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How many electrons could be held in the second shell of an atom if the spin quantum number $m_{s}$ could have three values instead of just two? (Hint: Consider the Pauli exclusion principle.)

Chemistry 101

Chapter 6

Electronic Structure and Periodic Properties of Elements

Electronic Structure

Periodic Table properties

Carleton College

University of Central Florida

Rice University

University of Toronto

Lectures

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In chemistry and physics, …

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Suppose two electrons in a…

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Two electrons in the same …

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What is the Pauli exclusio…

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What is the total number o…

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Apply the Pauli exclusion …

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How many electrons could o…

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How many electrons can hav…

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How many electrons in an a…

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How many electrons can be …

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What is the maximum number…

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An electron in a multielec…

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An electron in a certain a…

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problem. 43 from Chapter six is asking us how many electrons could be held in the second Shelvin Adam, if the spin quantum number would have three values instead of just two. So let's start by answering this question, withdrawing the second shell of an Adam So this will be to s and then this show will be to pee color. So normally we would do a upward facing Harrow and a downward facing arrow. Because we have, I spend value of plus half and minus half. But in this hypothetical situation, we could have 1/3 direction of spin. So I'm Sandra Day and let's do the same for the P orbital. No. Yeah, on more So given that we've drawn, all are electrons. In the second show, we have 369 and 12 total elections. That would be the answer as to how many electrons would be able to be held in the second show. If three electrons all right, the quantum could have three values incident

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