Question

For vibrational states, the Boltzmann equation can be written as

$\frac{N_1}{N_0}=exp(-∆E/kT)$

where $N_0$and data-custom-editor="chemistry" $N_1$are the populations of the lower and higher energy states, respectively, $∆E$is the energy difference between the states, $k$is Boltzmann’s constant, and $T$is the temperature in kelvins.

For temperatures of $20°\mathrm{C}$and data-custom-editor="chemistry" $40°\mathrm{C}$, calculate the ratios of the intensities of the anti-Stokes and Stokes lines for data-custom-editor="chemistry" ${\mathrm{CCl}}_4$at role="math" localid="1646297502302" $\left(\mathrm{a}\right)218{\mathrm{cm}}^{-1},\left(\mathrm{b}\right)459{\mathrm{cm}}^{-1},\left(\mathrm{c}\right)790{\mathrm{cm}}^{-1}$.

For each temperature and Raman shift, calculate how much more intense the Stokes line is compared to the anti-Stokes line.

Short answer

The ratio of intensities of the anti-Stokes and Stokes line for ${\mathrm{CCl}}_4$

(a) at frequency shift of role="math" localid="1646295813247" $218{\mathrm{cm}}^{-1}$is data-custom-editor="chemistry" $0.343$for temperature role="math" localid="1646296025354" $20°\mathrm{C}$and $0.363$for temperature role="math" localid="1646296072432" $40°\mathrm{C}$.

(b) at frequency shift of data-custom-editor="chemistry" $459{\mathrm{cm}}^{-1}$is data-custom-editor="chemistry" $0.104$for temperature data-custom-editor="chemistry" $20°\mathrm{C}$anddata-custom-editor="chemistry" $0.121$ for temperature data-custom-editor="chemistry" $40°\mathrm{C}$.

(c) at frequency shift of data-custom-editor="chemistry" $790{\mathrm{cm}}^{-1}$is data-custom-editor="chemistry" $0.0206$for temperature data-custom-editor="chemistry" $20°\mathrm{C}$and data-custom-editor="chemistry" $0.026$for temperature data-custom-editor="chemistry" $40°\mathrm{C}$.

Step-by-step solution

Part (a) Step 1. Given information

For temperatures of 20°C and 40°C, calculate the ratios of the intensities of the anti-Stokes and Stokes lines for ${\mathrm{CCl}}_4$at $\left(\mathrm{a}\right)218{\mathrm{cm}}^{-1},\left(\mathrm{b}\right)459{\mathrm{cm}}^{-1},\left(\mathrm{c}\right)790{\mathrm{cm}}^{-1}$.

The expression for the Boltzmann equation is:

localid="1646297558365" $\frac{N_1}{N_0}=exp(-∆E/kT)$

The expression for the intensity of Raman lines is directly proportional to the population of energy state is:

localid="1646297566520" $I\propto N$ ...... (I)

Here, the intensity of the energy state is localid="1646297576372" $N$.

The expression for the intensity of the stokes lines and the population at lower state using Equation (I) is:

localid="1646297587414" $I_S\propto N_0$...... (II)

Here, the intensity of the stokes lines is localid="1646297609580" $I_S$and the population at lower state is localid="1646297598703" $N_0$.

The expression or the intensity of the anti-stokes limes and the population at a higher energy state using Equation (I) is:

localid="1646297618683" $I_A\propto N_1$...... (III)

Here, the intensity of the anti-stokes lines is localid="1646297630331" $I_A$and the population at higher state is localid="1646297642012" $N_1$.

Divide the Equation (III) by Equation (II).

localid="1646297655590" $\frac{I_A}{I_S}=\frac{N_1}{N_0}$...... (IV)

Substitute localid="1646297665338" $exp\left(-∆E/kT\right)$for localid="1646297674768" $\frac{N_1}{N_0}$in Equation (IV).

localid="1646297682204" $\frac{I_A}{I_S}=exp\left(-∆E/kT\right)$...... (V)

The expression for temperature 1 using Equation (V) is:

localid="1646297693308" ${\left(\frac{I_A}{I_S}\right)}_1=exp\left(-∆E/k{T}_1\right)$...... (VI)

Write the expression for the temperature 2 using Equation (V).

localid="1646297701551" ${\left(\frac{I_A}{I_S}\right)}_2=exp\left(-∆E/k{T}_2\right)$...... (VII)

The expression for the change in energy between the states is:

$∆E=hc∆\overline{\nu }$...... (VIII)

The plank constant is $6.626\times {10}^{-34}\mathrm{J}.\mathrm{s}$, the Boltzmann’s constant is $1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}$, and speed of light is $3\times {10}^8\mathrm{m}/\mathrm{s}$.

Part (a) Step 2. The ratio of intensities of the anti-Stokes and Stokes line temperature at 20°C.

Substitute the values in Equation (VIII) as follows:

$$\begin{gathered} ∆E=6.626\times {10}^{-34}\mathrm{J}.\mathrm{S}\times 3\times {10}^8\mathrm{m}/\mathrm{s}\times 218{\mathrm{cm}}^{-1} \\ =19.878\times {10}^{-26}\mathrm{J}.\mathrm{m}\times 218{\mathrm{cm}}^{-1}\times \frac{{10}^2\mathrm{cm}}{1\mathrm{m}} \\ =4.33\times {10}^{-21}\mathrm{J} \end{gathered}$$

Substitute the value in Equation (VI) as follows:

$$\begin{gathered} {\left(\frac{I_A}{I_S}\right)}_1=\exp \left(-4.33\times {10}^{-21}\mathrm{J}/\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\times (20+273\mathrm{K})\right)\right) \\ =\exp \left(-3.13\times {10}^2/293\right) \\ =\exp \left(-1.06\right) \\ =0.343 \end{gathered}$$

Part (a) Step 3. The ratio of intensities of the anti-Stokes and Stokes line temperature at 40°C.

Substitute the value in Equation (VII) as follows:

$$\begin{gathered} {\left(\frac{I_A}{I_S}\right)}_2=\exp \left(-4.33\times {10}^{-21}\mathrm{J}/\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\times (40+273\mathrm{K})\right)\right) \\ =\exp \left(-3.13\times {10}^2/313\right) \\ =0.367 \end{gathered}$$

Part (b) Step 1. The ratio of intensities of the anti-Stokes and Stokes line temperature at 20°C.

Substitute the values in Equation (VIII) as follows:

$$\begin{gathered} ∆E=6.626\times {10}^{-34}\mathrm{J}.\mathrm{S}\times 3\times {10}^8\mathrm{m}/\mathrm{s}\times 459{\mathrm{cm}}^{-1} \\ =19.878\times {10}^{-26}\mathrm{J}.\mathrm{m}\times 459{\mathrm{cm}}^{-1}\times \frac{{10}^2\mathrm{cm}}{1\mathrm{m}} \\ =9.12\times {10}^{-21}\mathrm{J} \end{gathered}$$

Substitute the value in Equation (VI) as follows:

$$\begin{gathered} {\left(\frac{I_A}{I_S}\right)}_1=\exp \left(-9.12\times {10}^{-21}\mathrm{J}\mathrm{J}/\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\times (20+273\mathrm{K})\right)\right) \\ =\exp \left(-6.608\times {10}^2/293\right) \\ =\exp \left(-0.022\right) \\ =0.104 \end{gathered}$$

Part (b) Step 2. The ratio of intensities of the anti-Stokes and Stokes line temperature at 40°C.

Substitute the value in Equation (VII) as follows:

$$\begin{gathered} {\left(\frac{I_A}{I_S}\right)}_2=\exp \left(-9.21\times {10}^{-21}\mathrm{J}/\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\times (40+273\mathrm{K})\right)\right) \\ =\exp \left(-6.608\times {10}^2/313\right) \\ =\exp \left(-2.11\right) \\ =0.121 \end{gathered}$$

Part (c) Step 1. The ratio of intensities of the anti-Stokes and Stokes line temperature at 20°C.

Substitute the values in Equation (VIII) as follows:

$$\begin{gathered} ∆E=6.626\times {10}^{-34}\mathrm{J}.\mathrm{S}\times 3\times {10}^8\mathrm{m}/\mathrm{s}\times 790{\mathrm{cm}}^{-1} \\ =19.878\times {10}^{-26}\mathrm{J}.\mathrm{m}\times 790{\mathrm{cm}}^{-1}\times \frac{{10}^2\mathrm{cm}}{1\mathrm{m}} \\ =1.57\times {10}^{-20}\mathrm{J} \end{gathered}$$

Substitute the value in Equation (VI) as follows:

$$\begin{gathered} {\left(\frac{I_A}{I_S}\right)}_1=\exp \left(-1.57\times {10}^{-20}\mathrm{J}/\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\times (20+273\mathrm{K})\right)\right) \\ =\exp \left(-1.137\times {10}^2/293\right) \\ =\exp \left(-3.88\right) \\ =0.0206 \end{gathered}$$

Part (c) Step 2. The ratio of intensities of the anti-Stokes and Stokes line temperature at 40°C.

Substitute the value in Equation (VII) as follows:

$$\begin{gathered} {\left(\frac{I_A}{I_S}\right)}_2=\exp \left(-1.57\times {10}^{-21}\mathrm{J}/\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\times (40+273\mathrm{K})\right)\right) \\ =\exp \left(-1.137\times {10}^2/313\right) \\ =\exp \left(-3.63\right) \\ =0.026 \end{gathered}$$