The reduced mass (\(\mu\)) is given by:

\[
\mu = \frac{m_{\text{H}} \cdot m_{\text{Cl}}}{m_{\text{H}} + m_{\text{Cl}}}
\]

Let's break down the calculation:

- The mass of hydrogen (H) is \(1.008 \, \text{u}\), which is \(1.008 \times 1.66054 \times 10^{-27} \, \text{kg}\).
- The mass of chlorine (Cl) is \(35.45 \, \text{u}\), which is \(35.45 \times 1.66054 \times 10^{-27} \, \text{kg}\).

Substituting these into the formula:

\[
\mu = \frac{(1.008 \times 1.66054 \times 10^{-27}) \times (35.45 \times 1.66054 \times 10^{-27})}{(1.008 \times 1.66054 \times 10^{-27}) + (35.45 \times 1.66054 \times 10^{-27})}
\]

Let's calculate this value.The reduced mass (\(\mu\)) of the HCl molecule is approximately \(1.63 \times 10^{-27} \, \text{kg}\).

Would you like more details or have additional questions?

Here are some related questions:
1. How is the reduced mass of a molecule used in quantum mechanics?
2. What effect does a change in atomic mass have on the reduced mass?
3. How does reduced mass impact the infrared absorption of molecules?
4. Can you determine the reduced mass for polyatomic molecules?
5. What role does reduced mass play in molecular spectroscopy?

**Tip:** The reduced mass is always smaller than the mass of either of the two atoms involved, making it crucial for accurately calculating molecular vibrations.

Learn how to calculate the reduced mass (\( \mu \)) of the HCl molecule using the formula \( \mu = \frac{m_{\text{H}} \cdot m_{\text{Cl}}}{m_{\text{H}} + m_{\text{Cl}}} \). This problem involves determining the effective mass of a molecule in quantum mechanics, specifically focusing on the hydrogen chloride (HCl) molecule. Understand the significance of reduced mass in molecular vibrations and its application in quantum mechanics.

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