Maxwell Boltzmann distribution

In summary, the problem is asking for the number of atoms in the n=1 and n=2 states of a sample of mercury vaporized in a flame. The energy difference between these two states is 4.86 eV and the sample contains 10^20 atoms at a temperature of 1600K. The statistical weight of these states is assumed to be equal, and using the Maxwell-Boltzmann distribution, the number of atoms in each state is calculated to be 5*10^19. However, this may not be the correct answer as it does not take into account the actual statistical weight of the states.
  • #1
vincisonfire
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Homework Statement


The energy difference between the first excited state of mercury and the ground state is 4.86 eV.
(a) If a sample of mercury vaporized in a flame contains 10^20
atoms in thermal equilibrium at 1600K, calculate the number of atoms in the n=1 (ground) and n=2 (first-excited) states. (Assume the Maxwell-Boltzmann distribution applies and that the n=1 and n=2 states have equal statistical weights.)

Homework Equations


Maxwell Boltzmann Distribution

The Attempt at a Solution


I thought that since they have the same statistical weight, there must be 5*10^19 particles in each state. But I don't think it is the good answer since we use this number for another exercice and it doesn't yield the good answer.
I don't know how to figure out these number using Maxwell Boltzmann distribution.
 
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  • #2
Maxwell-Boltzmann refers to the distributions of atom velocities and is not relevant here.

What are relative populations of 2 states, separated by energy ΔE and at a temperature T?
 
  • #3
"Statistical weight" refers to the number of different states there are with n=1 and n=2, respectively. In reality the statistical weight is certainly not the same for these states, but never mind: all you need is the energy difference between the two states, and then answer Redbelly's question.

[It's a bit unfortunate that *sometimes* by "statistical weight" one means exactly the Boltzmann factor you will have to calculate: so by that definition you are correct to say the populations in the states are equal. But clearly, the person who phrased the question did not have that meaning of "statistical weight" in mind.]
 

What is the Maxwell Boltzmann distribution?

The Maxwell Boltzmann distribution is a probability distribution that describes the distribution of speeds of particles in a gas or liquid at a given temperature. It is named after James Clerk Maxwell and Ludwig Boltzmann, who independently developed the distribution in the late 19th century.

What factors influence the shape of the Maxwell Boltzmann distribution?

The shape of the Maxwell Boltzmann distribution is influenced by two main factors: temperature and mass of the particles. As temperature increases, the distribution shifts towards higher speeds, and as mass increases, the curve becomes broader and flatter.

What is the significance of the Maxwell Boltzmann distribution?

The Maxwell Boltzmann distribution is significant because it is a fundamental concept in thermodynamics and statistical mechanics. It helps us understand the behavior of particles in a gas or liquid and has applications in various fields, including chemistry, physics, and engineering.

How is the Maxwell Boltzmann distribution related to the ideal gas law?

The Maxwell Boltzmann distribution is related to the ideal gas law through the average kinetic energy of the particles. The distribution allows us to calculate the average kinetic energy of particles in a gas, which is one of the variables in the ideal gas law (PV = nRT).

Can the Maxwell Boltzmann distribution be applied to all types of particles?

Yes, the Maxwell Boltzmann distribution can be applied to all types of particles, as long as they are in a gas or liquid state. The distribution is valid for both classical particles, such as atoms and molecules, and quantum particles, such as electrons and photons.

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