Question

A hypothetical diatomic molecule of oxygen \((mass = 2.656 \times 10^{-26} kg)\) and hydrogen \((mass = 1.67 \times 10^{-27} kg)\) emits a photon of wavelength 2.39 \(\mu\)m when it makes a transition from one vibrational state to the next lower state. If we model this molecule as two point masses at opposite ends of a massless spring, (a) what is the force constant of this spring, and (b) how many vibrations per second is the molecule making?

Short answer

(a) The force constant is approximately 561 N/m. (b) The frequency of vibration is about 1.26 × 10^14 Hz.

Step-by-step solution

Calculate the reduced mass

The reduced mass \( \mu \) of a diatomic molecule with masses \( m_1 \) and \( m_2 \) is given by the formula: \[ \mu = \frac{m_1 m_2}{m_1 + m_2} \] where \( m_1 = 2.656 \times 10^{-26} \text{ kg} \) and \( m_2 = 1.67 \times 10^{-27} \text{ kg} \). Substitute these values to find \( \mu \): \[ \mu = \frac{(2.656 \times 10^{-26})(1.67 \times 10^{-27})}{2.656 \times 10^{-26} + 1.67 \times 10^{-27}} \approx 1.60 \times 10^{-27} \text{ kg} \]

Find the energy of the photon

The energy \( E \) of a photon is given by the equation: \[ E = \frac{hc}{\lambda} \] where \( h = 6.626 \times 10^{-34} \text{ J} \cdot \text{s} \) (Planck's constant), \( c = 3.00 \times 10^{8} \text{ m/s} \) (speed of light), and \( \lambda = 2.39 \times 10^{-6} \text{ m} \). Substitute these values into the equation:\[ E = \frac{6.626 \times 10^{-34} \cdot 3.00 \times 10^{8}}{2.39 \times 10^{-6}} \approx 8.32 \times 10^{-20} \text{ J} \]

Calculate the frequency of vibration

The relationship between energy \( E \) and frequency \( u \) is given by \( E = h u \). Rearrange the formula to find the frequency \( u \):\[ u = \frac{E}{h} = \frac{8.32 \times 10^{-20}}{6.626 \times 10^{-34}} \approx 1.26 \times 10^{14} \text{ Hz} \]

Determine the force constant

The formula to find the force constant \( k \) for a harmonic oscillator is \( k = 4\pi^2\muu^2 \). Use \( \mu = 1.60 \times 10^{-27} \text{ kg} \) and \( u = 1.26 \times 10^{14} \text{ Hz} \):\[ k = 4\pi^2 \times 1.60 \times 10^{-27} \times (1.26 \times 10^{14})^2 \approx 561 \text{ N/m} \]

Key concepts

Reduced Mass of a Diatomic Molecule

The concept of reduced mass is a key element in studying the dynamics of a diatomic molecule. In a simple model where two atoms are bonded together, we can think about the mass of one atom shifting some of its effects onto the other atom. This mix results in the so-called "reduced mass," a handy way to simplify complex problems where both atoms influence each other's motion.

The formula to calculate the reduced mass \( \mu \) is:

• \( \mu = \frac{m_1 m_2}{m_1 + m_2} \)

Here, \(m_1\) and \(m_2\) are the masses of the two atoms in the molecule. For the given oxygen-hydrogen molecule, you plug in the given masses and perform a straightforward calculation yielding a reduced mass of approximately \(1.60 \times 10^{-27} \text{ kg}\). This reduced mass is then used in further calculations, acting as a simplified representation of how the two atoms together influence the system's overall motion.

Photon Energy

Photon energy is the energy held by a single photon, and it plays a critical role when a molecule transitions between different states. In this context, a photon is emitted when the molecule jumps from a higher to a lower vibrational state.

To find the energy \( E \) of a photon, use the formula:

• \( E = \frac{hc}{\lambda} \)

Where \( h \) is Planck's constant \((6.626 \times 10^{-34} \text{ J} \cdot \text{s})\), \( c \) is the speed of light \((3.00 \times 10^8 \, \text{m/s})\), and \( \lambda \) is the wavelength of the emitted photon. In our case, the equation tells us that the energy of the photon is approximately \( 8.32 \times 10^{-20} \text{ J} \). This energy relates directly to the frequency and is useful for understanding the transitions occurring within the molecule.

Harmonic Oscillator Model

A harmonic oscillator model is often used to simplify the behavior of a vibrating diatomic molecule. This model views the two atoms as masses attached to a spring that can stretch and compress. This analogy helps to understand the vibrational states the molecule experiences.

In this model, the frequency of vibrations \( u \) can be directly connected to energy using the fundamental equation:

• \( E = h u \)

Where \( E \) represents the energy of the vibration, and \( h \) represents Planck's constant. By rearranging the formula, you can solve for frequency \( u \), giving you an insight into how many vibrations occur per second. For the oxygen-hydrogen molecule, the frequency is approximately \(1.26 \times 10^{14} \text{ Hz}\), meaning the molecule undergoes this many oscillations per second.

Force Constant of a Vibrational Mode

The force constant \( k \) is a measure of the stiffness of the bond between the atoms in a diatomic molecule. It relates to the energy required to stretch or compress the bond, similar to how a spring behaves in Hooke's law.

To calculate the force constant using the harmonic oscillator model, use the equation:

• \( k = 4\pi^2 \mu u^2 \)

Here, \( \pi \) is the mathematical constant representing the ratio of a circle's circumference to its diameter, \( \mu \) is the reduced mass, and \( u \) is the frequency of vibrations. For our example, substituting the known values yields a force constant of about \( 561 \text{ N/m} \). This constant reflects the "springiness" of the bond in the molecular system, highlighting how tightly the atoms are bonded together.