Question

A sample of germanium is doped to the extent of 10^14 donor atoms/cm3

and 7*10^13

acceptor atoms/cm3

. At the room temperature of the sample, the resistivity of pure (intrinsic)
germanium is 60 Ω-cm. If the applied electric field is 2V/cm, find the total conducting current
density.

Answer 1

52.24 mA/cm is the total conducting current  density.

Explanation:

In case of pure germanium,

$n_{1}$ = n = p

$n_{1}$ = б_{1}/(μ_{n} μ_{p})q

= 1/60(3800 + 1800) * 1.6 * 10^-19

= 1.86 * 10^13 electrons/cm^3

resistivity = 60 Ω-cm

E = 2V/cm,

J = I/A

= 2V/60 Ω * 1/8

= 1/30 A * 80.96 * 10^-3 * 2

= 52.24 mA/cm

Learn more: Current density

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