Question

1. A molecule has a ground state and two excited electronic energy levels all of which are not degenerate. The energies of the three states are E = 0, E1 = 1x10^-20 J and E2 = 2x10^-20 J. Calculate the partition functions at 298 and 1000K. What fraction of the molecules is in each of the three states at these temperatures?

Answer 1

Boltzmann distribution:

N(i)/N(0) = exp(-(E(i) - E(0))/kT)

Boltzmann constant k = 1.381 x 10^(-23) J/K

(a) T = 298 K

N(1)/N(0) = exp(-(E(1) - E(0))/kT)

= exp(-(1.0 x 10^(-20) - 0)/(1.381 x 10^(-23) x 298))

= 0.08804

N(2)/N(0) = exp(-(E(2) - E(0))/kT)

= exp(-(2.0 x 10^(-20) - 0)/(1.381 x 10^(-23) x 298))

= 7.75*10^-3

Fractional population for level 1 = N(1)/(N(0) + N(1) + N(2))

= 0.08804/(1 + 0.08804 + 7.75*10^-3)

= 0.081

(b) T = 1000 K

N(1)/N(0) = exp(-(E(1) - E(0))/kT)

= exp(-(1.0 x 10^(-20) - 0)/(1.381 x 10^(-23) x 1000))

= 0.48475

N(2)/N(0) = exp(-(E(2) - E(0))/kT)

= exp(-(3.0 x 10^(-20) - 0)/(1.381 x 10^(-23) x 1000))

= 0.11391

Fractional population for level 1 = N(1)/(N(0) + N(1) + N(2))

= 0.48475/(1 + 0.48475 + 0.11391)

= 0.303 = 0.30