# 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](HF_DF_Spectra.png)

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- **(a)** Compute the reduced mass:

  $$
  \mu = \frac{m_H \cdot m_F}{m_H + m_F}
  $$

  Use SI units.  
  *Answer:*

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- **(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:*

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- **(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:*