transition:

  $\Delta E=\frac{{\hslash }^2}{I}L$

Here, I is the moment of inertia, $\hslash$ is the reduced Planck’s constant and L is the upper energy state.

  $I=2m{\left(\frac{l}{2}\right)}^2$

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

  $\Delta E=\frac{hc}{\lambda }$

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

Calculations:

  $\begin{array}{l}I=2m{\left(\frac{l}{2}\right)}^2\\ I=2\left(\text{1u×1}{\text{.66×10}}^{\text{-27}}\text{kg/u}\right){\left(\frac{0.074\times {10}^{-9}\text{m}}{2}\right)}^2\\ I=4.54\times {10}^{-48}\text{kg}\cdot {\text{m}}^{\text{2}}\end{array}$

  $\begin{array}{l}\Delta E=\frac{{\hslash }^2}{I}L\\ \Delta E=\frac{{\left(1.054\times {10}^{-34}\text{J}\cdot \text{s}\right)}^2}{4.54\times {10}^{-48}\text{kg}\cdot {\text{m}}^{\text{2}}}\left(2\right)\\ \Delta E=4.9\times {10}^{-21}\text{J}\end{array}$

  $\begin{array}{l}\frac{hc}{\lambda }=4.9\times {10}^{-21}\text{J}\\ \lambda \text{=}\frac{hc}{4.9\times {10}^{-21}\text{J}}\\ \lambda =\frac{\text{6}{\text{.63×10}}^{\text{-34}}\text{J}\cdot {\text{s×3×10}}^{\text{8}}\text{m/s}}{4.9\times {10}^{-21}\text{J}}\\ \lambda =4.065\times {10}^{-5}\text{m}\end{array}$

Conclusion:

Thus, energy emitted in $L=2$ to $L=1$ transition is $4.9\times {10}^{-21}\text{J}$ and the wavelength is $4.065\times {10}^{-5}\text{m}$ .

(c)

To determine

To Calculate: The energy and wavelength emitted in $L=3$ to $L=2$ transition.

Explanation of Solution

Given:

The equilibrium separation between H atoms in hydrogen molecule is $l=0.074\text{nm}$

  $L=3$ to $L=2$ transition.

Formula Used:

Energy emitted during the transition:

  $\Delta E=\frac{{\hslash }^2}{I}L$

Here, I is the moment of inertia, $\hslash$ is the reduced Planck’s constant and L is the upper energy state.

  $I=2m{\left(\frac{l}{2}\right)}^2$

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

  $\Delta E=\frac{hc}{\lambda }$

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

Calculations:

  $\begin{array}{l}I=2m{\left(\frac{l}{2}\right)}^2\\ I=2\left(\text{1u×1}{\text{.66×10}}^{\text{-27}}\text{kg/u}\right){\left(\frac{0.074\times {10}^{-9}\text{m}}{2}\right)}^2\\ I=4.54\times {10}^{-48}\text{kg}\cdot {\text{m}}^{\text{2}}\end{array}$

  $\begin{array}{l}\Delta E=\frac{{\hslash }^2}{I}L\\ \Delta E=\frac{{\left(1.054\times {10}^{-34}\text{J}\cdot \text{s}\right)}^2}{4.54\times {10}^{-48}\text{kg}\cdot {\text{m}}^{\text{2}}}\left(3\right)\\ \Delta E=7.35\times {10}^{-21}\text{J}\end{array}$

  $\begin{array}{l}\frac{hc}{\lambda }=7.35\times {10}^{-21}\text{J}\\ \lambda \text{=}\frac{hc}{7.35\times {10}^{-21}\text{J}}\\ \lambda =\frac{\text{6}{\text{.63×10}}^{\text{-34}}\text{J}\cdot {\text{s×3×10}}^{\text{8}}\text{m/s}}{7.35\times {10}^{-21}\text{J}}\\ \lambda =2.71\times {10}^{-5}\text{m}\end{array}$

Conclusion:

Thus, energy emitted in $L=3$ to $L=2$ transition is $7.35\times {10}^{-21}\text{J}$ and the wavelength is $2.71\times {10}^{-5}\text{m}$ .