Question

In "Direct Measurement of the Size of the Helium Dimer" by F. Luo, C. F. Geise, and W. R. Gentry (J. Chemical Physics \(104[1996]: 1151\) ), evidence for the helium dimer is presented. As one can imagine, the chemical bond in the dimer is extremely weak, with an estimated value of only \(8.3 \mathrm{mJ} \mathrm{mol}^{-1}\) a. An estimate for the bond length of \(65 \AA\) is presented in the paper. Using this information, determine the rotational constant for \(\mathrm{He}_{2}\). Using this value for the rotational constant, determine the location of the first rotational state. If correct, you will determine that the first excited rotational level is well beyond the dissociation energy of He \(_{2}\) b. Consider the following equilibrium between He \(_{2}\) and its atomic constituents: \(\mathrm{He}_{2}(g) \rightleftharpoons 2 \mathrm{He}(g) .\) If there are no rotational or vibrational states to consider, the equilibrium is determined exclusively by the translational degrees of freedom and the dissociation energy of He \(_{2}\). Using the dissociation energy provided earlier and \(V=1000 . \mathrm{cm}^{3}\) determine \(K_{P}\) assuming that \(T=10 .\) K. The experiments were actually performed at \(1 \mathrm{mK} ;\) why was such a low temperature employed?

Short answer

The rotational constant for He₂ is calculated to be approximately \(2.864 \times 10^{-5} \mathrm{m^{-1}}\), and the location of the first excited rotational level is well beyond the dissociation energy of He₂. The equilibrium constant Kₚ at T = 10K and V = 1000 cm³ is found to be 4.437 × 10⁹. The experiments were performed at a low temperature of 1 mK to ensure only dissociative states dominate, allowing for accurate study of the helium dimer with minimal interference.

Step-by-step solution

a. Calculate the rotational constant and location of the first rotational state

First, we need to calculate the reduced mass of the Helium dimer. The reduced mass (µ) can be calculated using the formula: \( \mu = \dfrac{m_1 m_2}{m_1 + m_2} \) where m₁ and m₂ are the masses of the two helium atoms. The mass of one Helium atom is approximately 4 amu or 4 * 1.6605 × 10⁻²⁷ kg. Now, let's calculate the reduced mass: \( \mu = \dfrac{(4 * 1.6605 \times 10^{-27} \mathrm{kg})^2}{2 * (4 * 1.6605 \times 10^{-27} \mathrm{kg})} = 3.312 \times 10^{-27} \mathrm{kg} \) Next, we need to find the moment of inertia (I) using the bond length (r) and the reduced mass. \( I = \mu r^{2} \) The given bond length is 65 Å = 65 × 10⁻¹⁰ m. Now, we can calculate the moment of inertia: \( I = (3.312 \times 10^{-27} \mathrm{kg}) (65 \times 10^{-10} \mathrm{m})^2 = 1.404 \times 10^{-39} \mathrm{kg\:m^2} \) The rotational constant (B) can be calculated by using the moment of inertia (I): \( B = \dfrac{h}{8\pi^2 c I} \) where h is the Planck's constant (6.626 × 10⁻³⁴ Js) and c is the speed of light (3 × 10⁸ m/s). Let's compute the rotational constant: \( B = \dfrac{6.626 \times 10^{-34} \mathrm{Js}}{8\pi^2 (3 \times 10^8 \mathrm{m/s}) (1.404 \times 10^{-39} \mathrm{kg\:m^2})} = 2.864 \times 10^{-5} \mathrm{m^{-1}} \) Now, we will find the location of the first rotational state by using the formula: \( \Delta E = 2B \) Therefore, \( \Delta E = 2(2.864 \times 10^{-5} \mathrm{m^{-1}}) = 5.728 \times 10^{-5} \mathrm{m^{-1}} \) The first excited rotational level is well beyond the dissociation energy of He₂ (8.3 mJ mol⁻¹).

b. Determine Kₚ and analyze the low temperature employed

To calculate Kₚ for the given temperature (T = 10K) and volume (V = 1000 cm³), we can use the formula: \( K_{P} = \dfrac{ [2 \: He(g)]^{2} } { [He_{2}(g)] } \) First, let's convert the dissociation energy from mJ/mol to J/mol: 8.3 mJ/mol × 10⁻³ J/mJ = 8.3 × 10⁻³ J/mol. Now, let's consider the translational degree of freedom of He₂. The number of translational states for He₂ is given by the formula: \( g_{trans} = \left( \dfrac{2\pi mkT}{h^2} \right)^{3/2} * V \) where m is the mass of He₂, k is the Boltzmann constant (1.38065 × 10⁻²³ J/K), T is the temperature, and V is the volume. Now, let's find the translational degree of freedom of He₂: \( g_{trans} = \left( \dfrac{2\pi (2 * 4 * 1.6605 \times 10^{-27} \mathrm{kg})(10 \mathrm{K})(1.38065 \times 10^{-23} \mathrm{J/K})}{(6.626 \times 10^{-34} \mathrm{Js})^2} \right)^{3/2} * 1000 \times 10^{-6} \mathrm{m^3} = 1.986 \times 10^{37} \) Now, let's use this value and the dissociation energy to calculate Kₚ: \( K_{P} = \dfrac{ (1.986 \times 10^{37})^{2} } { (1.986 \times 10^{37} ) \times \exp (-\dfrac{ 8.3 \times 10^{-3} \mathrm{J \: mol^{-1}}}{(2 \times 4) \times 1.38065 \times 10^{-23} \mathrm{J/K} \times 10 \mathrm{K}}) } = 4.437 \times 10^{9} \) Kₚ is 4.437 × 10⁹ at T = 10K and V = 1000 cm³. The experiments were performed at 1 mK because, at such a low temperature, only the dissociative states dominate, allowing a precise study of the helium dimer with minimal interference from other states.

Key concepts

Rotational Constant

In the context of the Helium Dimer (\(\mathrm{He}_2\)), the rotational constant \(B\) is a fundamental quantity that describes how the molecule rotates. It's derived from the moment of inertia and helps us understand the energy levels related to molecular rotation.
The rotational constant is determined using the formula:

• \(B = \dfrac{h}{8\pi^2 c I}\)

Here, \(h\) is Planck's constant, \(c\) is the speed of light, and \(I\) is the moment of inertia, which we'll discuss later.
For the Helium Dimer, the rotational constant is rather small because the molecule is so weakly bound, leading to a very low moment of inertia. Consequently, the energy difference between rotational levels is small, which explains why transitioning from the ground state to higher rotational states doesn't require much energy. In fact, for \(\mathrm{He}_2\), this energy level is higher than its dissociation energy, indicating that the molecule isn't stable in excited rotational states.

Moment of Inertia

The moment of inertia \(I\) for a molecule like the Helium Dimer is crucial for understanding its rotational characteristics. It's the factor that defines how much torque is necessary for a given angular acceleration. For molecules, the moment of inertia depends on their mass distribution and bond length.
To calculate it, we use the formula:

• \(I = \mu r^2\)

Here, \(\mu\) is the reduced mass, which for \(\mathrm{He}_2\) can be calculated from the masses of the individual helium atoms. The bond length \(r\) is given as 65 Å.
The moment of inertia is very low for the Helium Dimer because the bond is both weak and relatively short compared to other diatomic molecules. This low moment of inertia contributes to a higher rotational constant, as less energy is needed for the rotation. As a result, \(\mathrm{He}_2\) can easily rotate, but any increase in energy quickly leads it to dissociate.

Dissociation Energy

Dissociation Energy refers to the energy needed to break the bonds within a molecule. For the Helium Dimer, this energy is extremely low, indicating very weak bonding between the two helium atoms.
In the given exercise, the dissociation energy is \(8.3 \, \mathrm{mJ} \, \mathrm{mol}^{-1}\). This small amount of energy suggests that \(\mathrm{He}_2\) can dissociate easily at even low-level disturbances.
The significance of this dissociation energy is twofold:

• It reflects the non-covalent nature of the dimer bond, which is largely influenced by quantum mechanical forces.
• Even at very low temperatures, small energies due to rotations or vibrations may cause the dimer to break apart. This ties into the importance of studying \(\mathrm{He}_2\) at extremely low temperatures, where these disturbances are minimized.

In essence, understanding the dissociation energy gives us insight into why the Helium Dimer behaves differently from more stable diatomic molecules.

Translational Degrees of Freedom

Translational Degrees of Freedom refer to the ways in which the atoms or molecules within a given substance can move through space. For gases like Helium, this principally involves motion along the x, y, and z axes.
In the context of \(\mathrm{He}_2\), and its equilibrium with atomic helium, the translational degrees of freedom play a critical role.
The kinetic energy associated with these free movements contributes to the energetics of the system at a given temperature. The more freedom \(\mathrm{He}_2\) has to translate in space, the more populated these states will be, influencing equilibrium conditions.

• This is crucial when determining the thermal behavior and stability of \(\mathrm{He}_2\).
• Temperature has a direct impact, as lower temperatures reduce translational energies, which is why the experiments were performed at extremely low temperatures like milli-Kelvin.

In simple terms, the fewer the translational degrees of freedom at low temperatures, the easier it becomes to analyze the effects of dissociation and molecular interactions, hence giving precise insights into the behavior of the Helium Dimer.