How many electrons have $n = 3$ $n = 3$ , $m_{l} = 0$ $m_l = 0$ , and $m_{s} = + \frac{1}{2}$ $m_s = +1/2$ ?

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How many electrons have $n = 3$ $n = 3$ , $m_{l} = 0$ $m_l = 0$ , and $m_{s} = + \frac{1}{2}$ $m_s = +1/2$ ?

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Answer 1 · 2015-10-04T16:35:20

Three electrons.

Explanation:

The idea here is that each electron in an atom has an unique set of quantum numbers, which describes exclusively that electron.

The four quantum numbers are

![http://socratic.org/questions/what-are-quantum-numbers]( useruploads.socratic.org )

In your case, the only quantum number that is not accounted for is the angular momentum quantum number, $l$ $l$ , which can take values from $0$ $0$ to $(n - 1)$ $(n-1)$ , where $n$ $n$ is the principal quantum number.

So, the question wants you to find the number of electrons that

can be located on the third energy level, since $n = 3$ $n=3$ ;

can be located in an orbital that has a specific orientation, since $m_{l} = 0$ $m_l=0$

have spin-up, since $m_{s} = + \frac{1}{2}$ $m_s = +1/2$

Start by finding how many orbitals can accomodate these electrons. SInce you know what the relationship of $l$ $l$ is to $n$ $n$ , you can say that

$l = 0, 1, ..., (n-1) implies l = {0; 1; 2}$

For $l=0$ , which corresponds to the s-orbital, you can have one electron located in the 3s-orbital that has spin-up.

For $l=1$ , which corresponds to the 3p-orbitals, you can have an electron located in the $3p_y$ orbital, for which $m_l=0$ , that has spin-up.

Finally, for $l=2$ , which corresponds to the 3d-orbitals, you can have an electron located in the $3d_(yx)$ , which would correspond to $m_l=0$ , that has spin-up.

Therefore, these quantum numbers can correspond to a total of three electrons

$n=3" " l=0" "m_; = 0" "m_s = +1/2$

$n=3" "l=1" "m_l = 0" "m_s = +1/2$

$n=3" "l=2" "m_l = 0" "m_s = +1/2$

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How many electrons have $n = 3$ $n = 3$ , $m_{l} = 0$ $m_l = 0$ , and $m_{s} = + \frac{1}{2}$ $m_s = +1/2$ ?

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Explanation:

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How many electrons have n = 3n=3n = 3, m_l = 0ml=0m_l = 0, and m_s = +1/2ms=+12m_s = +1/2?

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How many electrons have $n = 3$ $n = 3$ , $m_{l} = 0$ $m_l = 0$ , and $m_{s} = + \frac{1}{2}$ $m_s = +1/2$ ?