511014moleculescm3 usingequation12

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Unformatted text preview: 0 Q4: Jc = 1 nv a 4 Jc = 4.56 * 1018 = molecules/cm2‐sec (given in question) ⎛ 3RT ⎞1 / 2 va = ⎜ ⎟ ⎝M⎠ If we use nitrogen gas, then M = 28g/mole. va = 5.2 * 104 cm/sec Substitute Jc and va into the first equation and obtain n = 3.51 * 1014 mo...
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This note was uploaded on 03/04/2013 for the course MASE 426 taught by Professor Pirouz during the Spring '09 term at Case Western.

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