- #1
jeebs
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Here's the problem:
A GaAs quantum well of thickness 10 nm is held in a cryostat at a low temperature and is excited with a powerful ultrashort laser pulse at time t = 0. The laser pulse injects 1 × 1015 m−2 electrons and holes per unit area of the quantum well into the conduction and valence bands respectively. The band gap of GaAs is 1.519 eV, and the effective masses of the electrons and heavy holes are 0.067 me and 0.4 me
respectively.
(i) Calculate the Fermi energies of the electrons and holes at t = 0. (ii) Sketch the emission spectrum that you would expect to observe just after the crystal has been excited by the laser, identifying clearly the minimum and maximum photon energies that would
be emitted.
(iii) Calculate the modulus of the maximum k vector of the electron within the plane of the quantum well and compare it to the k vector inside the crystal of the photon that is emitted. (The refractive index of GaAs is 3.5.)
Throughout this question you may ignore the light hole band and assume that excitonic effects are negligible. The density of states per unit area for a free particle of mass m in two dimensions is given by [tex] g(E) = \frac{m}{\pi\hbar^2} [/tex]
So, the particles will have their free motion behaviour in 2 directions with a wavevector kxy, and be confined in the 3rd direction. This gives the electron energy [tex] E_e = E_g + \frac{\hbar^2 n^2 \pi^2}{2m_e^*d^2} + \frac{\hbar^2 k_x_y^2}{2m_e^2} [/tex] where Eg is the band gap, d is the well width, me* is the effective electron mass. Also the holes have energy
[tex] E_h = \frac{\hbar^2 n^2 \pi^2}{2m_h^*d^2} + \frac{\hbar^2 k_x_y^2}{2m_h^2} [/tex]
so, for the first bit, what I've said is that the Fermi energy is when all the electrons/holes occupy lowest states, the Fermi energy is the highest occupied state. Looking at the density of states, that is constant for any energy so when I calculate it it gives me an answer in the region of 1037 states m-2. Since there are only 1015 electrons and holes per square metre, I've said that when they all occupy the lowest states, they can all be in the n=n'=1 and kxy = 0 level.
This gives me a Fermi level of 0.94eV for the holes and 7.14eV for the electrons (is this on the high side?)
Is this allowed? I have my reservations about this, since electrons, being fermions, shouldn't be in the same energy state. I couldn't think of any other way to get the Fermi levels though...
anyway, for the next two parts, they involve finding the maximum energy of the confined particles. This I am having trouble with, because I am not told the height of the potential well. I do not know what the maximum energy is that a particle can have and still be confined, so I cannot get the maximum peak on the emission spectrum (part 2), or the maximum k vector of the electron within the well. I know I'd be able to find the maximum n and n' if I was told what the band gap of the material forming the walls of the well, but I'm not...
Not sure where to go from here...
A GaAs quantum well of thickness 10 nm is held in a cryostat at a low temperature and is excited with a powerful ultrashort laser pulse at time t = 0. The laser pulse injects 1 × 1015 m−2 electrons and holes per unit area of the quantum well into the conduction and valence bands respectively. The band gap of GaAs is 1.519 eV, and the effective masses of the electrons and heavy holes are 0.067 me and 0.4 me
respectively.
(i) Calculate the Fermi energies of the electrons and holes at t = 0. (ii) Sketch the emission spectrum that you would expect to observe just after the crystal has been excited by the laser, identifying clearly the minimum and maximum photon energies that would
be emitted.
(iii) Calculate the modulus of the maximum k vector of the electron within the plane of the quantum well and compare it to the k vector inside the crystal of the photon that is emitted. (The refractive index of GaAs is 3.5.)
Throughout this question you may ignore the light hole band and assume that excitonic effects are negligible. The density of states per unit area for a free particle of mass m in two dimensions is given by [tex] g(E) = \frac{m}{\pi\hbar^2} [/tex]
So, the particles will have their free motion behaviour in 2 directions with a wavevector kxy, and be confined in the 3rd direction. This gives the electron energy [tex] E_e = E_g + \frac{\hbar^2 n^2 \pi^2}{2m_e^*d^2} + \frac{\hbar^2 k_x_y^2}{2m_e^2} [/tex] where Eg is the band gap, d is the well width, me* is the effective electron mass. Also the holes have energy
[tex] E_h = \frac{\hbar^2 n^2 \pi^2}{2m_h^*d^2} + \frac{\hbar^2 k_x_y^2}{2m_h^2} [/tex]
so, for the first bit, what I've said is that the Fermi energy is when all the electrons/holes occupy lowest states, the Fermi energy is the highest occupied state. Looking at the density of states, that is constant for any energy so when I calculate it it gives me an answer in the region of 1037 states m-2. Since there are only 1015 electrons and holes per square metre, I've said that when they all occupy the lowest states, they can all be in the n=n'=1 and kxy = 0 level.
This gives me a Fermi level of 0.94eV for the holes and 7.14eV for the electrons (is this on the high side?)
Is this allowed? I have my reservations about this, since electrons, being fermions, shouldn't be in the same energy state. I couldn't think of any other way to get the Fermi levels though...
anyway, for the next two parts, they involve finding the maximum energy of the confined particles. This I am having trouble with, because I am not told the height of the potential well. I do not know what the maximum energy is that a particle can have and still be confined, so I cannot get the maximum peak on the emission spectrum (part 2), or the maximum k vector of the electron within the well. I know I'd be able to find the maximum n and n' if I was told what the band gap of the material forming the walls of the well, but I'm not...
Not sure where to go from here...
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