Sketch the energy levels of the hydrogen atom for n=1,2,3,4 and 5 versus n and l. Label the levels both spectroscopically (1s, 2s, 2p,…) and by their n and l values. Indicate the total degeneracy for each l level and add up the total degeneracy for each n level.

From the above

$ψ(r ⃗ )=ψ_{n l m} (r ⃗ )=R_{n l} (r)*Y_l^m (θ,φ)$

For each principal quantum number n there will be an energy $E_n$ they will get a small correction term from the angular quantum number $l$,where $l=0,..,n-1$. ($l$ takes $n$ values)

$l=0$ are the s orbitals,$l=1$ are the p orbitals, $l=2$ are the d orbitals,etc.

For each $l$ level there are possible $2l+1$ values of $m=-l,-l+1,…,0,…,l-1,l$ and for each value of m there are possible 2 values of the spin. So the m degeneracy of an l orbital is $2(2l+1)$ and the total degeneracy for each n level is $∑_{(l=0)}^{(n-1)} 2(2l+1)=2n^2$

The figure with how these levels are situated in energy is:

Make a composite sketch showing the l=0,1,2,and 3 effective potentials for the hydrogen atom. Add the locations of the n=1,2,3 and 4 energy levels to your effective potential sketch, and explain how the levels in the different wells are lined up.

Recall that the effective potential in the hydrogen atom is

$V_eff (r)=V(r)+\frac{(ℏ^2 l(l+1))}{(2μr^2)}=-\frac{(k e^2)}{r}+\frac{(ℏ^2 l(l+1))}{(2mr^2)}$ where $k=1/(4πϵ_0)$ and $m≈m_e$

For $l=0,1,2,3$ the graph of $V_eff (r)$ is presented. In black there are the energy levels $(n=1,2,3,4)$: Explanation (how the levels are lined up):

It can be easy seen that for example for $n=1$ there is possible just one level (for $l=0$) since the potential well corresponding to $l=1$ (light blue) is above the $n=1$ energy. For $n=2$ , since this energy is above the wells of $l=0$ and $l=1$ and below $l=2$ (green) and $l=3$ (violet) there are possible just two values of l (0 and 1). This is why for example for the first energy level (n=1) there is possible just one orbital (1s). The existence of the other superior orbitals (1p, 1d, etc) is forbidden by the fact that their energy wells (in which they need to lie) are above the $n=1$ energy.