Question #01fe2

1 Answer
Dec 10, 2016

Answer:

#l = 0, 1, 2, 3#.
#m_l = -3, -2, -1, 0, 1, 2, 3#
#m_s = +1/2# or #-1/2#

Explanation:

I assume that you're talking about the quantum numbers.

So, if you're not familiar with the concepts behind quantum numbers, I'd recommend you take a look at this video first. If you're good, let's go...

So to quickly recap here's what each quantum number describes:

#n# = energy level.
#l# = subshell; ranges from 0 up to #n-1#
#m_l# = orbital; ranges from #-l# to #l#
#m_s# = tells the spin of the electron; is either #+1/2# or #-1/2#

Now, your question asks what are the possible values for #l#, #m_l#, and #m_s# for #n = 4#. So, let's just work through this from top to bottom.

Firstly, if #n=4#, let's consider what #l# could be. We know that #l# ranges between 0 to #n-1#. Therefore, we can say that:

#l = 0, 1, 2, 3#.

Side note: this essentially means that the n=4 energy level can hold electrons in the s, p, d and f subshells

Okay, so now that we know all possible values of #l#, let's look at #m_l#. We know that #m_l# ranges between #-l# and #l#. But which of the above #l#'s do we use? Well, consider this: if we just looked at the highest value of #l# present, it would take care of all the lower ones. See what I mean below:

#m_l = -3, -2, -1, 0, 1, 2, 3#

See how values like #-2# and #1# are taken care of?

So, that's really it. #m_s# will only be either #+1/2# or #-1/2# regardless of the other 3 quantum numbers.

Hope that helped :)