1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Quantum Dot

  1. Mar 18, 2008 #1
    1. The problem statement, all variables and given/known data
    A “quantum dot” is a semiconductor device that may be modeled as an electron with an effective mass m that is confined in an infinite spherical well. Suppose that the spherical well has radius a and the electron has effective mass [itex] m = f m_e[/itex], where f is some real number and [itex]m_e[/itex] is the mass of the electron.

    What is the energy difference between the ground state and the first excited state, expressed as a function of a and m.

    2. Relevant equations
    The Radial Schrodinger equation:
    [tex] -\frac{\hbar ^2}{2m} \frac{d^2 u}{dr^2} + \displaystyle \left[ V + \frac{\hbar ^2}{2m} \frac{l (l+1)}{r^2} \right] u = E u [/tex]

    if [itex] l=0[/itex]
    [tex]E_{n0} = \frac{n^2 \pi^2 \hbar^2}{2ma^2}[/tex]

    if [itex] l \neq 0 [/itex]
    [tex] E_{nl} = \frac{\hbar^2}{2ma^2} \beta^2_{nl} [/tex]
    where [itex] \beta_{nl} [/itex] is the nth zero of the lth spherical Bessel function.

    3. The attempt at a solution

    Now the zeroth Spherical bessel function will give me the same energy solution as the l=0 case and so we're consistent. My issue is that I'm not terribly sure which is the proper ground state. Is [itex] l = n =0 [/itex] the ground state or is [itex] n = 0, l\neq 0 [/itex] the ground state for arbitrary l. I know that l refers to the orbital angular momentum of (in this case) the electron, so I would assume we would need to take arbitrary l into consideration but I'm really not too sure.
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you help with the solution or looking for help too?



Similar Discussions: Quantum Dot
Loading...