Homework 04

# Homework 04 - = 0.02 eV/nm Using a numerical procedure...

This preview shows page 1. Sign up to view the full content.

EEE 434/591—Fall 2010 Homework No. 4 13. For a finite potential well of width 20 nm and height 0.3 eV (use an effective mass appropriate to GaAs electrons: m = 0.067m 0 ), compute the energy needed to ionize an electron (move it from the lowest energy level to the top of the well). If a 0.4 eV photon excites the electron out of the well, what is its kinetic energy? What is its wavelength? We can compute β from (2.65) to be 14.51, so that there are a great many quantized levels in this well. The lowest level is found from Fig. 2.5, for which we estimate ξ 1 = 1.469 E 1 = 3.01 meV Hence, the ionization energy is 296.9 meV. The kinetic energy will be E kin = 0.4 0.2969 = 103 meV λ = h 2 mE kin = 14.8 nm 15. Consider a potential well with V 0 for x < 0, and V(x) = Fx for x > 0, with F
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: = 0.02 eV/nm. Using a numerical procedure, compute the ten lowest energy levels in the quantum well. From the bound states tool, the lowest 10 energy levels are at 0.143, 0.25, 0.337, 0.414, 0.485, 0.551, 0.613, 0.672, 0.728, 0.783 eV. 591 : The lowest four zeroes of the Airy function are approximately -2.338, -4.088, -5.52, and -6.787. These correspond to the energies s = − z = 2 mE 2 2 2 meE s Λ Ν Μ Ξ Π Ο 2 / 3 and the last factor has the value 9.33 × 10-18 for GaAs parameters and the given field. Hence, this can be used to give the comparison Level Airy Numerical Tool 1 0.143 0.143 2 0.25 0.25 3 0.337 0.337 4 0.415 0.414...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online