Question

1. In a system at equilibrium with evenly spaced energy levels, you measure the ratio of the population of molecules in ground state versus the first excited state at 54.6. Calculate the partition function (Q) assuming a Boltzmann distribution of the populated energy levels (use 4 energy levels). Calculate the partition function if all molecules were in the ground state.

Answer 1

Answer:

Let, No = population of ground state

      N1 = population of first excite state

      E = energy difference between two lwvwls

then according to Boltzmann distribution,

N1/No = exp(-E/kT)                   ....(1)

given,    No : N1 = 54.6 : 1

so (1) will become

1/54.6 = exp(-E/kT)

on solving we will get

E/kT = 4.00                  ...(2)

We know,

partition function (Q) = _i exp(-E_i/kT)

Given that there are 4 equally spaced energy levels, so energy values will be

E_i = 0, E, 2E and 3E

So,

Q = exp(-0/kT) + exp(-E/kT) + exp(-2E/kT) + exp(-3E/kT)   ....(3)

put E/kT = 4.00 (from(2)) then (3) will become

Q = exp(0) + exp(-4.00) + exp(-8.00) + exp(-12.00) = 1.0186 = Q

Even if all the molecules were in ground state then the partition function remains same, because the partition function tells us about the number of available state in which molecule can be found not the state(s) where molecules exactly is.