Maxwell Boltzman Distribution, Please help

In summary, the Franck-Hertz experiment showed that the energy difference between the first excited state of mercury and the ground state is 4.86 eV. Using the Maxwell-Boltzmann distribution and assuming equal statistical weights for the n=1 (ground) and n=2 (first excited) states, a sample of mercury vaporized in a flame containing 1.02×10^20 atoms at 1613 K would have 267884 atoms in the first excited state.
  • #1
inferno298
25
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 1.02×10^20 atoms in thermal equilibrium at 1613 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



n2/n1=g(E2)/g(E1)*e^((E1-E2)/(k T))
k=boltzman constant = (1.3807*10^-23), divide by 1.602*10^-19 to get in eV

The Attempt at a Solution



So I felt like it was almost a plug and chug. The distinct states at n=2, first excited state, is 8. For n=1 g(E1)= 2. Sooo :
n2/n1=(8/2)*e^((-4.86 eV)/(k*1613))
I got 2.626*10^-15, so that's the ratio of atoms in the first excited state compared to the ground state? So I just multiply it by the number of atoms 1.02*10^20 and I get 267884 atoms. Answer isn't correct though and I can't figure out what I may be missing.
 
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  • #2
inferno298 said:
Assume that the Maxwell-Boltzmann distribution applies and that the n = 1 (ground) and n = 2 (first excited) states have equal statistical weights.
Have you made use of the last part of this statement?
 
  • #3
I guess I am not too sure of what that means.
 
  • #4
nmv I got it Thanks!
 
  • #5



I would like to point out that your attempt at a solution is correct and your reasoning is sound. However, there may be a mistake in your calculation or conversion of units that is leading to the incorrect answer. I would recommend double-checking your calculations and units, and also considering the significant figures in your final answer. Additionally, it may be helpful to consult with your peers or instructor for further clarification. Remember, in science, it is important to always double-check our work and be open to seeking help when needed.
 

1. What is the Maxwell Boltzmann distribution?

The Maxwell Boltzmann distribution is a probability distribution that describes the speeds of particles in a gas at a given temperature. It shows the frequency of particles with different speeds in a gas and is derived from the kinetic theory of gases.

2. How is the Maxwell Boltzmann distribution related to temperature?

The Maxwell Boltzmann distribution is directly related to temperature. As the temperature increases, the distribution shifts towards higher speeds, meaning there is a higher proportion of particles with higher speeds. As the temperature decreases, the distribution shifts towards lower speeds.

3. What factors affect the shape of the Maxwell Boltzmann distribution?

The shape of the Maxwell Boltzmann distribution is affected by the temperature and mass of the particles in the gas. Higher temperatures and lighter particles result in a broader distribution, while lower temperatures and heavier particles result in a narrower distribution.

4. How is the Maxwell Boltzmann distribution used in real-world applications?

The Maxwell Boltzmann distribution is used in various fields, such as chemistry, physics, and engineering, to understand the behavior of particles in gases. It is also used in the design and optimization of gas-based technologies, such as combustion engines and gas sensors.

5. Are there any limitations to the Maxwell Boltzmann distribution?

The Maxwell Boltzmann distribution assumes that particles in a gas are in thermal equilibrium and do not interact with each other. In reality, this is not always the case, especially at high pressures or in dense gases. Additionally, it only applies to ideal gases, so deviations may occur in real gases.

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