Question

The spring (force) constant of HF is 970 N/m (1 N = 1 kg m s² ).

(A) Calculate the fundamental frequency (expressed in units of cm-1 ) and the zero point energy (in energy units, J).

(B) Earlier in the term we discussed the relationship between the energy and the position and momentum uncertainties. For the harmonic oscillator case, it would be

E ≥ ((Δp) ² / 2µ) + (1 / 2) µω² (Δx) ²

(Equation 1)

The ground state of the harmonic oscillator is an interesting case of the Heisenberg uncertainty principle, since it obeys the equality:

ΔxΔp_x = h_bar / 2

(Equation 2)

Therefore, for the ground state, the equality holds in Eq. 1 above. Calculate the uncertainty in position for the ground state given that the energy of the ground state is E₀ = (1 / 2) hν . You should obtain a value with units of length (i.e. meters, Å, etc.). Compare this width to the length of the HF bond of 0.91 Å and comment on the significance of the quantum delocalization. Hint 1: You will end up with a quartic equation, which will have two identical positive roots, and two other identical negative roots. The negative roots are meaningless since the uncertainty must be positive. It's easier to get the answer another way. Simplify your math by writing the equation in terms of χ = (Δx) ² , which makes it quadratic, and you'll just find the positive roots. You can do it! You learned all that math, so now use it! Hint 2: you will need to use the correct units for the frequency (remember that ω = 2πν ).

(C) Comment on how the mass influences the position uncertainty. How would the situation be different if you were considering DF instead of HF (D is deuterium)? Make sure to consider all of the mass dependencies.

Answer 1