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asambeis [7]
3 years ago
7

Assuming dopant atoms are uniformly distributed in a silicon crystal, how far apart are these atoms when the doping concentratio

ns are 5.0x1015 cm-3 and 5.0x1020 cm-3?
Chemistry
1 answer:
enyata [817]3 years ago
8 0

Answer:

d =~ 5.8μm

d =~ 0.13 μm

Explanation:

when the doping concentrations are 5 × 10^15 cm^-3

d = v^-1/3  ; where d represent the distance between the atoms , and v  represent the volume

d =1/ ∛v

d = 1/ ∛5 × 10^15

d = 1/ 170997.5

d = 5.85 × 10 ^ -6

d =~ 5.8μm

when the doping concentrations are 5 × 10^20 cm^-3

d = v^-1/3  ; where d represent the distance between the atoms , and v  represent the volume

d =1/ ∛v

d = 1/ ∛5 × 10^20

using the principle of surds and standard forms, we have

d = 1/ ∛0.5 × 10^21  

d = 1/7937005.26

d = 1.26 × 10 ^ -7

d = 0.126 × 10 ^ -6

d =~ 0.13 μm

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             since  pH = -log[H₃O⁺]

                           11.38 =  -log[H₃O⁺]

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                  pOH = -log[OH⁻]

                     2.62  = -log[OH⁻]

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                     [OH⁻] = 10^{-11.19}

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