A potential difference of 2V is applied between the opposite faces of a Ge crystal plate of area 1 cm2 and thickness 0.5 mm. If the concentration of electrons in Ge is 2×1019/m3 and mobilities of electrons and holes are 0.36 m2volt-secand 0.14 m2volt-sec,respectively, then the current flowing through the plate will be

σ=neμe+μh ∵ne=nh=n=2×1019×1.6×10−190.36+0.14=1.6Ω−m−1R=ρLA=LσA=0.5×10−31.6×10−4=258Ω ∴i=VR=225/8=1625A=0.64A

A potential difference of 2V is applied between the opposite faces of a Ge crystal plate of area 1 cm2 and thickness 0.5 mm. If the concentration of electrons in Ge is 2×1019/m3 and mobilities of electrons and holes are 0.36 m2volt-secand 0.14 m2volt-sec,respectively, then the current flowing through the plate will be

σ=neμe+μh ∵ne=nh=n=2×1019×1.6×10−190.36+0.14=1.6Ω−m−1R=ρLA=LσA=0.5×10−31.6×10−4=258Ω ∴i=VR=225/8=1625A=0.64A

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A potential difference of 2V is applied between the opposite faces of a Ge crystal plate of area 1 $c m^{2}$ and thickness 0.5 mm. If the concentration of electrons in Ge is $2 \times 10^{19} / m^{3}$ and mobilities of electrons and holes are 0.36 $\frac{m^{2}}{volt - \sec} and 0.14 \frac{m^{2}}{volt - \sec},$ respectively, then the current flowing through the plate will be

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0.56 A

0.64 A

0.25 A

0.45 A

answer is D.

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Detailed Solution

$\begin{array}{l} σ = ne (μ_{e} + μ_{h}) (∵ n_{e} = n_{h} = n) \\ = 2 \times 10^{19} \times 1 .6 \times 10^{- 19} (0 .36 + 0 .14) = 1 .6 {(Ω - m)}^{- 1} \\ R = ρ \frac{L}{A} = \frac{L}{σA} = \frac{0 .5 \times 10^{- 3}}{1 .6 \times 10^{- 4}} = \frac{25}{8} Ω \\ ∴ i = \frac{V}{R} = \frac{2}{25 / 8} = \frac{16}{25} A = 0 .64 A \end{array}$