Statistics physics problem -- atom is in the ground state or excited state?

[Oliver321]

I am learning for my exam in particle physics. One topic is statistical physics. There I ran into this question:
Consider an atom at the surface of the Sun, where the temperature is 6000 K. The
atom can exist in only 2 states. The ground state is an s state and the excited state at
1.25 eV is a p state. What is the probability to find the atom in the excited state?

My attempt to solve this problem was to use Maxwell-Bolzmann-energy-distribution (because the atoms are relatively far apart I may not use fermi-dirac or Bose-Einstein statistic): I take one mole of atoms. First calculate N(E), than determine the number of particles which are over 1.25 eV (integral from N(1.25eV) to N(infinity)). Than I can take the ratio and consequently I get the probability.

My concern with that method is, that i think it must be way easier to solve, cause in general the problems I get to solve are solvable with less calculating.

One other problem I have no solution to is followed:
In a metal the Fermi energy describes
(1) the highest occupied energy state of a free electron at zero temperature
(2) the minimum energy necessary to remove an electron from the metal
(3) the mean thermal energy of the atoms at temperature T
(4) the energy necessary to break the bonds between the metal atoms

I would say 1 and 2 are right. Normally only one awnser is possible. But I don’t know which should be wrong and why.

I appreciate every help! Thank you very much!

 

[Orodruin]

My attempt to solve this problem was to use Maxwell-Bolzmann-energy-distribution (because the atoms are relatively far apart I may not use fermi-dirac or Bose-Einstein statistic): I take one mole of atoms. First calculate N(E), than determine the number of particles which are over 1.25 eV (integral from N(1.25eV) to N(infinity)). Than I can take the ratio and consequently I get the probability.

This is not correct. You do not need to look at the distribution of kinetic energies, it has nothing to do with the internal state of the atoms. What you need to do is to find the relative occupancy of the ground state to the excited state, knowing that the occupancy is proportional to $e^{-E/k_BT}$.

Edit:

One other problem

Please limit your questions to one per thread. If you have several questions, open several threads.

 

[Oliver321]

This is not correct. You do not need to look at the distribution of kinetic energies, it has nothing to do with the internal state of the atoms. What you need to do is to find the relative occupancy of the ground state to the excited state, knowing that the occupancy is proportional to $e^{-E/k_BT}$.

Edit:

Please limit your questions to one per thread. If you have several questions, open several threads.

Thank you!
So I should calculate N(p state)/ N(ground state). Therefore I use N(E)= g*f. With f the distribution function you mentioned above and g the density of states ( degeneracy of 2n^2)?
But why do I use the relative occupancy? Shouldn’t I use N(excited)/(N(ground)+N(excited)) to get the probability? Otherwise I would only know how many excited states there are for every groundstate? In case one I get a result of 0.392 particles in excited state in relation to particles in ground state (degeneracy of theecited state is 8, degeneracy of groundstate is 2). In case two I get a 28% probability to find a particle in excited state.
Is this a realistic value? Seems very small

 

[Orodruin]

But why do I use the relative occupancy? Shouldn’t I use N(excited)/(N(ground)+N(excited)) to get the probability?

Well, they are related. Dividing both denominator and nominator by N(excited), your probability becomes 1/(1+r), where r = N(ground)/N(excited).

If the temperature is very large (relative to the energy difference), you would expect there to be no difference between the occupancy of the states and the probability would be 50%. If the temperature is very small, you would expect essentially zero probability to be in the excited state. Why do you think 28% is small? Have you looked at what $k_B T$ is for 6000 K and compared it to the energy difference between the states?

 

[Oliver321]

Well, they are related. Dividing both denominator and nominator by N(excited), your probability becomes 1/(1+r), where r = N(ground)/N(excited).

If the temperature is very large (relative to the energy difference), you would expect there to be no difference between the occupancy of the states and the probability would be 50%. If the temperature is very small, you would expect essentially zero probability to be in the excited state. Why do you think 28% is small? Have you looked at what $k_B T$ is for 6000 K and compared it to the energy difference between the states?

Why do I expect to approach 50%? I would rather think that all particles get enough energy to be excited if the temperature is high enough?

Another question regarding my attempt to solve this problem: Why does the Bolzmann energy distribution only tells me the kinetic energy? Can you give me a hint where in derivation of the energy distribution (picture) the kinetic energy gets in?

I appreciate your help!

 

[Orodruin]

Why do I expect to approach 50%? I would rather think that all particles get enough energy to be excited if the temperature is high enough?

No, this is not correct. You can never obtain a situation where the excited state is more occupied than the ground state just from thermal interactions. In thermal equilibrium, the state occupancy is proportional to $e^{-E/k_BT}$, this is always smaller for larger $E$.

Another question regarding my attempt to solve this problem: Why does the Bolzmann energy distribution only tells me the kinetic energy? Can you give me a hint where in derivation of the energy distribution (picture) the kinetic energy gets in?

The Boltzmann energy distribution just tells you that occupancy is proportional to $e^{-E/k_BT}$ - it is valid as long as your states have low occupancy so quantum effects do not come into play. The Maxwell-Boltzmann distribution is what you were trying to use. It is the distribution of velocities given that the kinetic energy is given by the classical mv^2/2. Changing some variables you can find the density of states expressed in different variables. However, in this problem, you are not at all interested in the distribution of velocities, you are interested in the internal degrees of freedom of the molecules. If you were interested in the full distribution function (in terms of both velocity and internal states), then it would be given by the product of the Maxwell-Boltzmann distribution and the distribution function of the internal states.

See https://en.wikipedia.org/wiki/Boltzmann_distribution versus https://en.wikipedia.org/wiki/Maxwell–Boltzmann_distribution

 

[Oliver321]

No, this is not correct. You can never obtain a situation where the excited state is more occupied than the ground state just from thermal interactions. In thermal equilibrium, the state occupancy is proportional to $e^{-E/k_BT}$, this is always smaller for larger $E$.The Boltzmann energy distribution just tells you that occupancy is proportional to $e^{-E/k_BT}$ - it is valid as long as your states have low occupancy so quantum effects do not come into play. The Maxwell-Boltzmann distribution is what you were trying to use. It is the distribution of velocities given that the kinetic energy is given by the classical mv^2/2. Changing some variables you can find the density of states expressed in different variables. However, in this problem, you are not at all interested in the distribution of velocities, you are interested in the internal degrees of freedom of the molecules. If you were interested in the full distribution function (in terms of both velocity and internal states), then it would be given by the product of the Maxwell-Boltzmann distribution and the distribution function of the internal states.

See https://en.wikipedia.org/wiki/Boltzmann_distribution versus https://en.wikipedia.org/wiki/Maxwell–Boltzmann_distribution

Thank you very much!

One last question: I don’t understand why I use the maxwell-Boltzmann-distribution and not the fermi-Dirac-distribution(because quantum effects are recent here)? Doesn’t quantisation count as quantum effect? What would count as quantum effect so that i know when to use another distribution?

 

[Orodruin]

The FD or BE distributions only become relevant when the occupation numbers (in the combined spin-phase space space) are close to one. As long as your atoms are sufficiently sparse, both the FD and BE distributions will be well approximated by the MB distribution.