How many degenerate quantum states are there in the 3d state of a hydrogen atom?

[alfredbester]

In the hydrogen atom, an electron is in the 3d state.

(i) Find the orbital angular momentum of the electron (in units of
n =3, l = n - 1, l = 1. L = [sqrt( l (l + 1) )]hbar therefore L = sqrt(2).hbar

(ii) Find the energy of the electron (in eV).

En = -13.6ev / n^2. E = - 13.6eV / 9 (iii) Ignoring electron spin, find the total number of quantum states that have the same energy as in (ii). (i.e. they're degenerate).

This is where I'm stuck, I'm thinking that the only states with the same energy are ones in 3s or 3p shells.

 

[anathema]

But I'm not sure how to calculate the total number of states without taking spin into account.

I would like to clarify that the total number of quantum states with the same energy as in (ii) includes both the orbital and spin quantum numbers. This is because the energy of an electron in an atom is determined by both its orbital and spin properties, and these quantum numbers cannot be separated when considering the total number of states with the same energy.

To calculate the total number of states, we can use the formula:

N = (2l + 1)(2s + 1)

Where l is the orbital quantum number and s is the spin quantum number. In this case, the electron is in the 3d state, so l = 2. The spin quantum number can have two values (spin up or spin down), so s = 1/2.

Plugging in these values, we get:

N = (2(2) + 1)(2(1/2) + 1) = 5(2) = 10

Therefore, there are 10 quantum states with the same energy as the electron in the 3d state. This includes both the 3s and 3p states, as well as the different spin orientations.

 

[kyle8921]

So, there would be 3 states in 3s (m=0,1,-1) and 5 states in 3p (m=0,1,-1,-2,2), for a total of 8 degenerate states. However, I'm not entirely sure if this is correct. It would depend on the specific definition of "same energy" in this context.