advertisement

Your Name:________________ 3.225 Quiz 2007 e=1.602x10-19 C mo=9.11x10-31 kg c=2.998x108 m/sec εo=8.854x10-12 F/m kb=1.38x10-23 J/K h=6.626x10-34 J-sec ħ=1.054x10-34 J-sec A=6.022x1023 mole-1 1. 1­D Silicon Silicon has a valence 4, a=0.5428nm (a) Using the Drude model, calculate the estimated conductivity along the 1-D wire n= Z 4 ne2 τ = , σ= → calculate using τ = 10-14 s a a m0 (b) What is the plasma frequency? ωp = ne 2 m 0ε 0 c) Now let us have electron waves instead of electron particles in our 1-D Si; determine kF and EF nπ 4 2π , n = ⇒ k F 1−d = 2 a a 2 2 2 2 ℏ kF 2π ℏ = = * 2 2m ma k F 1−d = EF 1−d 1 (d) Derive the density of states for the 1-D Si for free electron waves g (E) = 1-d dN dk 1 2kL 2mE , N= , k= dk dE L π ℏ ⇒ g( E ) = 1-d 2m 1 πℏ E (e) If the band gap of interest is (ћ2π2/(2m0a2)), draw the E vs. band structure for nearly free electron silicon 2 2. 2­D Semiconductor (a) Derive the density of states for a 2-D semiconductor, g(E)=m*/(πћ2) dN dk 1 k 2 L2 2m*E , N= , k= dk dE L 2π ℏ * m 2-d ⇒ g (E) = 2 πℏ g (E) 2-d = (b) Assume that Eg = 1.1eV for this intrinsic 2-D semiconductor (i.e. same as for 3-D Si). Calculate n at room temperature and compare to 3-D Si, where n = 1010 cm-3 for intrinsic material. Assume m*=m0. Number of e- per area in conduction band = n (as 2-d semi conductor) ∞ n= ∫ f ( E ) g ( E )dE, f ( E ) ∼ 2-d EC g 2-d ( E ) = ⇒ n= m* , ⇒ n= πℏ ∞ Eg E-E F exp and E C = E g (as intrinsic) , EF = 2 k bT 2-d ∫ f ( E ) g ( E )dE = EC ∞ Eg E m* exp + dE ∫ πℏ Eg k b T 2k b T Eg m* k b T exp πℏ 2k b T n2-d < n3-d as unlike g3-d(E), g2-d (E) is not an increasing function of E (in fact it is a constant). 3. Dielectric Properties Describe ε(ω) for 1-D Si in the context of the sources of polarization. Just electronic polarization…draw ε(ω) vs ω. Show ωoe etc. 3