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 (. per m^3 ) semiconductor of conductivity 500 mho / m is

Correct answer: A. 8 × 10^ 21 U sing, σ=n_e q μ_e Here, σ=500 mho / m e=1.6 × 10^ -19 C μ_e=0.4 ~m^2 ~V^ -1 ~s^ -1 n_e= 500 1.6 × 10^ -19 × 0.4 =7.8 × 10^ 21 8 × 10^ 21 ~m^ -3

AIIMS 2013

Exam: AIIMS
Year: 2013
Subject: Physics
Chapter: Current Electricity

AIIMS 2013 · Physics · Current Electricity

Given that the mobility of electrons in Ge is 0.4 ~m^2 ~V^ -1 ~s^...

Given that the mobility of electrons in Ge is

0.4 \mathrm{~m}^2 \mathrm{~V}^{-1} \mathrm{~s}^{-1} and electronic charge is

1.6 \times 10^{-19} \mathrm{C}

. The number of donor atom \left(\right.

per

\mathrm{m}^3

) semiconductor of conductivity

500 \mathrm{mho} / \mathrm{m}

Options

A $8 \times 10^{21}$
B $8 \times 10^{15}$
C $5 \times 10^{21}$
D $8 \times 10^{16}$

Correct answer

A. $8 \times 10^{21}$

Step-by-step solution

\mathrm{U} \operatorname{sing}, \sigma=n_e q \mu_e

Here,

\sigma=500 \mathrm{mho} / \mathrm{m}

e=1.6 \times 10^{-19} \mathrm{C}

\mu_e=0.4 \mathrm{~m}^2 \mathrm{~V}^{-1} \mathrm{~s}^{-1}

\therefore \quad n_e=\frac{500}{1.6 \times 10^{-19} \times 0.4}

=7.8 \times 10^{21} \simeq 8 \times 10^{21} \mathrm{~m}^{-3}

Practice Current Electricity on Quantrex Academy →

More from Current Electricity