The energy of the first excited state of the system

In summary: I think you are misunderstanding the question. The question is asking for the energy of the 7th electron in the ground state, not the 7th energy level. The 7th energy level would correspond to the 49th electron. Also, the formula given in the question is incorrect, it should be ##(n_x^2+n_y^2)\frac{h^2}{8mL^2}## not ##(n_x^2+n_y^2)\frac{h^2}{8mL}##. Here is a summary of the conversation:In summary, five electrons are in a two-dimensional square potential energy well with sides of length L. The potential energy is infinite at the sides and zero inside
  • #1
hidemi
208
36
Homework Statement
I used the eq. below, but I can't find the answer.
Relevant Equations
Σ (nx^2+ ny^2) (h2/8mL^2)
Five electrons are in a two-dimensional square potential energy well with sides of length L. The potential energy is infinite at the sides and zero inside. The single-particle energies are given by (h^2/8mL^2) (nx^2+ ny^2) where nx and ny are integers. The energy of the first excited state of the system is
(A) 13 (h^2/8mL^2) (B) 22 (h^2/8mL^2) (C) 24 (h^2/8mL^2) (D) 25 (h^2/8mL^2) (E) 27 (h^2/8mL^2)

the correct answer is B
 

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  • #2
What did you do?
 
  • #3
PeroK said:
What did you do?
I know the formula, but I've no idea how to start.
 
  • #4
hidemi said:
I know the formula, but I've no idea how to start.
In post #1 you wrote that you used the formula. If you meant that, show how you used it.
 
  • #5
haruspex said:
In post #1 you wrote that you used the formula. If you meant that, show how you used it.

I calculate this way as per attached and gives me a matching answer. However, I use the same way to calculate the question below, I don’t have any answer.
 

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  • #6
hidemi said:
I calculate this way as per attached and gives me a matching answer. However, I use the same way to calculate the question below, I don’t have any answer.
What principles are you using to calculate these answers? You haven't said anything about why you are choosing these system configurations.
 
  • #7
hidemi said:
I calculate this way as per attached and gives me a matching answer. However, I use the same way to calculate the question below, I don’t have any answer.
So it seems you have now answered the question in post #1, right? but can't see how to use the same principles on the question in the attachment in post #5.
Now, this is a subject about which I know next to nothing, yet I think I see how it works.
For the six in ground state, what do you think their energy levels are? What is the next available level?
 
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  • #8
haruspex said:
So it seems you have now answered the question in post #1, right? but can't see how to use the same principles on the question in the attachment in post #5.
Now, this is a subject about which I know next to nothing, yet I think I see how it works.
For the six in ground state, what do you think their energy levels are? What is the next available level?
 

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  • #9
I'll name the questions to avoid confusion:
Q1: the original 5-electron question in post #1;
Q2: the '7-th electron' question in post #5's attachment.

From your postings it is not clear (to me at least) if you have now understood and correctly answered both questions!

Note that Q1 asks for the total energy of the system in its 1st excited state. But Q2 asks for the energy of only the 7-th electron in its ground state. Q1 and Q2 want different things - but you find them in similar ways.

Do you know which (who's) principle you are using?

As an additional note, multi-electron systems - such as the ones in the questions - have potential energy due to the mutual repulsion of the electrons. But the expression ##(n_x^2+n_y^2)\frac{h^2}{8mL}## only applies to a single electron with zero potential energy inside the 'well'. So the question is flawed. But you are meant to ignore that.

EDIT: Typo' corrected.
 
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  • #10
Steve4Physics said:
I'll name the questions to avoid confusion:
Q1: the original 5-electron question in post #1;
Q2: the '7-th electron' question in post #5's attachment.

From your postings it is not clear (to me at least) if you have now understood and correctly answered both questions!

Note that Q1 asks for the total energy of the system in its 1st excited state. But Q2 asks for the energy of only the 7-th electron in its ground state. Q1 and Q2 want different things - but you find them in similar ways.

Do you know which (who's) principle you are using?

As an additional note, multi-electron systems - such as the ones in the questions - have potential energy due to the mutual repulsion of the electrons. But the expression ##(n_x^2+n_y^2)\frac{h^2}{8mL}## only applies to a single electron with zero potential energy inside the 'well'. So the question is flawed. But you are meant to ignore that.

EDIT: Typo' corrected.
Thanks your comments. I'll look into it more.
 
  • #11
hidemi said:
(Attachment in post #8)
Those numbers lead to answer C, as required: 7th & 8th, 8 each.
 
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  • #12
haruspex said:
Those numbers lead to answer C, as required: 7th & 8th, 8 each.
So, my thoughts written in the post #8 attachment are correct?
If so, I still don't quite understand the calculation behind it. Do you have a better explanation?
 
  • #13
hidemi said:
So, my thoughts written in the post #8 attachment are correct?
If so, I still don't quite understand the calculation behind it. Do you have a better explanation?
As I wrote, I have never studied the subject, so my understanding is based on this thread, and a vague understanding that electrons have no individuality.
No two electrons can be in the same state in the well, and such a state is defined by spin and an ordered pair (m,n) of natural numbers, spin being binary. The energy is characterized by the sum of squares of m, n.
So for the lowest total energy we fill in two with opposite spins at (1,1), two at (1,2), two at (2,1). After that it gets a bit trickier because you need to figure out that e.g. 22+22<12+32.
 
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  • #14
hidemi said:
So, my thoughts written in the post #8 attachment are correct?
If so, I still don't quite understand the calculation behind it. Do you have a better explanation?
How is it that you are studying QM and basic kinematics at the same time?

The QM square well, Pauli Exclusion Principle and electron energy/spin states are a lot more advanced than SUVAT equations!
 
  • #15
haruspex said:
As I wrote, I have never studied the subject, so my understanding is based on this thread, and a vague understanding that electrons have no individuality.
No two electrons can be in the same state in the well, and such a state is defined by spin and an ordered pair (m,n) of natural numbers, spin being binary. The energy is characterized by the sum of squares of m, n.
So for the lowest total energy we fill in two with opposite spins at (1,1), two at (1,2), two at (2,1). After that it gets a bit trickier because you need to figure out that e.g. 22+22<12+32.
Yes, this is the Pauli Exclusion Principle. It's the same basis on which atomic orbitals fill up, with at most two electrons occupying each distinct energy state.
 
  • #16
PeroK said:
Yes, this is the Pauli Exclusion Principle. It's the same basis on which atomic orbitals fill up, with at most two electrons occupying each distinct energy state.
It just dawned on me that if you consider the points with integer coordinates in the first quadrant then the energy order matches the order of distance from the origin,
 
  • #17
haruspex said:
It just dawned on me that if you consider the points with integer coordinates in the first quadrant then the energy order matches the order of distance from the origin,
Yes, for a large number of electrons you can map the occupied states to an octant in k-space, were $$\vec k = (\frac{\pi n_x}{L_x},\frac{\pi n_y}{L_y}, \frac{\pi n_z}{L_z})$$.
This leads to the concept of the Fermi Energy and Fermi Surface etc.:

https://en.wikipedia.org/wiki/Fermi_energy
 
  • #18
PeroK said:
How is it that you are studying QM and basic kinematics at the same time?

The QM square well, Pauli Exclusion Principle and electron energy/spin states are a lot more advanced than SUVAT equations!
I am reviewing all these topics.
 

Related to The energy of the first excited state of the system

1. What is the first excited state of a system?

The first excited state of a system refers to the energy level of the system when one of its electrons has been excited from its ground state to a higher energy level.

2. How is the energy of the first excited state calculated?

The energy of the first excited state is calculated using the formula E = -13.6 eV/n2, where n is the principal quantum number of the energy level.

3. What factors affect the energy of the first excited state?

The energy of the first excited state is affected by the atomic number, the number of electrons in the system, and the distance between the nucleus and the electron.

4. Why is the first excited state important in understanding the behavior of atoms?

The first excited state is important because it is the lowest energy level above the ground state, and it is from this state that electrons can transition to higher energy levels. This helps us understand the emission and absorption of light by atoms, as well as their chemical properties.

5. Can the energy of the first excited state be measured experimentally?

Yes, the energy of the first excited state can be measured experimentally using techniques such as spectroscopy, which involves studying the interaction between light and matter to determine the energy levels of atoms.

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