How many electrons are in n = 2? What about n = 4, l = 3? What about n = 6, l = 2, m_l = -1?
1 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.
figures.boundless.com
1^"st" set-> n=2
The principal quantum number,
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 amaximum 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,
Now, the subshell is given by the angular momentum quantum number,
l=0 -> the s-subshelll=1 -> the p-subshelll=2 -> the d-subshelll=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,
Since you know the exact orbital, it follows that only two electrons can share these three quantum numbers, one having spin-up,