Orbital Quantum Numbers And Total Electron Energy

[BOAS]

Hello

Homework Statement

The orbital quantum number for the electron in the hydrogen atom is l = 4. What
is the smallest possible value (in eV) for the total energy of this electron? (Use the
quantum mechanical model of the hydrogen atom.)

Homework Equations

The Attempt at a Solution

I know that the angular momentum of the electron is given by;

$L = \sqrt{l(l + 1)}\frac{h}{2 \pi}$

$L = \sqrt{20} \frac{h}{2 \pi}$

L = 4.64x10^-33 Kgm²s^-1

My textbook doesn't really discuss the QM picture of the atom, so I don't know how to relate this to the energy of the electron.

I know how to do it for the Bohr model, but clearly that's no good.

I appreciate any help you can give,

thanks!

<EDIT>

Oops.

" In fact, calculating the energy from the quantum mechanical wave function gives the expression Bohr derived for the energy:"

This thread can be ignored/deleted. sorry.

 

[DrClaude]

The energy of the electron is given by the Bohr equation
$$
E_n = - \frac{\mathrm{Ry}}{n^2}
$$
where $\mathrm{Ry} \approx 13.6\ \mathrm{eV}$ is the Rydberg constant (expressed in units of energy).

This energy is independent of the angular momentum quantum number $l$. However, there is the constraint that $l < n$. Therefore, if $l = 4$, the lowest value of $n$ is 5, and hence the lowest energy is
$$
E_5 = - \frac{13.6\ \mathrm{eV}}{5^2} = 0.544\ \mathrm{eV}
$$