The length of a germanium rod is 0.58 cm and its area of cross-section is 1 mm2. if for Germaniumni=2⋅5×1019m−3,μh=0⋅19m2/V−s μe=0⋅9m2/V−s, then the resistance of the rod will be

R=ρLA=Lnieμe+μhA =0⋅58×10−22⋅5×10191⋅6×10−19[0⋅19+0⋅39]×1×10−6 =2⋅5kΩ

The length of a germanium rod is 0.58 cm and its area of cross-section is 1 mm2. if for Germaniumni=2⋅5×1019m−3,μh=0⋅19m2/V−s μe=0⋅9m2/V−s, then the resistance of the rod will be

R=ρLA=Lnieμe+μhA =0⋅58×10−22⋅5×10191⋅6×10−19[0⋅19+0⋅39]×1×10−6 =2⋅5kΩ

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The length of a germanium rod is 0.58 cm and its area of cross-section is 1 mm². if for Germanium $n_{i} = 2 \cdot 5 \times 10^{19} m^{- 3}, μ_{h} = 0 \cdot 19 m^{2} / V - s$ $μ_{e} = 0 \cdot 9 m^{2} / V - s$ , then the resistance of the rod will be

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$2 \cdot 5 kΩ$

$5 \cdot 0 k Ω$

$7 \cdot 5 kΩ$

$10 k Ω$

answer is A.

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Detailed Solution

$R = \frac{ρL}{A} = \frac{L}{n_{i} e (μ_{e} + μ_{h}) A} = \frac{0 \cdot 58 \times 10^{- 2}}{(2 \cdot 5 \times 10^{19}) (1 \cdot 6 \times 10^{- 19}) [0 \cdot 19 + 0 \cdot 39] \times (1 \times 10^{- 6})} = 2 \cdot 5 kΩ$

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