Chapter 11, Problem 11.57E

Interpretation Introduction

Interpretation:

The wavelength of light necessary to cause the transition from state $l=5$ to $l=6$ and $l=7$ to $l=8$ is to be calculated and the validation of the statement that this model can be used to describe $C_{60}$’s electronic absorptions is to be stated.

Concept introduction:

The energy for the 3-D rotational motion is given by,

$E=\frac{l\left(l+1\right){\hslash }^2}{2I}$

The energy of the particle depends on the moment of inertia, quantum number and Planck’s constant. The total energy is quantized.

Answer to Problem 11.57E

The wavelength of light necessary to cause the transition from state $l=5$ to $l=6$ and $l=7$ to $l=8$ is $331\text{nm}$ and $250\text{nm}$ respectively. There is error of about $0-3\%$ in the calculation of wavelength of transition using the given model. Thus, it can be used for describing the $C_{60}$’s electronic absorptions.

Explanation of Solution

The moment of inertia for the electron in the spherical $C_{60}$ molecule is $1.12\times {10}^{-49}\text{kg}\cdot {\text{m}}^2$

The energy for the 3-D rigid rotor for quantum number $l$ is,

$E_l=\frac{l\left(l+1\right){\hslash }^2}{2I}$ …(1)

Where,

• $l$ is the quantum number.

• $I$ is the moment of inertia.

• $\hslash$ has the value $\frac{h}{2\pi }$, $h$ is the Planck’s constant.

For the calculation of wavelength of the transition of states from $l=5$ to $l=6$, the energy of the states is calculated first.

Substitute the value of $l=5$ in equation (1) as shown below.

$\begin{array}{l}E\left(5\right)=\frac{5\left(5+1\right){\hslash }^2}{2I}\\ E\left(5\right)=\frac{5\left(5+1\right){\left(\frac{6.626\times {10}^{-34}\text{kg}\cdot {\text{m}}^2\cdot {\text{s}}^{-1}}{2\pi }\right)}^2}{2\times 1.12\times {10}^{-49}\text{kg}\cdot {\text{m}}^2}\\ E\left(5\right)=1.49\times {10}^{-18}\text{J}\end{array}$

Similarly for $l=6$,

$\begin{array}{l}E\left(6\right)=\frac{6\left(6+1\right){\hslash }^2}{2I}\\ E\left(6\right)=\frac{6\left(6+1\right){\left(\frac{6.626\times {10}^{-34}\text{kg}\cdot {\text{m}}^2\cdot {\text{s}}^{-1}}{2\pi }\right)}^2}{2\times 1.12\times {10}^{-49}\text{kg}\cdot {\text{m}}^2}\\ E\left(6\right)=2.09\times {10}^{-18}\text{J}\end{array}$

The energy difference is calculated by $E\left(6\right)-E\left(5\right)$ as shown below.

$\begin{array}{rll}E\left(6\right)-E\left(5\right)& =& 2.09\times {10}^{-18}\text{J}-1.49\times {10}^{-18}\text{J}\\ & =& 6\times {10}^{-19}\text{J}\end{array}$

Thus, the energy difference $E\left(6\right)-E\left(5\right)$ is $6\times {10}^{-19}\text{J}$.

The wavelength transition is calculated by the formula,

$\Delta E=\frac{hc}{\lambda }$

Where,

• $\Delta E$ is the energy difference.

• $h$ is the Planck’s constant.

• $\lambda$ is the wavelength.

• $c$ is the speed of light.

Substitute the values of energy difference,

$\begin{array}{rll}6\times {10}^{-19}\text{J}& =& \frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times \text{3}\times {10}^8\text{m}/\text{s}}{\lambda }\\ \lambda & =& \frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times \text{3}\times {10}^8\text{m}/\text{s}}{6\times {10}^{-19}\text{J}}\\ & =& 3.31\times {10}^{-7}\text{m}\end{array}$

Thus, the wavelength of the transition of states from $l=5$ to $l=6$ is $3.31\times {10}^{-7}\text{m}$ or $331\text{nm}$.

Similarly for the calculation of wavelength of the transition of states from $l=7$ to $l=8$, the energy of the states is calculated first.

Substitute the value of $l=7$ in the formula (1) as shown below.

$\begin{array}{l}E\left(7\right)=\frac{7\left(7+1\right){\hslash }^2}{2I}\\ E\left(7\right)=\frac{7\left(7+1\right){\left(\frac{6.626\times {10}^{-34}\text{kg}\cdot {\text{m}}^2\cdot {\text{s}}^{-1}}{2\pi }\right)}^2}{2\times 1.12\times {10}^{-49}\text{kg}\cdot {\text{m}}^2}\\ E\left(7\right)=2.78\times {10}^{-18}\text{J}\end{array}$

Similarly for $l=8$,

$\begin{array}{l}E\left(8\right)=\frac{8\left(8+1\right){\hslash }^2}{2I}\\ E\left(8\right)=\frac{8\left(8+1\right){\left(\frac{6.626\times {10}^{-34}\text{kg}\cdot {\text{m}}^2\cdot {\text{s}}^{-1}}{2\pi }\right)}^2}{2\times 1.12\times {10}^{-49}\text{kg}\cdot {\text{m}}^2}\\ E\left(8\right)=3.58\times {10}^{-18}\text{J}\end{array}$

The energy difference is calculated by $E\left(7\right)-E\left(8\right)$ as shown below.

$\begin{array}{rll}E\left(6\right)-E\left(5\right)& =& 3.58\times {10}^{-18}\text{J}-2.78\times {10}^{-18}\text{J}\\ & =& 8\times {10}^{-19}\text{J}\end{array}$

Thus, the energy difference $E\left(6\right)-E\left(5\right)$ is $8\times {10}^{-19}\text{J}$.

The wavelength transition is calculated by the formula,

$\Delta E=\frac{hc}{\lambda }$

Where,

• $\Delta E$ is the energy difference.

• $h$ is the Planck’s constant.

• $\lambda$ is the wavelength.

• $c$ is the speed of light.

Substitute the values of energy difference,

$\begin{array}{r}8\times {10}^{-19}\text{J}\begin{array}{}\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times \text{3}\times {10}^8\text{m}/\text{s}}{\lambda }\end{array}\\ \lambda \begin{array}{}\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times \text{3}\times {10}^8\text{m}/\text{s}}{8\times {10}^{-19}\text{J}}\end{array}\\ \begin{array}{}2.5\times {10}^{-7}\text{m}\end{array}\end{array}$

Thus, the wavelength of the transition of states from $l=7$ to $l=8$ is $2.5\times {10}^{-7}\text{m}$ or $250\text{nm}$.

The experimentally measured absorptions at wavelengths of $331\text{nm}$ and $256\text{nm}$. The percentage error can be calculated by,

$\%\text{error}=\left|\frac{\text{Experimental}\text{value}-\text{Calculated}\text{value}}{\text{Experimental}\text{value}}\right|\times 100$

The percentage error in wavelength of the transition of states from $l=5$ to $l=6$ is,

$\begin{array}{rll}\%\text{error}& =& \left|\frac{\text{328}-331}{\text{328}}\right|\times 100\\ & =& 0.91\%\end{array}$

The percentage error in wavelength of the transition of states from $l=5$ to $l=6$ is $0.91\%$.

The percentage error in wavelength of the transition of states from $l=7$ to $l=8$ is,

$\begin{array}{rll}\%\text{error}& =& \left|\frac{\text{256}-250}{256}\right|\times 100\\ & =& 2.34\%\end{array}$

The percentage error in wavelength of the transition of states from $l=7$ to $l=8$ is $2.34\%$.

There is only error of about $0-3\%$ in the calculation of wavelength of transition using the given model. Thus, it can be used for the describing the $C_{60}$’s electronic absorptions.

Conclusion

The wavelength of light necessary to cause the transition from state $l=5$ to $l=6$ and $l=7$ to $l=8$ is $331\text{nm}$ and $250\text{nm}$. There is only error of about $0-3\%$ in the calculation of wavelength of transition using the given model. Thus, it can be used for the describing the $C_{60}$’s electronic absorptions.