Question

An oscillator with frequency f = 2.1×1012 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. 1)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)?

Answer 1

Answer:

T = 55.39 K

Explanation:

Given data:

frequency $F = 2.1\times 10^{12} Hz$

Plank constant $H = 6.626 ×10^{-34} J s$

$\frac{P_1}{P_o} = \frac{1}{2}$

we know that expression for calculating temperature is given as

$\frac{P_1}{P_o} = e^{\frac{h F}{k_b t}$

here, P_1 probability that oscillator has $E = \epsilon$ . P_o is probability  that has  E = 0 and Boltzman constant has a value$k_b = 1.3807\times 10^{-23} J/k$

from above expression

$ln\frac{1}{2} = \frac{6.626\times 10^{-34} \times 8\times 10^{11}}{1.3807\times 10^{-23} \times T}$

SOLVING FOR T WE HAVE$T = \frac{5.3008\times 10^{-22}}{9.57\times 10^{-24}}$

T = 55.39 K