The characteristic energy of rotation for HCl is 1.33 times 10 to the negative third electron volts, assuming a value of 35.4 𝑢 for the atomic mass of chlorine and 1.0 𝑢 for the atomic mass of hydrogen. Determine the reduced mass 𝜇 for the HCl molecule. Find the separation distance between the H and Cl atoms.
We can call this separation distance 𝑑 and start off with a diagram of this rotating molecule. We have an HCl molecule, which is an atom of hydrogen joined to an atom of chlorine. We’re told that the mass of the chlorine atom is 35.4 𝑢, where 𝑢 is an atomic mass unit, the average mass of a nuclear element, a proton or neutron. While the hydrogen atom has a mass of just 1.0 𝑢. This lopsided molecule rotates about an axis with an energy of 1.33 times 10 to the negative third electron volts.
As a first step, we want to solve for what’s called the reduced mass of this molecule. To calculate the reduced mass, what that means is taking a two-body rotational system, where the masses that we’ve called 𝑚 one and 𝑚 two might be unequal, and simplifying it into a representative system that just has one mass, the reduced mass. We’ve called it 𝑚 sub 𝑟.
These two systems rotate in equivalent ways. But we use reduced mass to simplify our system and our calculations. The reduced mass that’s created from a system with two bodies is equal to the product of the mass of those two bodies divided by their sum. In our case then, the reduced mass 𝜇 is equal to the mass of the hydrogen atom times the mass of the chlorine atom all divided by the sum of the masses of these two atoms.
Plugging in for these masses, leaving them in terms of atomic mass units, we find a result of 0.97 𝑢. That’s the reduced mass of our hydrogen chlorine molecule which interestingly is less than the mass of either atom. Knowing this reduced mass, we can now redraw our two-body system as a one-body system with that mass 𝜇. We now have a one-body system of reduced mass 𝜇 rotating about the axis a distance 𝑑 away. It’s that distance 𝑑 we want to solve for next. And to do it, we’ll use the characteristic rotational energy 𝐸 sub 𝑟. That rotational energy is equal to Planck’s constant squared divided by eight 𝜋 squared times the moment of inertia of our system.
In our case, our system can be modelled as a point mass. And when we look up the moment of inertia of a point mass, we see it’s equal to the mass of the system multiplied by the distance between the mass and its rotation axis squared. Bringing these two equations together, we can say that the rotational energy of our particular molecule is equal to Planck’s constant squared over eight 𝜋 squared times 𝜇 times 𝑑 squared.
Rearranging to solve for 𝑑, we see it’s equal to Planck’s constant divided by 𝜋 times the square root of eight times 𝐸 sub 𝑟 times 𝜇. We’ll assume that Planck’s constant is exactly 6.626 times 10 to the negative 34th joule seconds. That means that when we enter our value for 𝐸 sub 𝑟 in our equation, we’ll want to convert its energy units from electron volts to joules.
To make this conversion, we’ll recall that one joule is approximately equal to 6.242 times 10 to the 18th electron volts. We’ll also want to convert the units of 𝜇 which right now have atomic mass units. One atomic mass unit is approximately equal to 1.66 times 10 to the negative 27th kilograms. So we’ll use that conversion when we enter 𝜇.
We’re now ready to plug in for ℎ, 𝐸 sub 𝑟, and 𝜇 — converting the units of 𝐸 sub 𝑟 and 𝜇 — to solve for the distance 𝑑. When we enter all these values into our expression for 𝑑, we see, looking under the square root sign, that our units of electron volts cancel out as due our atomic mass units. Calculating 𝑑, we find that it is 0.13 nanometers. That’s the distance between the hydrogen atom and the chlorine atom in this molecule.