Question

Page lo Show that at high temperatures, q for a simple harmonic oscillator (Eq. 5-6) can be expanded in the form hv kT 24 For what value of kT/hv is the approximation q kT/hv accurate to 0.01%? Show that Δε/kT av/ difference between successive energy levels. (Page 116.) eT for a harmonie oscillator, where Ae is the energy Roann energy distribution from Ea. (4-11)

Answer 1

Equation (s - 6) q = e^-theta/2T/1 - e^-theta/T = e^h nu/2kT/e^h nu/kT - 1 times e^-h nu/kT/e^-h nu/kT = e^-h nu/2kT/1 - e^-h nu/kT let x = h nu/kT q = e^-x/2/1 - e^-x = (1 - x/2 + x^2/2 times 4 +---)/[1 - {1 - x + x^2/2 - x^3/6 + ---}] = (1 - x/2 + x^2/8 + ---)/1 - 1 + x - x^2/2 + x^3/6 + --- = (1 - x/2 + x^2/8 + ---)/x (1 - x/2 + x^2/6 + ---) = 1/x [1 - x/2 + x^2/8 + ---][1 - (x/2 - x^2/6)]^-1 = 1/x [1 - x/2 + x^2/8 + ---][1 + (x/2 - x^2/6) + (x/2 - x^2/6)^2 + ---] = 1/x [1 - x/2 + x^2/8 + ---][1 + x/2 - x^2/6 + x^2/4 + ----] \r\n q = 1/x [1 - x/2 + x^2/8 + ---][1 + x/2 + x^2/12 + ----] = 1/x [1 + x/2 + x^2/12 - x/2 - x^2/4 + x^2/8 + ---] = 1/x [1