The energy difference $\Delta$E between the J = 1 and J = 2 rotational levels of NO is 1.24 × 10⁻²⁴ J. Given that the degeneracy of the rotational levels is 2J + 1, calculate the ratio of populations of these levels at 298 K.

Question

Quizplus · Accepted Answer

The ratio of populations of the J = 1 and J = 2 rotational levels is given by the Boltzmann distribution:$$\frac{N_2}{N_1} = e^{-\Delta E/kT}$$where:* $N_2$ is the population of the J = 2 level* $N_1$ is the population of the J = 1 level* $\Delta E$ is the energy difference between the two levels* $k$ is the Boltzmann constant* $T$ is the temperaturePlugging in the given values, we get:$$\frac{N_2}{N_1} = e^{-(1.24 	imes 10^{-24} 	ext{ J})/(1.38 	imes 10^{-23} 	ext{ J/K})(298 	ext{ K})}$$$$\frac{N_2}{N_1} = 0.60$$Therefore, the ratio of populations of the J = 1 and J = 2 rotational levels at 298 K is 0.60.

The Energy Difference Δ\DeltaΔ E Between the J = 1 and J = 2

The Energy Difference $\Delta$ E Between the J = 1 and J = 2