Question

The vibrational frequency for \(D_{2}\) expressed in wave numbers is \(3115 \mathrm{cm}^{-1} .\) What is the force constant associated with the bond? How much would a classical spring with this force constant be elongated if a mass of \(1.50 \mathrm{kg}\) were attached to it? Use the gravitational acceleration on Earth at sea level for this problem.

Short answer

The force constant associated with the bond in \(D_2\) is approximately \(516 \, \text{N/m}\). When a \(1.50 \, \text{kg}\) mass is attached to a classical spring with this force constant, the spring would be elongated by approximately \(0.0286 \, \text{m}\) or \(2.86 \, \text{cm}\).

Step-by-step solution

Write down the given information

The vibrational frequency for \(D_2\) in wavenumbers is \(3115 \,\text{cm}^{-1}\) (Recall that 1 wavenumber = \(1 \,\text{cm}^{-1}\)). The mass to be attached to the spring is \(1.50 \, \text{kg}\).

Calculate the vibrational frequency in Hz

First, we need to convert the vibrational frequency given in wavenumbers to Hz. To do this, we can use the conversion factor that relates the speed of light \(c\) in vacuum to frequency and wavenumber: \(ν = c·\bar{ν}\) Where \(ν\) is the frequency in Hz, \(\bar{ν}\) is the wavenumber in cm^{-1}, and \(c\) is the speed of light in cm/s. Since \(c = 2.998 × 10^{10} \, \text{cm/s}\), we get: \(ν = (2.998 × 10^{10} \, \text{cm/s})× (3115 \, \text{cm}^{-1})\) \(ν = 9.336 × 10^{13} \, \text{Hz}\)

Calculate the force constant

The formula for vibrational frequency derived from Hooke's Law is: \(ν = \frac{1}{2\pi} \sqrt{\frac{k}{μ}}\) Where \(k\) is the force constant, \(μ\) is the reduced mass of the diatomic molecule, and \(ν\) is the frequency in Hz. The reduced mass of \(D_2\) can be calculated using the formula: \(μ = \frac{m_1 × m_2}{m_1 + m_2}\) Since both deuterium atoms are identical, we have \(m_1 = m_2 = m_D\), where \(m_D\) is the mass of a deuterium atom: \(μ = \frac{m_D × m_D}{m_D + m_D} = \frac{m_D}{2}\) Now, we can rearrange the vibrational frequency formula to solve for the force constant: \(k = 4\pi^2·μ·ν^2\) The mass of a deuterium atom is \(m_D = 2 \, \text{amu}\), which is approximately \(3.34 × 10^{-27} \, \text{kg}\). We can substitute this and the frequency into the equation to find the force constant: \(k = 4\pi^2· \frac{3.34 × 10^{-27} \, \text{kg}}{2} · (9.336 × 10^{13} \, \text{Hz})^2\) \(k ≈ 516 \, \text{N/m}\)

Calculate the elongation of the spring

Using Hooke's Law, we can now find the elongation of the spring when a \(1.50 \, \text{kg}\) mass is attached to it. The equation is: \(F = k \Delta L\) Where \(F\) is the force exerted by the mass on the spring, \(k\) is the force constant, and \(\Delta L\) is the elongation of the spring. Since we are considering the gravitational force acting on the mass, we have: \(F = m·g\) Where \(m\) is the mass and \(g\) is the gravitational acceleration on Earth at sea level (\(9.81 \, \text{m/s}^2\)). Now, we can rearrange the equation to find the elongation: \(\Delta L = \frac{F}{k} = \frac{m·g}{k}\) Plugging in the values, we get: \(\Delta L = \frac{(1.50 \, \text{kg} )· (9.81 \, \text{m/s}^2)}{516 \, \text{N/m}}\) \(\Delta L ≈ 0.0286 \, \text{m}\) The elongation of the spring when a \(1.50 \, \text{kg}\) mass is attached to it is approximately 0.0286 meters or 2.86 centimeters.

Key concepts

Force Constant

The force constant, often represented by the symbol 'k', is a fundamental parameter in the study of oscillating systems such as springs and molecular vibrations. It quantifies the stiffness of a spring or the strength of a bond within a diatomic molecule. A higher force constant indicates a stiffer spring or a stronger bond, which would require more force to either compress or extend. This concept is critical when we try to understand how molecular bonds behave under vibrational motion.

In the context of the exercise, calculating the force constant provides us insight into the bond strength of the molecule in question. For a classical spring, this force constant allows us to predict how much a spring would elongate or compress under a certain force.

Hooke's Law

Hooke's Law is a principle of physics that relates the force needed to extend or compress a spring to the distance it is stretched or compressed. It can be stated as: \[ F = -k \times \text{spring elongation} \], where 'F' stands for the force applied to the spring, 'k' is the force constant, and 'spring elongation' is the distance the spring's length has been changed from its resting position.

This principle can also be applied to understand the vibrational frequencies of molecules. In the exercise, Hooke's Law helps us determine the elongation of a classical spring when a known mass is attached. By equating the gravitational force exerted by the mass to the restorative force of the spring, we can deduce the change in length for the spring.

Reduced Mass

Reduced mass is the 'effective' inertial mass that appears in the two-body problem of Newtonian mechanics. It is a key concept when dealing with the vibrational motion of diatomic molecules. The formula to calculate the reduced mass '\(μ\)' of a diatomic molecule, when masses of the two atoms are '\(m_1\)' and '\(m_2\)', is: \[ μ = \frac{m_1 \times m_2}{m_1 + m_2} \].

When considering isotopes like deuterium (\(D_2\)) the atoms have the same mass, which simplifies the calculation. Understanding reduced mass is essential for calculating the vibrational frequencies and interpreting bonding properties within molecules, as seen in the exercise.

Spring Elongation

Spring elongation is the change in length of a spring when a force is applied. It is directly related to Hooke's Law, which describes that the extension of a spring is in direct proportion with the applied force, provided the limit of proportionality is not exceeded. The elongation of the spring, often denoted as '\(\Delta L\)', can be expressed in meters or centimeters. Measuring the elongation of a spring is crucial for applications ranging from simple mechanics to complex molecular dynamics. In the exercise, the elongation was calculated to understand how much a spring with a certain force constant would stretch when subjected to the weight of an attached mass.