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Mobility of electrons in a semiconductor...

Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an n-type semiconuctor, the density of electrons is `10^(19)m^(-3)` and their mobility is `1.6m^(2)//(V.s)` then the resistivity of te semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to :

Text Solution

Verified by Experts

As we know current density
`j=sigma E=n e v_(d)`
`sigma n e (v_(d))/E=n e mu`
`1/(sigma) = p 1/(n_(e)e mu_(e))=` Resistivity
`=1/(10^(19)xx1.6xx10^(19)-19xx1.6)`
or `P=0.4Omegam`
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Knowledge Check

  • In an intrinsic semiconductor, the density of conduction electrons is 7.07 xx 10^(15)m ^(-3) . When it is doped with indium, the density of holes becomes 5xx 10^(22)m^(-3) . Find the density of conduction electron is doped semiconductor

    A
    Zero
    B
    `1 xx 10^(9)m^(-3)`
    C
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    D
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  • A pure semiconductor has equal electron and hole concentration of 10^(16) m^(-3) . Doping by indium increases na to 5 xx 10^(22)m^(-3) . Then, the value of n_(e) in the doped semiconductor is

    A
    `10^(6)//m^(3)`
    B
    `10^(22)//m^(3)`
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    `2xx10^(9)//m^(3)`
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  • A potential difference of 5V is applied across a conductor of length 10 cm. IF drift of electron is 2.5 xx 10^(4) m//s , then electron mobility is SI unit is

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    B
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    C
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    D
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