3.155J6.152J Microelectronic Processing Fall Term. 2003 Bob handley Martin schmidt Problem set 1 solutions Out Sept. 8. 2003 Due Sept 17. 2002 1. a) Mean free path 2= kBI d-p where k=1.38×102,T=293K,p=10mTor=133Pa,d=1A=100m 38×10-3(293 6.9cm ford=1A Jr(m)(332)017mfrd=2A b) Use the ideal gas law pV=nkB T 1.33Pa vk7(138×102X90)33×10m-33×10cm3 P c) Flux on a surface P 2IkBT where m=mass of particle =(28 amu)(1.67x 10-kg/amu)=4.76 x 10-kg J, =3.86x 1022 molecules 39×108cm2s- m2s d) We are trying to find the average molecular speed of n2 molecules impinging a surface in the chamber. two ways to do this 2.3=2.3×104 L 2351 3.155J/6.152J Microelectronic Processing Fall Term, 2003 Bob O'Handley Martin Schmidt Problem set 1 Solutions Out Sept. 8, 2003 Due Sept.17, 2002 1. a) Mean free path l = kBT 2pd2 p where kB = 1.38 ¥ 10-23, T = 293 K, p = 10 m Torr =133 Pa, d ª 1A = 10-10 m l = 1.38 ¥10-23 ( )( ) 293 K 2p 10-10 ( ) m 2 ( ) 1.33Pa = 6.9cm 1.7cm Ï Ì Ô Ó Ô b) Use the ideal gas law pV = N kB T c) Flux on a surface: Jx = p 2pkBTm where m = mass of particle = ( 28 amu) ( 1.67 ¥ 10 –27 kg/amu) = 4.76 ¥ 10 -26 kg Jx = 3.86 ¥1022 molecules m2 s = 3.9 ¥1018 cm-2 s -1 d) We are trying to find the average molecular speed of N2 molecules impinging a surface in the chamber. Two ways to do this: vx = 2J n = 2.3 m s = 2.3¥104 cm s or vx = 2kBT pm = 235 m s for d = 1 A for d = 2 A N V = p kBT = 1.33Pa (1.38¥10-23 ) 293 ( ) K = 3.3¥1020 m-3 = 3.3¥1014 cm-3