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