Problem Set 2

Problem Set 2#

Problem 1 (40 points evenly divided)#

The fundamental absorption feature of hydrogen fluoride (HF) occurs at
\(\bar{\nu} = 3960 \ \text{cm}^{-1}\) (see Figure 1), where the wavenumber \(\bar{\nu}\) is related to the frequency \(\nu\) by:

\[ \bar{\nu} = \frac{\nu}{c} \]

where \(c\) is the speed of light (\(2.99 \times 10^{10} \ \text{cm/s}\)).
Assuming the Harmonic Oscillator is a good model for the vibrational motion of HF, we can relate this IR absorbance peak to the force constant and reduced mass of HF as follows:

\[ \bar{\nu} = \frac{1}{2\pi c} \sqrt{\frac{k}{\mu}} \]

Figure 1: IR Spectra of hydrogen fluoride (HF, top) and deuterium fluoride (DF, bottom)
IR Spectrum of HF and DF

  • (a) Compute the reduced mass:

    \[ \mu = \frac{m_H \cdot m_F}{m_H + m_F} \]

    Use SI units.
    Answer:

  • (b) Using \(\bar{\nu} = 3960 \ \text{cm}^{-1}\) and the reduced mass calculated in part (a), compute the force constant \(k\) for the HF bond. Express your answer in SI units.
    Answer:

  • (c) The reduced mass of deuterium fluoride (DF) can be calculated using:

    \[ \mu = \frac{m_D \cdot m_F}{m_D + m_F} \]

    Can the change in reduced mass alone account for the fundamental absorption of DF occurring at
    \(2900 \ \text{cm}^{-1}\) as compared to \(3960 \ \text{cm}^{-1}\) for HF?
    Explain your reasoning and provide any calculations to support your answer.
    Answer:

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