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(a) What is the lowest possible value of the principal quantum number ( $n$ ) when the angular momentum quantum number $(\ell)$ is $1 ?$ (b) What are the possible values of the angular momentum quantum number (\ell) when the magnetic quantum number $\left(m_{\ell}\right)$ is 0 given than $n \leq 4 ?$

The lowest possible value for $n=2$Possible $\ell$ values are 0,1,2,3 For a given value of $\ell,$ there are $(2 \ell+1)$ integral values of $m_{\ell}$ which range from $-\ell$ to $+\ell$

Chemistry 101

Chapter 7

Quantum Theory and the Electronic Structure of Atoms

Electronic Structure

Carleton College

University of Central Florida

Rice University

University of Kentucky

Lectures

04:49

In chemistry and physics, electronic structure is the way the electrons of an atom are arranged in relationship to the nucleus. It is determined by the subshells the electrons are bound to, which are in turn determined by the principal quantum number ("n") and azimuthal quantum number ("l"). The electrons within an atom are attracted to the protons in the nucleus of that atom. The number of electrons bound to the nucleus is equal to the number of protons in the nucleus, which is called the atomic number ("Z"). The electrons are attracted to the nucleus by this mutual attraction and are bound to the nucleus. The electrons within an atom are attracted to each other and this attraction determines the electron configuration. The electron configuration is described by the term symbol, which is the letter used to identify each subshell.

16:45

In physics, the wave–particle duality is the concept that every object or process, no matter how large or how small, behaves as both a wave and a particle. The wave–particle duality is one of the central concepts in quantum mechanics.

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So this question we're talking about potential situations involving quantum numbers. So in the first situation, they want us to figure out the lowest possible principal quantum number. So n when l is equal to one. Okay, so when l is equal to one, that means we have a P orbital so and the rules for P or bills, you can only have a P orbital when n is greater than or equal to to. So by that logic, the lowest possible principal quantum number would be to considering that El is one. The second scenario they want us to figure out how many. Excuse me. Just a second. They want us there. They want to know what are the possible angular momentum Quantum numbers? L When m sub l, the magnetic resonance quarter number is equal to zero and in is less than or equal to four. So when m sub l is equal to zero, you know, for the rules EMC Bell is equal to for is from negative Bell to l. That's the range. So if l If am Sabella's zero, we have a zero in the ranges When l is one to three in zero. So when l a zero We have an s or Bill when they haven't s orbital, the Onley m sub l is zero because l A zero when we have a P orbital, that means l is gonna be one. And M Sevele can be negative 10 or one when l is equal to two. That means M Isabelle could be negative to negative 101 or two and the same thing for when Ellis three. So the reason we have to stop it three is because when n is less than or equal to four. So the highest orbit we can have would be when n is equal to four and that is would be an F orbital. And that is because the rule for F horribles is and must be greater than or equal to four. So we have to stop it for But we can go all the way from yes e d and F. So any of these four options will have an M sub l equal to zero when the principal corner number end is less than or equal to four

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