What is the orbital angular momentum quantum number for the orbital 5#p#?

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I don't understand anything about it. I don't know where to start.

1 Answer
Oct 11, 2016

Answer:

#l=1#

Explanation:

figures.boundless.com

The first thing to note here is that you don't need to know the value of the principal quantum number, #n#, to answer this question.

In other words, you don't need to know the energy level on which the electron resides, which in this case would be #n=5#, to say that the angular momentum quantum number, #l#, for a #p# subshell is equal to #1#.

The #5p# subshell given to you is described by two quantum numbers

  • #n=5 -># this means that the electron is located on the fifth energy level

  • #p -># the means that it is located in the p subshell

Now, the angular momentum quantum number designates the identity of the subshell in which the electron is located

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

In this case, the angular momentum quantum number must be equal to #1# because #1# is the value that describes the #p# subshell for any electron located on an energy level that is #n > 1#.

As you can see from the table, for #n=1#, #l# can only take one value, #l=0#. This means that the first energy level holds #1# subshell, the #s# subshell.

By comparison, for #n=5#, #l# can take

#n = 5 implies l = {0, 1, 2, 3,4 }#

The fifth energy level holds #5# subshells, and #l=1# describes the #p# subshell regardless of the value of #n > 1#.