The energy difference between an excited state and the ground state of a system is 0.2 eV.If the population ratio of the excited state to the ground state is 2.9 * 10^-4,what is the temperature of the system? A)300 K B)330 K C)285 K D)315 K

Question

Quizplus · Accepted Answer

We can use the formula for population ratio at thermal equilibrium: 
N2/N1 = g2/g1 * exp(-E21/kT)
where N2/N1 is the population ratio, g2/g1 is the degeneracy ratio (usually 1 for electronic states), E21 is the energy difference between the excited state and ground state, k is the Boltzmann constant, and T is the temperature.

Substituting the given values, we get:
2.9 = 1 * exp(-0.2/1.38x10^-23T)
Taking natural logarithms on both sides:
ln 2.9 = -0.2/1.38x10^-23T
Solving for T:
T = -0.2/1.38x10^-23ln2.9
T = 285 K (approximately)

Therefore, the best choice is C: 285 K.

The Energy Difference Between an Excited State and the Ground