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Archived Orbital Quantum Numbers And Total Electron Energy

  1. Mar 24, 2014 #1
    Hello :smile:

    1. The problem statement, all variables and given/known data

    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.)

    2. Relevant equations



    3. The attempt at a solution

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

    [itex]L = \sqrt{l(l + 1)}\frac{h}{2 \pi}[/itex]

    [itex]L = \sqrt{20} \frac{h}{2 \pi}[/itex]

    L = 4.64x10-33 Kgm2s-1

    My text book 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.
     
    Last edited: Mar 24, 2014
  2. jcsd
  3. Feb 5, 2016 #2

    DrClaude

    User Avatar

    Staff: Mentor

    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
    $$
    E_5 = - \frac{13.6\ \mathrm{eV}}{5^2} = 0.544\ \mathrm{eV}
    $$
     
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