Question

1. A single molecule has three possible energy states with the following energy values: Ej = 0 eV, E2 = 10 x 10-3 eV and E3 = 20 x 10-3 eV (a) Determine the value of the partition function Z at room temperature (T = 295 K) (b) Determine the free energy of this molecule at room temperature. (c) Calculate the probability of the molecule being in each state.

Answer 1

The values given in the question are:

Energies of the three possible energy states of molecule E1 =0eV , E2 =10*10-3 eV , E3=20*10-3 eV

Temperature= 295 K

(a)Partition Function Z is defined as

Z=$\small \sum_{j=0}^{n}$ e-Ej / KT

K=Boltzmann constant= 8.617×10−5 eV⋅K−1

Z=Therefore substituting the value of energies

Z=e-0+ e-10*10-3 eV / (8.617×10−5 eV⋅K−1 * 295K) + e-20*10-3 eV / (8.617×10−5 eV⋅K−1 * 295K)

Z=1+e-0.393+e-0.786

Z=1+0.675+0.455

Z=2.13

(c)

Probability of molecule being in a particular state is defined by P(Ej) = e-Ej / KT / $\small \sum_{j=0}^{n}$ e-Ej / KT

where $\small \sum_{j=0}^{n}$ e-Ej / KT = Z

Probability of molecule being in E1 state=e-0 / Z

=1/2.13

=0.469

Probability of molecule being in E2 state= e-10*10-3 eV / (8.617×10−5 eV⋅K−1 * 295K)/ Z

=0.675/2.13

=0.317

Probability of molecule being in E3 state= e-20*10-3 eV / (8.617×10−5 eV⋅K−1 * 295K)/ Z

=0.455/2.13

=0.214