# How many electrons are in #n = 2#? What about #n = 4, l = 3#? What about #n = 6, l = 2, m_l = -1#?

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

Since the question is a bit ambiguous, I will assume that you're dealing with *three distinct sets* of quantum numbers.

In addition to this, I will also assume that you're fairly familiar with quantum numbers, so I won't go into too much details about what each represents.

#1^"st"# set# -> n=2#

The *principal quantum number*, **orbitals** you have in this energy level.

The number of orbitals you get *per energy level* can be found using the equation

#color(blue)("no. of orbitals" = n^2)#

Since each orbital can hold a**maximum** of two electrons, it follows that as many as

#color(blue)("no. of electrons" = 2n^2)#

In this case, the second energy level holds a total of

#"no. of orbitals" = n^2 = 2^2 = 4#

orbitals. Therefore, a maximum of

#"no. of electrons" = 2 * 4 = 8#

electrons can share the quantum number

#2^"nd"# set#-> n=4, l=3#

This time, you are given both the *energy level*, **subshell**,

Now, the **subshell** is given by the *angular momentum quantum number*,

#l=0 -># *the s-subshell*#l=1 -># *the p-subshell*#l=2 -># *the d-subshell*#l=3 -># *the f-subshell*

Now, the number of **orbitals** you get *per subshell* is given by the *magnetic quantum number*,

#m_l = -l, ..., -1, 0, 1, ..., +l#

#m_l = {-3; -2; -1; 0; 1; 2; 3}#

So, the f-subshell can hold total of **seven** orbitals, which means that you have a maximum of

#"no. of electrons" = 2 * 7 = 14#

electrons that can share these two quantum numbers,

#3^"rd"# set#-> n=6, l=2, m_l = -1#

This time, you are given the energy level, **exact orbital**,

Since you know the exact orbital, it follows that only **two electrons** can share these three quantum numbers, one having spin-up,