Question

Suppose a pure si crystal has 5 1028 atoms m3 . It is doped by 1 ppm concentration of pentavalent as. Calculate the number of electrons and holes. Given that ni =1.5 1016 m3 .

Answer 1

The number of electrons are ne = 4.5 × 10^9 m^−3.

Explanation:

1 ppm = 1 part per million = 1106.

so no. of pentavalent As atoms doped in given Si

crystal, = 5×10^28 / 10^6 = 5 × 10^22 m^−3.

As one pentavalent atoms donates one free electron to the crystal structure, so no. of free electrons in the given Si crystal.

ne = 5 × 10^22 m^−3

Number of holes, nh = ni^2 / ne = (1.5 × 10^16)^2 / 5 × 10^22

= 4.5 × 10^9 m^−3

Thus the number of electrons are ne = 4.5 × 10^9 m^−3.

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Which has more number of electrons Na and Na+? why?

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Answer 2

$\huge\boxed{\underline{\mathcal{\red{A}\green{N}\pink{S}\orange{W}\blue{E}\purple{R}}}}$.

• ⭐.Note that thermally generated electrons ${(n1 ~ 10^6 m^-3)}$are negligibly small as compared to those prodyced by doping.
• ⭐. Therefore , ${ne = ND}$.
• ⭐. Since ${nenh=n1^2}$,The number of holes , ${nh}$=${(1.5×10^16)^2/5×10^28×10^6}$
• ⭐. ${nh}$=${(2.25×10^32)/(5×10^22)}$
• ⭐.${~ 4.5× 10^9m^-3}$