First slide

Electrons and holes in semiconductors

Question

The length of germanium rod is 1 cm and its area of cross-section is $1 m m^{2}$ . If for germanium $n_{i} = 2.5 \times 10^{19} m^{- 3}, μ_{e} = 0.39 m^{2} v^{- 1} s^{- 1}, μ_{h} = 0.19 m^{2} v^{- 1} s^{- 1}$ . Its resistance in ohms is

Moderate

Solution

$R e s i s \tan c e = R = ρ \frac{l}{A} = \frac{l}{n_{i} e (μ_{e} + μ_{h}) A}$
$= \frac{1 \times 10^{- 2}}{2.5 \times 10^{19} \times 1.6 \times 10^{- 19} (0.39 + 0.19) \times 10^{- 6}}$
$= 4310.34 Ω$

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