Question

6. Consider a quantum system of N particles with only three possible states to oc cupy for each particle. The energy values of these states are equal to 0, E and 3, respectively. (a) (10 points) You observe that the probability to sample eash state is p=0.9, P2 = 0.09 and p = 0.01 at T = 300 K. What are the energies ez and ez? Recall that the probability to occupy 4th state is proportional to e«/T where k = 1.38 x 10-19 J/K. (b) (10 points) What is the average energy (total) of this system?

Answer 1

Part (a).

p2 = e^{-$\epsilon$2/kT}

i.e. 0.09 = e^{-$\epsilon$2/1.38*10^-19 J/K *300 K}

i.e. $\epsilon$2/1.38*10^-19 J/K * 300 K = 1.0942

i.e. $\epsilon$2 = 4.53*10^-17 J

p3 = e^{-$\epsilon$3/kT}

i.e. 0.01 = e^{-$\epsilon$3/1.38*10^-19 J/K *300 K}

i.e. $\epsilon$3/1.38*10^-19 J/K * 300 K =

i.e. $\epsilon$2 = 4.18*10^-17 J

Part (b).

The average energy of the system = (4.53+4.18)*10^-17/2 J = 4.355*10^-17 J