The length of germanium rod is $$1$$ cm and its area of $$cross-section$$ is $$1mm2$$. If for germanium $$ni=2.5$$×$$1019m$$−$$3$$ , μ$$e=0.39m2v$$−$$1s$$−$$1$$ , μ$$h=0.19m2v$$−$$1s$$−$$1$$. Its resistance in ohms is

Resistance $$=$$ $$R=$$ρ$$lA=lnie$$μ$$e+$$μ$$hA=1$$×$$10$$−$$22.5$$×$$1019$$×$$1.6$$×$$10$$−$$190.39+0.19$$×$$10$$−$$6$$ $$=4310.34$$ Ω

Electrons and holes in semiconductors

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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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