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  1. In a semiconductor material. The hole concentration is found to be 2 x 2.5 x 1015 cm-3. If mobility of carriers is 0.13 m2/ v-s. Then find the current density if electric field intensity is 3.62 x 10-19

    2. A.
    7.6237 x 10-4 A/cm2
    B.
    7.6237 x 10-5 A/cm2
    C.
    7.6237 x 10-3 A/cm2
    D.
    none of these

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    Answer : Option A

    Explanation :

    Current density J = σE

    Where σ = conductivity

    Given -> μ = 0.13 m2/v-s = 0.13 x 104 cm2/V sec

    P = 2.25 x 1015/cm3

    We have, ∴ ni = 1.5 x 1010

    Also n.p. =

    ∴ n = /p

    = (1.6 x 10-19 x 0.13 x 104 x 2.25 x 1015) x

    = (0.468) (4.5 x 1015)

    σ = 2.106 x 1015 μ/cm

    J = σE

    ∴ Current density = 2.106 x 1015 x 3.620 x 10-19

    = 7.6237 x 10-4 A/m2.

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