Question #7c3c0

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Question #7c3c0

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Answer 1 · 2016-12-08T11:23:06

Here's what I got.

Explanation:

I'm assuming that you're supposed to figure out how many electrons can share the two quantum numbers given to you

$n = 4$ $n =4" "$ and $l = 0$ $" "l=0$

As you know, we use four quantum numbers to describe the location and spin of an electron in an atom.

![figures.boundless.com]( useruploads.socratic.org )

The principal quantum number, $n$ $n$ , describes the energy level on which the electron resides. In other words, the value of $n$ $n$ tells you the energy shell on which you can find the electron.

In your case, $n = 4$ $n=4$ means that the electron is located on the fourth energy level.

The angular momentum quantum number, $l$ $l$ , describes the subshell in which the electron resides. More specifically, you have

$l = 0 \to$ $l = 0 ->$ the s subshell

$l = 1 \to$ $l=1 ->$ the p subshell

$l = 2 \to$ $l=2 ->$ the d subshell

$l = 3 \to$ $l = 3 ->$ the f subshell

In your case, $l = 0$ $l=0$ means that the electron is located in the s subshell.

Now, the maximum number of electrons that can occupy the s subshell, regardless of the energy level on which they're located, is equal to $2$ $2$ .

This is the case because the s subshell can only hold one orbital, as given by the magnetic quantum number, $m_{l}$ $m_l$ , which for $l = 0$ $l=0$ is equal to $0$ $0$ .

Moreover, each individual orbital can hold a maximum of $2$ $2$ electrons, one having spin-up, or $m_{s} = + \frac{1}{2}$ $m_s = +1/2$ , and the other having spin-down, or $m_{s} = - \frac{1}{2}$ $m_s = -1/2$ .

Therefore, you can say that a maximum of $2$ $2$ electrons can share the quantum numbers

$n = 4$ $n = 4" "$ and $l = 0$ $" "l = 0$

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Question #7c3c0

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