Ground state of 7 electrons in infinite square well

AI Thread Summary
The discussion centers on calculating the ground state energy of seven electrons in a one-dimensional infinite square well. The Pauli exclusion principle requires that all electrons occupy different quantum states, leading to two electrons in the first energy level (n=1) and five in the second (n=2). The calculated total energy is expressed as a multiple of h²/8mL², yielding a result of 22. However, there is confusion as a reference suggests the answer should be 44, prompting participants to identify the mistake. The clarification emphasizes the absence of angular momentum in a one-dimensional well, affecting the quantum state distribution.
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Homework Statement



Seven electrons are trapped in a one dimensional infinite square well of length L. What is the ground state energy of this system as a multiple of h2 / 8mL2?

Homework Equations


Energy of a single electron in state n is n2h2 / 8mL2

The Attempt at a Solution



Pauli exclusion principle says all 7 must have different quantum numbers.
starting from n = 1, we have L = 0 and mL = 0, and ms = -1/2 and 1/2, so there are two electrons in n = 1.
For n = 2, we have two electrons for L = 0
for L = 1, we have mL = -1, 0, 1, which means this subshell can hold 6 electrons. The remaining 3 electrons go into this subshell then.

Final tally: 2 electrons for n = 1 and 5 electrons for n = 2.

Total energy as a multiple of the given term then = 2*1^2 + 5*2^2 = 22.

Halliday Resnick says 44 for some reason. Can anybody spot my mistake?
 
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Hi sabinscabin,

sabinscabin said:

Homework Statement



Seven electrons are trapped in a one dimensional infinite square well of length L. What is the ground state energy of this system as a multiple of h2 / 8mL2?


Homework Equations


Energy of a single electron in state n is n2h2 / 8mL2


The Attempt at a Solution



Pauli exclusion principle says all 7 must have different quantum numbers.
starting from n = 1, we have L = 0 and mL = 0, and ms = -1/2 and 1/2, so there are two electrons in n = 1.
For n = 2, we have two electrons for L = 0
for L = 1, we have mL = -1, 0, 1, which means this subshell can hold 6 electrons. The remaining 3 electrons go into this subshell then.

Final tally: 2 electrons for n = 1 and 5 electrons for n = 2.

Total energy as a multiple of the given term then = 2*1^2 + 5*2^2 = 22.

Halliday Resnick says 44 for some reason. Can anybody spot my mistake?

For the 1-D infinite well, there are only two states for each energy level (corresponding to the two spin values).
 
ah thanks. That was incredibly stupid of me. It's in a square well so there's no angular momentum number since there's nothing to orbit.

D'oh!
 
No, not stupid. The important thing is to understand why something is wrong. If the test of stupidity is making a mistake, I'm in a lot of trouble.
 
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