How do I do this?

The vibrational properties of a diatomic molecule can often be described by Mie's pair potentiall}: winnerer], (4) where U (r) is the potential energy between the two atoms, r is the distance between the two atoms, 0,0 are positive constants, and A > [i is a constant. The case with, A : 12 is the LennardJones potential and was mueral in Lecture #2. For the purposes of this problem, assume there is a mass 'm ' that represents the dynamical mass of the molecule TL TR: (a) Find a combination of C, m, a that has the units offrequency. (10 pts} (b) Derive the ratio of the equilibrium positions and of the frequencies of small oscillations for A \"- 10, 14. Show that the units for the equilibrium position and for the frequency of the oscillations are consistent. (4 0 pts) The parameter A is not 12 in most substances. It is determined by measuring some of the properties of the substance. For simplicity, let's suppose that we can measure the separation of the two atoms and some characteristic frequencies of the system, usually called resonant frequencies, see the slides from lecture #2 for more info about the frequency range and meaning. We want to know which of these two measure- ments is more sensitive to a change of A because it could help decide which kind of experiment we will conduct in order to measure A. Let's begin . . . (1) A is dimensionless because it's an exponent. Otherwise, changing the units would change its value and the functional dependence of U(r). A molecule doesn't depend on the units we use to describe it. What are the units of C,o? (2) Find the particular combination of C, o, m that has units of the inverse of time. That is, frequency. Sort of method that Cliff showed with the 'bomb explosion in Ph la. Or, (3) Find the exponents of Coolmy such that the result has the units of 1/s. Sub- stitute the SI units of C, o, m and solve the equations. Sort of the method I did in Ph la with the Planck units. Either way, you have to get the same combination. This result provides a first estimation of the characteristic frequency of the sys- tem. However, since A is dimensionless, its role in the a frequency cannot be estimated by this method. We usually say that there's an unknown dimension- less factor (some say of ~ order 1, i. e. between 0.1 and 10, but it could be any number). The role of A can be found out by solving the small oscillations of the system. It matters because even if it is between 0.1 or 10, the actual frequency could take us from microwaves to UV! So we would rather find it out before placing an order for the detectors! In the following you can take two possible paths: either you solve the question for A = 10 and > = 14, or you find the general expression for any A. Choose the one you prefer! I'll be giving some hints for the arbitrary > case. The reason is that the general case can be used to fit experimental data. Although, one could use numerical methods all the way from U(r), having an analytic expression of the quantity to be measured, when available, can speed up the fitting process and the estimation of the instrumental error considerably, requiring many less computing resources than MCMC simulations, or similar. (4) I think U(r) = C()/r -6/76), as I did in the lecture notes, it's a better starting point to compute derivatives than equation .(5) Find the equilibrium position, To as a function of A and o. That is, find F(X) where ro = F(Do. Check that F(X) is dimensionless. Hint: the solution has (x/6) and (X - 6) combined in some way. Do not worry about the second derivative. Just the first derivative: there's only one solution and we know that the shape of U(r) for > > 6 must have a minimum because the repulsion wins the attraction as ro -> 0. We'll look at the second derivative later.(6) If you derive the general expression for any ), verify that you get the same result as in Lecture #2 for > = 12. (7) Find the ratio F(14)/F(10). Hint: the ratio is close to just a 2% difference!, with ro(14) > 6, we have A -6 > 0, and U"(ro) > 0. Just what we needed for ro to be an equilibrium point. So far, so good! (10) Finally, apply equation , or simply its ratio for two values of A to get the result. The result will not depend on C, o, m because it's a ratio. Nor on the famous 2n! Hint: the result shows that the two frequencies differ by about 54%! With w(A = 14) > w(X = 10). After all, we have not derived the actual value of the frequencies in Hz. If we had to, we would need the values of C, o, m. These parameters can be estimated from other properties of the substance, or being fitted simultaneously if we have enough independent measurements: Mie's pair potential can predict quite accu- rately surface tension, vapor pressure and other thermodynamic properties that can be measured in the lab. That's all! Quite a bit indeed.3Mie's pair potential can also be applied to symmetric Im'ilecules= not only diatomic molecules. 4It is the reduced mass of the system. It is a common parameter when solving twobody systems. 5It's; possible to prove that a 6 if you like solving 00" and 190 limits with 111 and L'Hpital. A pretty Small range to distinguish r\" from different values (If A > 6. 1. If a system is in equilibrium at a certain location 1' : To, the frequency of the small oscillations around that point is given by: 1 Mg 05c : ' 1 f 2W m l l where U' > 0 is the value of the second derivative of the potential energy: U (r), evaluated at the equilibrium point r : m. and m is the dynamical mass of the system.