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I do not get how the notes get 8•10^15. can you please show me how it got this? Thank you.

I do not get how the notes get 8•10^15. can you please show me how it got this? Thank you.

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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I do not get how the notes get 8•10^15. can you please show me how it got this? Thank you.
Example 18-7 (Semiconductor charge carrier analysis) A silicon wafer (Group IVA) is doped with 2 x 10 phosphorus (Group VA) atoms/cm, 4 x101° boron (Group IlIIA) atoms/cm° and 3 x 101 arsenic (Group VA) atoms/cm. Calculate the electron and hole concentrations (carriers per cm.) Given: n = 1.5 x 10/m. EX 18-7 P ATOMS/cc 16 AS ATOMS/Cc 16 2x1o SILI CON N= 4 x10 %3D B ATOMS h = 1,5 x10/cc+3×1o Think of acceptor and donor atoms canceling each other: N.- Na 4 x 101 -3.2 x 10-0.8 x 10/cm It looks like 0.8 x 10 acceptor atoms/cm are added to silicon. NET RESULT: 16 Na = 0.8x/0 I ATOMS si GET P-7YPE %3D CC SEMI CONDLU C7OR p>>n Nd=0 LG IS NEU TRALITY: Na + p →pãN, = |8x 10 HOLES NEGLECT Majority carrier concentration MASS ACTION LAW: np = n." Quick check: Solve for n: りミり。 (1,5 x10 2.81 X 10 e- p>>n %3D 8x10ʻ5/cc Minority carrier concentration
Transcribed Image Text:Example 18-7 (Semiconductor charge carrier analysis) A silicon wafer (Group IVA) is doped with 2 x 10 phosphorus (Group VA) atoms/cm, 4 x101° boron (Group IlIIA) atoms/cm° and 3 x 101 arsenic (Group VA) atoms/cm. Calculate the electron and hole concentrations (carriers per cm.) Given: n = 1.5 x 10/m. EX 18-7 P ATOMS/cc 16 AS ATOMS/Cc 16 2x1o SILI CON N= 4 x10 %3D B ATOMS h = 1,5 x10/cc+3×1o Think of acceptor and donor atoms canceling each other: N.- Na 4 x 101 -3.2 x 10-0.8 x 10/cm It looks like 0.8 x 10 acceptor atoms/cm are added to silicon. NET RESULT: 16 Na = 0.8x/0 I ATOMS si GET P-7YPE %3D CC SEMI CONDLU C7OR p>>n Nd=0 LG IS NEU TRALITY: Na + p →pãN, = |8x 10 HOLES NEGLECT Majority carrier concentration MASS ACTION LAW: np = n." Quick check: Solve for n: りミり。 (1,5 x10 2.81 X 10 e- p>>n %3D 8x10ʻ5/cc Minority carrier concentration
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