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Homework Help: Bohr Model applied to Excitons

  1. Oct 22, 2016 #1
    An exciton is a bound electron-hole pair (in a semiconductor). For this problem, think of an exciton as a hydrogen-like atom, with a negatively charged electron and positively charged hole orbiting each other.

    The permittivity of free space (ε0) is replaced with permittivity of the semiconductor (ε = 12).
    The mass of the electron is replaced with the effective mass of the electron-hole pair.

    1. The problem statement, all variables and given/known data (bold below is what I really need help on)

    A) Estimate the radius in nm and the ground state energy in eV for an exciton in Si.

    B) Approximately how large is the separation between atoms in a crystal of silicon? How does the radius compare with this number?

    C) Silicon atoms have an average kinetic energy of T*kB. How does the exciton binding energy (E1) compare with this number? What does this mean?

    D) All this is about electrostatic potential energy. Prove that it's reasonable to neglect the gravitational potential energy.

    me = 9.1*10-31
    eV = 1.602×10−19 J (N*m)
    h = 6.626*10-34
    ħ = 1.055*10-34
    a0 = 0.0529 nm
    ε*ε0 = 1.0359*10-10

    permittivity of silicon = εSi = kSiε0 where k = dielectric constant

    effective masses
    me* = 0.26me
    mh* = 0.36me

    2. Relevant equations
    r = mek2e4/(πħ3)

    me*mh*/(me* + mh*) = 0.15me

    r = n2h2*1.0359*10-10/(z*π*meffectivee2)

    3. The attempt at a solution

    12*a0/0.15 = 4.2nm (n2/z)


    Ry = -13.6eV

    B) I do not remember chemistry much. How is the separation between silicon atoms found?

    C) 300*1.38*10-23 m2kg/s2 = 4.14*10-23
    E1 = ?

    D) PEelectrostatic = kqQ/r
    FE = qE
    Fg = mg
    ∴ qE = mg
    & as long as qE/m is much larger than g, gravity can be ignored.

  2. jcsd
  3. Oct 22, 2016 #2


    User Avatar

    Staff: Mentor

    You get the same Rydberg constant as for hydrogen?

    Google :smile: There is no way to find that by first principles, so just Google "silicon interatomic distance".

    That should be the binding energy. Do you remember how to find that for hydrogen?
  4. Oct 22, 2016 #3
    Oh yes, sorry, with ε:
    The energy (for part A) is mee4/[8(h*1.0359*10-10)2]
    = (9.1*10^-31)(e^4)/[8*((6.626*10^-34)*(1.0359*10^-10))^2]
    = 1.32*1057
    Was that suppose to be a different mass?

    Thank you. I did Google it. Although, I either searched the wrong phrase or thought it wasn't simple from the results that came up earlier.
    atomic radius = 0.132 nm
    lattice parameter = 0.543 nm
    nearest neighbor distance = 0.235 nm
    Looks like the calculation before (4.2 nm) is wrong?

    The binding energy of an electron to the nucleus in the hydrogen atom is 13.6 eV.
    So the binding energy is the absolute value of Rydberg energy??
  5. Oct 23, 2016 #4
    BE = (938MeV/c2 + 0.511MeV/c2 - 938.3MeV/c2)*c2 =
    (938.484+ 0.511 - 938.783)*10^6
    = 212keV

    938.783 MeV (mass of hydrogen)
    1.673e-27 kg (mass of proton)
    >> 1 eV = 1.602e-19 J <<
    >> 1 J = m3 kg / s2 <<
    = 9.3989e8
    = 939.89e6
    = 939.89 MeV
  6. Oct 24, 2016 #5

    A) The 4.2nm (n2/z) is correct. The energy is found this way: E = m*e4/(8h2ε2), but
    B) 0.235 nm or 0.543 nm means that there are multiple atoms in-between each
    C) This is the same E1, but can be modified from more general E1 = me4/(8h2ε02) = -13.6 eV... E = E1*(mT/m)/[(ε/ε0)2]
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