Question

1.5 The ground level of an atom is split into two Zeeman states of equal statistical weight separated by 10,000 MHz. An assembly of such atoms is in thermal equilibrium at temperature TK. What is the fractional population difference of the Zeeman states when T = 300K, 20K. 4K, 1 - 5K?

Answer 1

We have a system of N atoms in thermal equilibrium. And we know that the ground level of each of these atoms is being split into two levels (these levels are told to be Zeeman levels, but don't worry about it at this point. )

Let these levels (Zeeman levels) have energies E₀ and E₁, E₁ > E₀

The atoms are distinguishable and are fermions, so the statistics we need to use in our problem would be the Fermi-Dirac statistics.

We need to calculate the population of the two levels, E₀ and E₁

The population of a level means how many no. of particles (atoms in our case) would be there in these levels. Another name for the population is the occupation no. The occupation number for any energy level E in Fermi-Dirac statistics is:

where N is the total no. of particles. (atoms in our case)

k is Boltzmann constant = 1.38 * 10^-23 J/K

T is the Temperature in Kelvin

If you are comfortable with Partition function formalism, one way of arriving at above formula is the following:

If Q is the partition function,

Where the summation is over all the states i and E_i is the energy of ith state.

Then the occupation number of ith level (state) is given by:

...(i)

So now in our problem, Partition function of a single atom is:

Q =

=

where = E₁ - E₀

So the occupation no. for the level E₀:

and the fractional occupation no for level E₀:

...(iii)

Similarly, the occupation no. for level E1:

and the fractional occupation no. for level E₁ :

....(iv)

Notice that instead of using the partition function formalism, you can equivalently use formula (i) to reach at formula (iii) and (iv).

Fractional population difference of these states:

The difference between above two, (iv) - (iii) would give you the required answer, i .e.

....(v)

Calculations:

hv

where v = 10,000 MHz = 10¹⁰ s^-1

  h = 6.626 x 10^-34 J-s = Planck's constant

So,    (6.626 x 10^-34 J-s ) x (10¹⁰ s^-1)

= 6.626 x 10^-14 J

Substituting the values of  , k, and T, we will get the required difference.

Inference:

1. For >> kT,

In this case, is nearly 0,

So, the fractional population difference reduces to 1

Hence, n₀ is very nearly equal to N and n₁ is very nearly equal to 0.

i.e. for sufficiently low temperatures, all the particles (atoms) are essentially in the level with lower energy, E₀.

2. For << kT,

In this case, is nearly 1

So, the fractional population difference reduces to 0

Hence, n₀ nearly equal to n₁

i.e. for sufficiently high temperatures, both the levels are nearly equally populated.