Suppose a pure si crystal has 5 × 10^28 atoms m–3. It is doped by 1 pp...
Introduction:
In this scenario, we have a pure silicon crystal that is doped with a concentration of 1 ppm (parts per million) of pentavalent arsenic (As). We need to calculate the number of electrons and holes in the crystal, given that the intrinsic carrier concentration (ni) is 1.5 × 10^16 m^−3.
Calculating the Doping Concentration:
To calculate the doping concentration, we need to convert the given ppm concentration to a number density. Since 1 ppm is equivalent to 1 part in 1 million, we can calculate the number of dopant atoms per unit volume as follows:
Doping concentration = (1 ppm) × (5 × 10^28 atoms m^−3)
= (1 × 10^−6) × (5 × 10^28 atoms m^−3)
= 5 × 10^22 atoms m^−3
Calculating the Number of Electrons:
In an n-type semiconductor, like the one we have here, the dopant introduces excess electrons. The number of electrons can be calculated by subtracting the number of dopant atoms from the number of pure silicon atoms:
Number of electrons = (Number of silicon atoms) - (Number of dopant atoms)
= (5 × 10^28 atoms m^−3) - (5 × 10^22 atoms m^−3)
= 4.9995 × 10^28 atoms m^−3
Calculating the Number of Holes:
To calculate the number of holes, we first need to determine the intrinsic carrier concentration (ni) of the pure silicon crystal. Given that ni is 1.5 × 10^16 m^−3, we can calculate the number of holes as follows:
Number of holes = (ni)^2 / Number of electrons
= (1.5 × 10^16 m^−3)^2 / (4.9995 × 10^28 atoms m^−3)
= 4.5 × 10^32 m^−6 / (4.9995 × 10^28 atoms m^−3)
= 9 × 10^3 m^−6
Summary:
In summary, the number of electrons in the doped silicon crystal is approximately 4.9995 × 10^28 atoms m^−3, while the number of holes is approximately 9 × 10^3 m^−6. These calculations were based on the given doping concentration of 1 ppm and the intrinsic carrier concentration of the pure silicon crystal.
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