How many orbitals are available with the following combination of quantum
numbers?
1) n = 2; l = 0
2) n = 3; l = 2
3) n = 4; l = 3
I have figured that it is
1) 2s
2) 3d
3) 4f
but i don't know what to do after this.
the number of orbitals in energy level n is n^2
a) n=2, n^2=4 ( 2s,2px,2py,2pz)
b) n=3, n^2=9 (3s,3px,3py,3pz,3d1,3d2,3d3,3d4,3d5)
To determine the number of orbitals available with a given combination of quantum numbers, you need to use the formula:
Number of orbitals = (2l + 1)
Here's how you can apply the formula to each of the given combinations:
1) For n = 2 and l = 0:
Using the formula, the number of orbitals would be:
Number of orbitals = (2 × 0 + 1) = 1
So, with n = 2 and l = 0, there is only one orbital available, which is 2s.
2) For n = 3 and l = 2:
Using the formula, the number of orbitals would be:
Number of orbitals = (2 × 2 + 1) = 5
So, with n = 3 and l = 2, there are five orbitals available, labeled as 3d.
3) For n = 4 and l = 3:
Using the formula, the number of orbitals would be:
Number of orbitals = (2 × 3 + 1) = 7
So, with n = 4 and l = 3, there are seven orbitals available, labeled as 4f.
Therefore, your initial identification of the orbitals is correct:
1) n = 2; l = 0 → 2s orbital
2) n = 3; l = 2 → 3d orbitals
3) n = 4; l = 3 → 4f orbitals
You were on the right track! The next step was to apply the formula (2l + 1) to calculate the number of orbitals for each combination of quantum numbers.