Question

Suppose a pure si crystal has 5 × 1028 atoms m–3. It is doped by 1 ppm concentration of pentavalent as. Calculate the number of electrons and holes. Given that ni

Answer 1

Hii dear,

# Answer-

a)Number of eletrons = 5×10^22 /cu.m

b)Number of holes = 4.5×10^9 /cu.m

# Explanation-

a)Each Arsenic atom will provide one electron.

Hence,

Number of electrons = Number of As atoms

Number of electrons = 5×10^28/10^6

Number of electrons = 5×10^22 per cu.m.

b)We know

ni^2 = number of electrons × number of holes

Number of holes = ni^2 / number of elecrons

Number of holes = (1.5×10^16)/5×10^22

Number of holes = 2.25×10^32/5×10^22

Number of holes = 4.5×10^9 per.cu.m

Thanks for asking...

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 , ${nh}$=${(1.5×10^16)^2/5×10^28×10^6}$
• ⭐. ${nh}$=${(2.25×10^32)/(5×10^22)}$
• ⭐.${~ 4.5× 10^9m^-3}$