- #1

- 92

- 0

## Homework Statement

You will recall from our discussion of the Franck-Hertz experiment that the energy difference between the first excited state of mercury and the ground state is 4.86 eV. If a sample of mercury vaporized in a flame contains 1.06×1020 atoms in thermal equilibrium at 1563 K, calculate the number of atoms in the first excited state. Assume that the Maxwell-Boltzmann distribution applies and that the n = 1 (ground) and n = 2 (first excited) states have equal statistical weights.

## Homework Equations

The Maxwell-Boltzmann distribution:

[itex]f(E)=A e^{\frac{E}{k_{b}T}}[/itex]

where A is a constant

k_b is the boltzmann constant = 8.617*10^-5 eV/K

and T is the temperature

n(E)dE = g(E) * f(E) dE

where n(E) = # per unit energy

g(E) = # of states per energy E

f(E) = MB distribution

## The Attempt at a Solution

I'm still having trouble learning how to use distributions, so bear with me (or help me understand it

I started off by normalizing the distribution by:

[tex]\int\limits_0^\infty f(E)^2 dE [/tex]

then I set it equal to 1 and solved for A. I found A=3.854.

I think that was the right first step.

and I know that g(E) for n=1 is 2 possible states

and n=2 has 8.

This is as far as I have gotten, as my professor hasn't really gone over how to use these distributions.

My guess is that I set

[itex] 1.06*10^{20} = n_{1}(E) dE + n_{2} (E) dE [itex]

after this I know I have to integrate, but I'm not sure how to set up the limits, among other things.