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Standard VIII

Chemistry

Covalent Bond

The electron ...

Question

The electron mobility in N-type germanium is $3900 {cm}^{2} / v - s$ and its conductivity is $6.24 mho / cm$ ; If the effect of cotters is negligible, then the impurity concentration will be

10^{15} {per cm}^{3}

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10^{13} {per cm}^{3}

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10^{12} {per cm}^{3}

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10^{16} {per cm}^{3}

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Solution

The correct option is D $10^{16} {per cm}^{3}$
The electrical conductivity is given by,

$σ = e μ_{e} n_{e} \Rightarrow n_{e} = \frac{σ}{e μ_{e}}$
$n_{e} = \frac{6.24}{3900 \times 1.6 \times 10^{- 19}} = 10^{16} {per cm}^{3}$

Hence, $(D)$ is the correct answer.

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Similar questions

Q. The electron mobility in N-type germanium is

3900 {cm}^{2} / V.s

and its conductivity is

6.24 mho/cm

, then impurity concentration is
(Assume that the effect of cotters is negligible)

Q. On doping germanium with donor atoms density of

10^{17} {c m}^{- 3}

its conductivity in mho/cm will be
(Given

μ_{e} = 3800 {c m}^{2} / V - s

and

n_{i} = 2.5 \times 10^{13} {c m}^{- 13}

)

Q. What is the number density of donor atoms which must be added to a pure germanium semiconductor to produce an N-type semiconductor of conductivity

6.4 X 10^{2} o h m^{- 1} m^{- 1}

? The mobility of electrons in N-type germanium is

4 X 10^{- 1} m^{2} V^{- 1} s^{- 1}

. Neglect contribution of holes to conductivity.

The product of the hole concentration and the conduction electron concentration turns out to be independent of the amount of any impurity doped. The concentration of conduction electrons in germanium is $6 \times 10^{19}$ per cubic metre. When some phosphorus impurity is doped into a germanium sample, the concentration of conduction electrons increases to $2 \times 10^{23}$ per cubic metre. Find the concentration of the holes in the doped germanium.

Q. The intrinsic conductivity of germanium at

27

^{\circ} C

2.13

m h o / m

and mobilities of electrons and holes are

0.38

and

0.18

m

^{2} / V s

respectively. The density of charge carriers is

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The electron mobility in N-type germanium is 3900 cm2/v−s and its conductivity is 6.24 mho/cm; If the effect of cotters is negligible, then the impurity concentration will be

The correct option is D 10 16 per cm3The electrical conductivity is given by, σ=e μe ne ⇒ ne=σe μe ne=6.243900×1.6×10−19=1016 per cm3 Hence, (D) is the correct answer.

The electron mobility in N-type germanium is $3900 {cm}^{2} / v - s$ and its conductivity is $6.24 mho / cm$ ; If the effect of cotters is negligible, then the impurity concentration will be

The correct option is D $10^{16} {per cm}^{3}$
The electrical conductivity is given by,

$σ = e μ_{e} n_{e} \Rightarrow n_{e} = \frac{σ}{e μ_{e}}$
$n_{e} = \frac{6.24}{3900 \times 1.6 \times 10^{- 19}} = 10^{16} {per cm}^{3}$

Hence, $(D)$ is the correct answer.