Given that the mobility of electrons in Ge is 0.4 m 2 V -1 s -1 and electronic charge is 1.6 × 10-19 C. The number of donor atom ( textper m 3) semiconductor of conductivity 500 mho / m is

Question

Given that the mobility of electrons in Ge is 0.4 m 2 V -1 s -1 and electronic charge is 1.6 × 10-19 C. The number of donor atom ( textper m 3) semiconductor of conductivity 500 mho / m is (A) 8 × 10 21 (B) 8 × 10 15 (C) 5 × 10 23 (D) 8 × 10 16

Tardigrade Admin · Accepted Answer

Using, σ = neqμe Here, σ = 500 mho/m e = 1.6 × 10-19 C μe=0.4 m2 V -1 s -1 ∴ ne=(500/1.6×10-19×0.4) = 7.8 × 1021 ≃ 8 × 1021 m-3

Q. Given that the mobility of electrons in Ge is $0.4 m^{2} V^{- 1} s^{- 1}$ and electronic charge is $1.6 \times 1 0^{- 19} C$ . The number of donor atom $(per m^{3})$ semiconductor of conductivity $500 mh o / m$ is

$8 \times 1 0^{21}$

$8 \times 1 0^{15}$

$5 \times 1 0^{23}$

$8 \times 1 0^{16}$

Using, $σ = n_{e} q μ_{e}$
Here, $σ = 500 mh o / m$
e = 1.6 $\times 1 0^{- 19} C$
$μ_{e} = 0.4 m^{2} V^{- 1} s^{- 1}$
$∴ n_{e} = \frac{500}{1.6 \times 1 0 ^{- 19} \times 0.4}$
$= 7.8 \times 1 0^{21} ≃ 8 \times 1 0^{21} m^{- 3}$

Q. Given that the mobility of electrons in Ge is 0.4m2V−1s−1 and electronic charge is 1.6×10−19C. The number of donor atom (perm3) semiconductor of conductivity 500mho/m is

8×1021

8×1015

5×1023

8×1016