a) The E-k relationship about the GaAs conduction-band minimum becomes non- parabolic at energies only slightly...

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a) The E-k relationship about the GaAs conduction-band minimum becomes non- parabolic at energies only slightly removed from Ec and is more accurately described by E- E = ak² bk4 where a and b are both greater than zero. What effect will this have on the effective mass of electrons in the GaAs conduction band? Substantiate your conclusion. b) Electrons in GaAs can transfer from the minimum to the L minima at sufficiently high electric fields. If electrons were to transfer from the minimum to the L minimum shown in Figure 3, would their effective mass increase of decrease? Explain. (The constant-energy surfaces about the L minima are actually ellipsoidal, but for simplicity assume the surfaces to be spherical in answering this question). asd E-E, (eV) 1 -1 T-300 K 0 (V1) Heavy holes Light holes (V2) 1.71 eV L A o To 1.42 eV T Ts 0.40 eV 1.90 eV 0.34 eV r k (wave vector) Figure 3. GaAs bandstructure. A X₂ Xs Split-off band (V3) 4, Ec E₂ a) The E-k relationship about the GaAs conduction-band minimum becomes non- parabolic at energies only slightly removed from Ec and is more accurately described by E- E = ak² bk4 where a and b are both greater than zero. What effect will this have on the effective mass of electrons in the GaAs conduction band? Substantiate your conclusion. b) Electrons in GaAs can transfer from the minimum to the L minima at sufficiently high electric fields. If electrons were to transfer from the minimum to the L minimum shown in Figure 3, would their effective mass increase of decrease? Explain. (The constant-energy surfaces about the L minima are actually ellipsoidal, but for simplicity assume the surfaces to be spherical in answering this question). asd E-E, (eV) 1 -1 T-300 K 0 (V1) Heavy holes Light holes (V2) 1.71 eV L A o To 1.42 eV T Ts 0.40 eV 1.90 eV 0.34 eV r k (wave vector) Figure 3. GaAs bandstructure. A X₂ Xs Split-off band (V3) 4, Ec E₂

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a The effective mass is given by Given So dE dk dE dk2 2ak4bk dE Thus m 2a 12bk ... View the full answer