243S2008-08-soln

# 243S2008-08-soln - EE243 Sp 2008 HW#8 Solutions Problem 1(a...

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EE243, Sp 2008 HW#8 Solutions Problem 1 (a) The flow rate at 1250 °C(1523K) = 1.5 liter/min × (1523/300 ) = 0.127 l/sec Gas stream velocity U = flow rate/tube cross-sectional area = 0.127 × 10 3 /50 = 2.54 cm/sec (b) Density of SiCl 4 = 170 gm/mole 22.4 × 10 3 cm 3 /mole × (1523/300) = 1.36 × 10 -3 gm/cm 3 Density of H 2 = 2 gm/mole 22.4 × 10 3 cm 3 /mole × (1523/300) = 1.6 × 10 -5 gm/cm 3 Density of gas mixture = 0.02 × 1.36 × 10 -3 + 0.98 × 1.6 × 10 -5 = 4.29 × 10 -5 gm/cm 3 The Reynold Number R e = ρ UL µ = 4.29 × 10 -5 gm/cm 3 × 2.54 cm/sec × (3 × 2.54)cm 3 × 10 -4 gm/cm-sec = 2.76 (viscous flow regime) (c) At 273K, 1 mole (=6.02 × 10 23 molecules) occupies 22.4 litres. At 1523K , N/V= 6.02 × 10 23 × 273 22.4 × 10 3 × 1523 = 4.8 × 10 18 molecules/cm 3 C T (SiCl 4 ) = 0.02 × 4.8 × 10 18 = 9.6 × 10 16 molecules/cm 3 (d) δ ~ 2 3 L R e L = 3.05 cm h G = D G / δ = 2.62 cm/sec Since 1 molecule of SiCl 4 gives 1 Si atom and the growth rate is mass-transfer limited, dy dt = = 1 ( 1 k s + 1 h G ) C T 5 × 10 22 h G C T 5 × 10 22 = 2.62 × 9.6 × 10 16 5 × 10 22 = 5.03 × 10 -6 cm/sec = 500 Å/sec Problem 2 The Grove CVD model states the growth rate : dy dt = = 1 ( 1 k s + 1 h G ) C T ρ Y (a) dy dt

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243S2008-08-soln - EE243 Sp 2008 HW#8 Solutions Problem 1(a...

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