Question

An oscillator with frequency f = 2.1×10^(12) Hz (about typical for a greenhouse gas molecule) is in equilibrium with a thermal reservoir at temperature T. The spacing between the energy levels of the oscillator is given by ε = hf, where h = 6.626×10^(-34) J*sec.
(a) For what temperature does P1/P0 = 1/2, where P1 is the probability that the oscillator has E = ε (the first excited state) and P0 is the probability that it has E = 0 (the lowest energy state)?
(b) If T is 10% of the value you calculated in part 1), what is the ratio P1/P0?
(c) At the temperature of part (b), what is the ratio P2/P1, where P2is the probability that the oscillator is in the second excited state (E = 2ε)?

Answer 1

Answer:

a) The temperature is 48.41 K

b) P₁/P₀ = 1

c) PL/P₀ = 1

Explanation:

a) Given:

P₁/P₀ = 1/2

According the expression:

$\frac{P_{1} }{P_{o} } =\frac{e^{-\beta E_{1} } }{e^{-\beta E_{o} } } \\\frac{P_{1} }{P_{o} }=\frac{e^{-\beta E_{1} } }{e^{-\beta *0 } }\\\frac{P_{1} }{P_{o} }=e^{-\beta E} \\\beta =\frac{1}{kt} \\\frac{P_{1} }{P_{o} }=e^-{\frac{\epsilon }{kt} }$

$\epsilon =hf=6.626x10^{-34} *2.1x10^{12} =1.39x10^{-21} J$

k = Boltzmann constant = 1.38x10⁻²³

$\frac{1}{2} =e^{\frac{-4.63x10^{-22} }{1.38x10^{-23} T} } \\\frac{1}{2}=e^{33.55T} \\ln(1/2)=-33.55/T\\-0.693=-33.55/T\\T=48.41K$

b) If T = 10%

$T_{2} =0.1*48.41 =4.841K$

$\frac{P_{1} }{P_{o} } =e^{\frac{-4.63x10^{-22} }{1.38x10^{-23} *4.841} }=1$

c) If

$\frac{P_{L} }{P_{1} } =e^{-\beta (E_{2}-E_{1} } =ex^{-\frac{\epsilon }{kT} } \\E_{2}-E_{1} =2\epsilon -\epsilon = \epsilon \\Then\\\frac{P_{L} }{P_{1} } = 1 (same)$