A pure Si crystal has 5 × 1022 atoms m-3. It is doped by 1 ppm concentration of pentavalent As. The number of holes is (ni2, = npne) (Take ni = 1.5 × 1016 m-3)

Question

A pure Si crystal has 5 × 1022 atoms m-3. It is doped by 1 ppm concentration of pentavalent As. The number of holes is (ni2, = npne) (Take ni = 1.5 × 1016 m-3) (A) 4.5 × 109 m-3 (B) 4 .5 × 106 m-3 (C) 2.5 × 109 m-3 (D) 2.5 × 106 m-3

Tardigrade Admin · Accepted Answer

The electron and holes concentration in a semiconductor in thermal equilibrium is ne nh = ni2 or nh = (ni2/ne) Here, ni = 1.5 × 1016m-3, ne = 5× 1022 m-3 ∴ nh = ((1.5× 1016)2/(5× 1022)) = (2.25× 1032/5× 1022) = 4.5 × 109 m-3

Q. A pure $S i$ crystal has $5 \times 1 0^{22}$ atoms $m^{- 3}$ . It is doped by $1$ ppm concentration of pentavalent As. The number of holes is $(n_{i}^{2}, = n_{p} n_{e})$ (Take $n_{i} = 1.5 \times 1 0^{16} m^{- 3})$

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

$4.5 \times 1 0^{6} m^{- 3}$

$2.5 \times 1 0^{9} m^{- 3}$

$2.5 \times 1 0^{6} m^{- 3}$

Q. A pure Si crystal has 5×1022 atoms m−3. It is doped by 1 ppm concentration of pentavalent As. The number of holes is (ni2​,=np​ne​) (Take ni​=1.5×1016m−3)

4.5×109m−3

4.5×106m−3

2.5×109m−3

2.5×106m−3

The electron and holes concentration in a semiconductor in thermal equilibrium is ne​nh​=ni2​ or nh​=ne​ni2​​ Here, ni​=1.5×1016m−3, ne​=5×1022m−3 ∴nh​=(5×1022)(1.5×1016)2​ =5×10222.25×1032​ =4.5×109m−3

Q. A pure $S i$ crystal has $5 \times 1 0^{22}$ atoms $m^{- 3}$ . It is doped by $1$ ppm concentration of pentavalent As. The number of holes is $(n_{i}^{2}, = n_{p} n_{e})$ (Take $n_{i} = 1.5 \times 1 0^{16} m^{- 3})$

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

$4.5 \times 1 0^{6} m^{- 3}$

$2.5 \times 1 0^{9} m^{- 3}$

$2.5 \times 1 0^{6} m^{- 3}$