In a system where the energy difference between the excited state and the ground state is 0.2 eV,what is the population ratio of the excited state relative to the ground state at a temperature of 300 K? A)4.4 * 10^-4 B)2.9 * 10^-4 C)1.02 D)0.001

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Quizplus · Accepted Answer

The population ratio of the excited state relative to the ground state is given by the Boltzmann distribution:$$N_e/N_g = e^{-\Delta E/k_BT}$$where:* $N_e$ is the population of the excited state* $N_g$ is the population of the ground state* $\Delta E$ is the energy difference between the excited state and the ground state* $k_B$ is the Boltzmann constant* $T$ is the temperatureIn this case, $\Delta E = 0.2 eV = 3.2 imes 10^{-19} J$, $k_B = 1.38 imes 10^{-23} J/K$, and $T = 300 K$. Plugging these values into the equation, we get:$$N_e/N_g = e^{-(3.2 imes 10^{-19} J)/(1.38 imes 10^{-23} J/K)(300 K)} = 4.4 imes 10^{-4}$$Therefore, the population ratio of the excited state relative to the ground state is 4.4 * 10^-4.

In a System Where the Energy Difference Between the Excited