For a pure Si crystal has 5 × 1028 atom m -3 It is doped by 1 PPM concentration of pentavalent As. Calculate the number of electron holes. Given that n i=1.5 × 1016 m -3 )

Question

For a pure Si crystal has 5 × 1028 atom m -3 It is doped by 1 PPM concentration of pentavalent As. Calculate the number of electron holes. Given that n i=1.5 × 1016 m -3 ) (A) 5.4 × 109 m -3 (B) 4.5 × 109 m -3 (C) 4.5 × 10-9 m -3 (D) 5.4 × 10-9 m -3

Tardigrade Admin · Accepted Answer

Each As atom will provide 1 free electron total no. of free electrons ne=(1/106) × 5 × 1028=5 × 1022 ne nh=ni2 ⇒ n h=(ni2/ne)

Q. For a pure $S i$ crystal has $5 \times 1 0^{28}$ atom $m^{- 3}$ It is doped by $1 PPM$ concentration of pentavalent As. Calculate the number of electron $&$ holes. Given that $n_{i} = 1.5 \times 1 0^{16} m^{- 3}$ )

$5.4 \times 1 0^{9} m^{- 3}$

$4.5 \times 1 0^{9} m^{- 3}$

$4.5 \times 1 0^{- 9} m^{- 3}$

$5.4 \times 1 0^{- 9} m^{- 3}$

Each As atom will provide $1$ free electron total no. of free electrons
$n_{e} = \frac{1}{1 0 ^{6}} \times 5 \times 1 0^{28} = 5 \times 1 0^{22}$
$n_{e} n_{h} = n_{i}^{2}$
$\Rightarrow n_{h} = \frac{n _{i}^{2}}{n _{e}}$

Q. For a pure Si crystal has 5×1028 atom m−3 It is doped by 1PPM concentration of pentavalent As. Calculate the number of electron & holes. Given that ni​=1.5×1016m−3 )

5.4×109m−3

4.5×109m−3

4.5×10−9m−3

5.4×10−9m−3