Finding the g-value in the Boltzmann distribution

In summary, the transition from the ground state to the first excited state occurs when an electron transitions from a 1s orbital to a 4s orbital.
  • #1
JoJoQuinoa
17
0
Hello,

I was wondering if someone could show me how to determine the number of orbitals available for a state and the number of electrons in that state. For calcium in the ground state, the electron config is 1s2 2s2 2p6 3s2 3p6 4s2. For the first excited state I assumed 1s2... 4s1 3d1.
From the solution for g_g, why the number of available orbitals is one when 4s is filled? Shouldn't it be zero?

Thanks!
AAS.PNG
 
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  • #2
If there were zero available orbitals, where would the electrons go? It doesn't mean "available for other electrons". It means "available for the valence electrons". (All filled subshells have #e = 2#o, so g = 1; you only need to consider the valence electrons.) There is one 4s orbital into which 2 electrons can go to give the configuration 4s2.
 
  • #3
@mjc123, thank you for your explanation.

That is what I thought originally but when I applied it to the excited state, it didn't work. But I think my electron configuration for the first excited state is wrong. I think it should be ##1s^2...4s^1 4p^1## based on the selection rules. If that's true then #o is 3 in ##4p^1## and there are 2 valence electrons.
 
  • #4
JoJoQuinoa said:
I think it should be ##1s^2...4s^1 4p^1## based on the selection rules.
Note that even without considering selection rules, ##4s^1 4p^1## is lower in energy than ##4s^1 3d^1##. (The Aufbau rules when filling orbitals do not necessarily apply to finding excited states).
 
  • #5
I'm not sure what excited state is being referred to - certainly not the first excited state, as 226.51 nm is much higher energy. The expression for ge would suggest a configuration like 4p2, for which #o is 3 and #e is 2. (4s4p would have g = 12; 2 for the 4s electron x 6 for the 4p.) And the 15 microstates of 4p2 don't all have the same energy - they are split into 1S, 3P and 1D states (and 3P is further split into J = 0, 1 and 2 states). So using the g value in the Boltzmann distribution would be problematic.

Can you provide more context for this problem?
 
  • #6
mjc123 said:
I'm not sure what excited state is being referred to - certainly not the first excited state, as 226.51 nm is much higher energy. The expression for ge would suggest a configuration like 4p2, for which #o is 3 and #e is 2. (4s4p would have g = 12; 2 for the 4s electron x 6 for the 4p.) And the 15 microstates of 4p2 don't all have the same energy - they are split into 1S, 3P and 1D states (and 3P is further split into J = 0, 1 and 2 states). So using the g value in the Boltzmann distribution would be problematic.

Can you provide more context for this problem?
Certainly

Problem: Use the Boltzmann distribution equation to calculate the percentage of calcium atoms that are in the first excited state in a hydrogen -air flame at 2,250K. The line associated with that transition has a wavelength of 226.51 nm.

Further Question: Is it not true that first excitation occurs when 'one' valence electron transition to a higher electronic state?

Thanks!
 
  • #7
DrClaude said:
Note that even without considering selection rules, ##4s^1 4p^1## is lower in energy than ##4s^1 3d^1##. (The Aufbau rules when filling orbitals do not necessarily apply to finding excited states).

Hi DrClaude,

Is this a common knowledge or is there a set of rules for this? I'm not a chemistry student so I've been using what I can vaguely remember from general chemistry 10 years ago o_O
 
  • #8

Related to Finding the g-value in the Boltzmann distribution

1. What is the Boltzmann distribution?

The Boltzmann distribution is a statistical distribution that describes the distribution of particles in a system at thermal equilibrium. It is used to calculate the probability of a particle having a certain energy level in a system at a given temperature.

2. What is the g-value in the Boltzmann distribution?

The g-value, also known as the degeneracy factor, is a constant that represents the number of ways a particle can occupy a certain energy level in a system. It takes into account the different possible states a particle can be in, such as spin or position, and is used in the Boltzmann distribution equation to calculate the probability of a particle being in a specific energy state.

3. How is the g-value calculated?

The g-value can be calculated by determining the number of possible states a particle can be in at a given energy level. For example, if a particle can have two different spin states at a certain energy level, the g-value would be 2. In more complex systems, the g-value can be calculated by considering all the possible combinations of states that a particle can have.

4. Why is the g-value important in the Boltzmann distribution?

The g-value is important because it allows for a more accurate calculation of the probability of a particle being in a specific energy state. Without taking into account the different possible states a particle can have, the Boltzmann distribution would not accurately describe the distribution of particles in a system at thermal equilibrium.

5. How is the g-value related to entropy?

The g-value is related to entropy because it is a measure of the disorder or randomness in a system. A higher g-value means that there are more possible states for a particle to be in, which corresponds to a higher entropy. In other words, a system with a higher g-value has more disorder and is less organized.

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