The characteristic energy of the N2 molecule is 2.48 times 10 to the negative fourth electron volts. Determine the separation distance between the nitrogen atoms.
We can call the 2.48 times 10 to the negative fourth electron volts, the characteristic energy, 𝐸. We want to solve for the separation distance between the two nitrogen atoms, a distance we’ll call 𝑑. If we draw a sketch of the N2 molecule, we can represent the molecule as the two nitrogen atoms separated by some distance we’re calling 𝑑. The energy that we’re given in the statement is a rotational energy for this molecule. If the molecule is modelled as a solid barbell, rotating about a central axis drawn in here, then that rotational energy 𝐸 is given by ℎ squared over eight 𝜋 squared times 𝐼, the moment of inertia.
If we solve this equation for 𝐼, we see it’s ℎ squared over eight 𝜋 squared times 𝐸, where Planck’s constant ℎ, we treat as exactly 6.626 times 10 to the negative 34th joule seconds. So we found the moment of inertia of the molecule in terms of its rotational energy. But we can also solve for it by considering the molecule as two masses separated from an axis of rotation. We can consider each of the two atoms to be a point mass rotating about an axis of rotation a distance we can call 𝑟 away.
To solve for the moment of inertia of these point masses, we can look the mathematical form of that relationship up in a table. For a point mass, the moment of inertia equals 𝑚 times 𝑟 squared. In our case, since we have two point masses, we can multiply this equation by two. In which case, we’ll have a second equation for the moment of inertia of our system. Setting these two equations equal to one another, we see that ℎ squared over eight 𝜋 squared times 𝐸 equals two 𝑚𝑟 squared. In this equation, we have a radius 𝑟. But we’re looking to solve for a distance 𝑑.
Looking back at our diagram now, we can see that 𝑑, the distance from one nitrogen atom to the other, is simply twice the radius 𝑟. Knowing that 𝑑 equals two 𝑟, we can make that substitution in for 𝑟 in our equation. With that done, we’re now ready to rearrange and solve for the distance 𝑑. When we make that rearrangement and simplify, we find that 𝑑 is equal to ℎ over two 𝜋 times the square root of 𝑚 times 𝐸. We know all of these variables except for 𝑚. Recall that 𝑚 came from the moment of inertia calculation of our nitrogen atoms treating them as point masses.
So 𝑚 is simply the mass of one nitrogen atom. Nitrogen typically has 14 nucleons, seven protons and seven neutrons. If we define one atomic mass unit, one 𝑎𝑚𝑢, as 1.66 times 10 to the negative 27th kilograms, then the mass of one nitrogen atom 𝑚 would be equal to 14 of those atomic mass units, again because nitrogen has 14 nucleons. Taking this fact, we can write: 𝑚 equals 14 times 1.66 times 10 to the negative 27th kilograms.
Now we’re ready to plug in and solve for 𝑑. When we do plug in for these values, we include a conversion factor between energy units of joules and electron volts, for the purpose of making units consistent across this expression. When we enter these values on our calculator, we find that 𝑑, the distance between the two nitrogen atoms, to three significant figures, is 0.110 nanometers. That’s The separation distance between these two atoms.