Question

The effective force constant of the molecular “spring” in HCL is $\mathbf{480}\mathbf{ }\mathbf{\text{N/m}}$, and the bond length is $\mathbf{0}.\mathbf{13} \mathbf{nm}$.

(a) Determine the energies of the two lowest-energy vibrational states.

(b) For these energies, determine the amplitude of vibration if the atoms could be treated as oscillating classical particles.

(c) For these energies, by what percentages does the atomic separation fluctuate?

(d) Calculate the classical vibrational frequency${\mathit{\omega }}_{\mathbf{v}\mathbf{h}}\mathbf{=}\sqrt{\mathbf{k}\mathbf{/}\mathbf{\mu }}$and rotational frequency for the rotational frequency${\mathit{\omega }}_{\mathbf{r}\mathbf{o}\mathbf{t}}\mathbf{=}\mathit{L}\mathbf{/}\mathit{I}$, assume that L is the its lowest non zero value, $\sqrt{\mathbf{1}\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{1}\mathbf{)}}\mathit{h}$and that the moment of inertia $\mathit{I}$is $\mathit{\mu }{\mathit{a}}^{\mathbf{2}}$.

(e) Is is valid to treat the atomic separation as fixed for rotational motion while changing for vibrational?

Short answer

(a) First two vibrational states have energy $0.179 \text{eV}$and $0.537 \text{eV}$.

(b) Amplitude of vibration are $A_{0,0}\approx 0.011 \text{m}$and $A_{1,0}\approx 0.019 \text{nm}$.

(c) Percentage of fluctuation is $\approx 8.5\%$and $\approx 14\%$.

(d) Classical vibrational frequency$5.4\times {10}^{14}{s}^{-1}$ and Rotational frequency $5.4\times {10}^{12}{s}^{-1}$.

(e) As rotational frequency is much smaller than the vibrational frequency, it is reasonable to treat the separation between atoms fixed for rotational movement but changing for vibrational movement.

Step-by-step solution

Identification of given data

Effective force constant$k=480 \text{N/m}$

Bond length$a=0.13 \text{nm}$

Significance of molecular vibrations

Movements of one atom within a molecule in relation to other atoms are known as molecular vibrations. The smallest vibrations are lengthening and shortening of a single link. To understand this, picture the bond as a spring and the molecular vibration as the spring's simple harmonic motion.

(a) Determining the energies of two lowest energy vibrational states.

To determine effective (reduced mass) of the hydrogen chloride molecule use the following formula,

$\mu =\frac{m_1{m}_2}{m_1+m_2}$

Here, $m_1$is atomic mass of hydrogen, and atomic mass of chlorine.

Substitute $1.008u$for $m_1$,$35.453u$for $m_2$in the above equation determine reduced or effective mass.

$$\begin{gathered} \mu =\frac{m_1{m}_2}{m_1+m_2} \\ =\frac{1.007\text{u}\times 35.453\text{u}}{1.007\text{u}+35.453\text{u}} \\ =0.979u \end{gathered}$$

First vibrational state has energy,

$E_{0.0}=\frac{1}{2}h\sqrt{\frac{k}{\mu }}$

Here,k is constant, and $\mu$is reduced mass or effective mass.

Substitute $1.055\times {10}^{-34}J.s$for plank constant h,$480 \text{N/m}$for k, $0.979\text{u}$for $\mu$in the above equation to solve for energy.

$$\begin{gathered} E_{0.0}=\frac{1}{2}h\sqrt{\frac{k}{\mu }} \\ =\frac{1}{2}(1.055\times {10}^{-34}\text{J.s})\sqrt{\frac{480 \text{N/m}}{(0.979\text{u})\left(\frac{1.66\times {10}^{-27}\text{kg}}{\text{u}}\right)}} \\ =2.87\times {10}^{-20}\text{J} \\ =(2.87\times {10}^{-20}\text{J})\left(\frac{\text{eV}}{1.6\times {10}^{-19}\text{J}}\right) \\ =0.179 \text{eV} \end{gathered}$$

The second vibrational state has energy

_{$E_{1.0}=\left(1+\frac{1}{2}\right)h\sqrt{\frac{k}{\mu }}$}

Substitute $1.055\times {10}^{-34}J.s$for plank constant h,$480 \text{N/m}$for k, $0.979\text{u}$for$\mu$ in the above equation to solve for energy.

$$\begin{gathered} E_{1.0}=\left(1+\frac{1}{2}\right)h\sqrt{\frac{k}{\mu }} \\ =\frac{3}{2}(1.055\times {10}^{-34}\text{J.s})\sqrt{\frac{480 \text{N/m}}{0.979\text{u}\times 1.66\times {10}^{-27}\text{kg}}} \\ =\frac{3}{2}(1.055\times {10}^{-34}\text{J.s})\sqrt{\frac{480\text{N/m}}{(0.979\text{u})\left(\frac{1.66\times {10}^{-27}\text{kg}}{\text{u}}\right)}} \\ =(8.61\times {10}^{-20}\text{J})\left(\frac{1 \text{eV}}{1.6\times {10}^{-19}\text{J}}\right) \\ =0.537\text{eV} \end{gathered}$$

Therefore, first two vibrational states have energy $0.179 \text{eV}$and $0.537\text{eV}$.

(b) Determining the amplitude of vibration.

If atoms were to be treated as oscillating classical particles, then its energy is given by

$E=\frac{1}{2}k{A}^2$

Where k= spring constant and A= amplitude

$A=\sqrt{\frac{2E}{\kappa }}$

So for, $E_{0,0}$,

$$\begin{gathered} A_{0,0}=\sqrt{\frac{2E_{0,,0}}{\kappa }} \\ =\sqrt{\frac{2\times 2.87\times {10}^{-20} \text{J}}{480 \text{N/m}}} \\ =0.0109\times {10}^{-9}\text{m} \\ {A}_{0,0}\approx 0.011 \text{nm} \end{gathered}$$

Similarly, for $E_{1,,0}$

$$\begin{gathered} A_{1,0}=\sqrt{\frac{2E_{1,,0}}{k}} \\ =\sqrt{\frac{2\times 8.61\times {10}^{-20} \text{J}}{480 \text{N/m}}} \\ =0.0189\times {10}^{-9}\text{m} \\ {A}_{0,0}\approx 0.019 \text{nm} \end{gathered}$$

(c) Determining the percentage of atomic separation fluctuate.

Given that bond length $a=0.13 \text{nm}$for $n=0$, the amplitude $A_{0,0}=0.011 \text{nm}$.

The fluctuation in the atomic separation in percentage is given by

$$\begin{gathered} \frac{A_{0,0}}{a}\times 100=\frac{0.011}{0.13}\times 100 \\ =8.46\% \\ \approx 8.5\% \end{gathered}$$

Similarly, for $n=1$,

$$\begin{gathered} \frac{A_{1,0}}{a}\times 100=\frac{0.019}{0.13}\times 100 \\ =14.61\% \\ \approx 14\% \end{gathered}$$

(d) Determining the frequency for classical vibrational and rotational frequency.

The classical vibrational frequency can be obtained using the following relation,

$W_{vib}=\sqrt{\frac{k}{\mu }}$

Substitute $480 \text{N/m}$for k, and $0.979\text{u}$for $\mu$in the equation to determine classical vibrational frequency.

$$\begin{gathered} W_{vib}=\sqrt{\frac{k}{\mu }}\backslash \mathrm{hfill} \\ {W}_{vib}=\sqrt{\frac{480 \text{N/m}}{1.625\times {10}^{-27} \text{kg}}}\backslash \mathrm{hfill} \\ {W}_{vib}=5.43\times {10}^{14}{\text{s}}^{\text{- 1}}\backslash \mathrm{hfill} \end{gathered}$$

Similarly, classical rotational frequency is given by an expression

$$\begin{gathered} W_{rot}=\frac{L}{I} \\ L=\sqrt{1(1+1)}h \\ I=\mu a^2 \end{gathered}$$

Now, by putting equation in that formula we get,

$$\begin{gathered} W_{rot}=\frac{L}{I} \\ =\frac{\sqrt{1(1+1)}h}{\mu a^2} \\ =\frac{\sqrt{2}h}{1.625\times {10}^{-27} \text{kg}\times {(0.13\times {10}^{-9} \text{m})}^2} \\ {W}_{rot}=5.42\times {10}^{12}{\text{s}}^{\text{- 1}} \end{gathered}$$

(e) Determining the validation to treat the atomic separation as fixed for rotational motion while changing for vibrational.

We see that:

$$\begin{gathered} W_{vib}=5.4\times {10}^{14}{\text{s}}^{\text{- 1}} \\ {W}_{rot}=5.42\times {10}^{12}{\text{s}}^{\text{- 1}} \end{gathered}$$

so, $W_{rot}\ll W_{vib}$

Thus, atomic separation is considered to be fixed for rotational motion owing to its smaller frequency when compared to larger frequency of vibrational motion.