# Finding the g-value in the Boltzmann distribution

• JoJoQuinoa
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.
JoJoQuinoa
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!

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.

@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.

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).

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?

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!

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

## 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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