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The force constant of \(^{79} \mathrm{Br}^{79} \mathrm{Br}\) is \(240 \mathrm{N} \cdot \mathrm{m}^{-1}\). Calculate the fundamental vibrational frequency and the zero-point energy of \(^{79} \mathrm{Br}^{79} \mathrm{Br}\)

Short Answer

Expert verified
Frequency is approximately \(9.6 \times 10^{12}\) Hz; zero-point energy is about \(3.18 \times 10^{-21}\) J.

Step by step solution

01

Determine Reduced Mass

The reduced mass \( \mu \) is calculated using the formula:\[ \mu = \frac{m_1 \cdot m_2}{m_1 + m_2} \]where \( m_1 \) and \( m_2 \) are the masses of the bromine atoms. Since it's \( ^{79} \mathrm{Br}^{79} \mathrm{Br} \), both atoms have the same mass. The molar mass of \( ^{79} \mathrm{Br} \) is approximately 79 amu, which is about \( 79 \times 1.66054 \times 10^{-27} \) kg. Therefore, the reduced mass is:\[ \mu = \frac{79 \times 1.66054 \times 10^{-27} \times 79 \times 1.66054 \times 10^{-27}}{79 \times 1.66054 \times 10^{-27} + 79 \times 1.66054 \times 10^{-27}} = \frac{79 \times 1.66054 \times 10^{-27}}{2} \approx 65.8 \times 10^{-27} \mathrm{kg} \]
02

Calculate Vibrational Frequency

The vibrational frequency \( u \) is determined using the formula:\[ u = \frac{1}{2\pi} \sqrt{\frac{k}{\mu}} \]where \( k = 240 \mathrm{N}\!\cdot\!\mathrm{m}^{-1} \) is the force constant. Plugging in the values:\[ u = \frac{1}{2\pi} \sqrt{\frac{240}{65.8 \times 10^{-27}}} \]Calculating this gives:\[ u \approx \frac{1}{2\pi} \sqrt{3.646 \times 10^{28}} \approx \frac{1}{2\pi} \times 6.035 \times 10^{14} \approx 9.6 \times 10^{12} \mathrm{Hz} \]
03

Calculate Zero-Point Energy

The zero-point energy (ZPE) is given by:\[ E_{\text{ZPE}} = \frac{hu}{2} \]where \( h = 6.626 \times 10^{-34} \mathrm{J\cdot s} \) is Planck's constant. Thus:\[ E_{\text{ZPE}} = \frac{6.626 \times 10^{-34} \times 9.6 \times 10^{12}}{2} \approx 3.18 \times 10^{-21} \mathrm{J} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reduced Mass
To understand vibrational frequency, the concept of reduced mass is crucial. The reduced mass (\( \mu \)) is calculated to simplify the analysis of two-body systems. It allows us to look at two interacting masses as a single effective mass. In the formula \( \mu = \frac{m_1 \cdot m_2}{m_1 + m_2} \), \(m_1\) and \(m_2\) are the masses of the two objects or atoms involved in the system.

For diatomic molecules like \(^{79}\mathrm{Br}^{79}\mathrm{Br}\), both atoms are identical with the same mass. This makes the calculations easier and the reduced mass value approximate to half of the mass of one atom since both are nearly equal:
  • Molar mass of bromine (\(^{79}\mathrm{Br}\)) is about 79 amu.
  • Conversion to kilograms results in \(79 \times 1.66054 \times 10^{-27}\) kg.
  • Reduced mass thus simplifies to \(\frac{2}{2} \approx 65.8 \times 10^{-27} \text{ kg}\).
The reduced mass concept is key in deriving properties like vibrational frequency in molecules, which heavily influences how the atoms in a molecule vibrate.
Zero-Point Energy
Even at their lowest energy state, or ground state, vibrations are intrinsic in molecules. Zero-point energy (ZPE) is the minimum possible energy that a quantum mechanical system may possess, distinct from classical intuition where the lowest energy would be zero.

The formula for zero-point energy is:\[ E_{\text{ZPE}} = \frac{hu}{2} \]where \( h \) is Planck's constant and \( u \) is the vibrational frequency. This formula shows that even at absolute zero (0 Kelvin), a molecule retains some energy, which means:
  • Molecules are never completely still.
  • Energy transitions cannot occur without this fundamental level.
It's essential for understanding molecular behavior because it affects how molecules interact, bond, and react in various states. The presence of zero-point energy implies that chemical reactions have a base energy level that must have overcome for the reaction to proceed.
Planck's Constant
Planck's constant (\( h \)) plays a critical role in quantizing energy levels in different physics and chemistry phenomena. The constant is denoted by \( h = 6.626 \times 10^{-34} \text{ J}\cdot\text{s} \).

Planck's constant essentially bridges wave-like and particle-like properties, enabling us to calculate:
  • Energy of photons and other micro-particles, under the formula \( E = hu \).
  • Discreteness in energy levels in quantum systems highlighting why energy in these systems isn’t continuous but comes in quanta.
In vibrational and quantum mechanics calculations like those involving zero-point energy, Planck's constant is critical. It reminds us that at a microscopic scale, energy transactions aren't fluid but occur in discreet, quantized steps. This foundational principle is what allows us to calculate zero-point energies and understand the nature of light, energy, and material properties at an atomic level.

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Most popular questions from this chapter

Prove that the derivative of an even (odd) function is odd (even).

This problem is similar to Problem 3-35. Show that the harmonic-oscillator wave functions are alternately even and odd functions of \(x\) because the Hamiltonian operator obeys the relation \(\hat{H}(x)=\hat{H}(-x)\). Define a reflection operator \(\hat{R}\) by $$ \hat{R} u(x)=u(-x) $$ Show that \(\hat{R}\) is linear and that it commutes with \(\hat{H}\). Show also that the eigenvalues of \(\hat{R}\) are \(\pm 1\). What are its eigenfunctions? Show that the harmonic-oscillator wave functions are eigenfunctions of \(\hat{R}\). Note that they are eigenfunctions of both \(\hat{H}\) and \(\hat{R}\). What does this observation say about \(\hat{H}\) and \(\hat{R}\) ?

Show that the moment of inertia for a rigid rotator can be written as \(I=\mu r^{2},\) where \(r=r_{1}+r_{2}(\text { the fixed separation of the two masses) and } \mu\) is the reduced mass.

The general solution for the classical harmonic oscillator is \(x(t)=C \sin (\omega t+\phi) .\) Show that the displacement oscillates between \(+C\) and \(-C\) with a frequency \(\omega\) radian \(\cdot \mathrm{s}^{-1}\) or \(v=\omega / 2 \pi\) cycle \(\cdot \mathrm{s}^{-1} .\) What is the period of the oscillations; that is, how long does it take to undergo one cycle?

It turns out that the solution of the Schrödinger equation for the Morse potential can be expressed as \\[ \tilde{E}_{v}=\tilde{v}\left(v+\frac{1}{2}\right)-\tilde{v} \tilde{x}\left(v+\frac{1}{2}\right)^{2} \\] where \\[ \tilde{x}=\frac{h c \tilde{v}}{4 D} \\] Given that \(\tilde{v}=2886 \mathrm{cm}^{-1}\) and \(D=440.2 \mathrm{kJ} \cdot \mathrm{mol}^{-1}\) for \(\mathrm{H}^{35} \mathrm{Cl}\), calculate \(\tilde{x}\) and \(\tilde{v} \tilde{x}\)

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