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Physics Question on Semiconductor electronics: materials, devices and simple circuits

A pure Ge specimen is doped with Al. The number density of acceptor atoms is approximately 1021m−3^{21} m^{-3}. If density of electron hole pair in an intrinsic semiconductor is approximately 1019m−3^{19} m^{-3}, the number density of electrons in the specimen is

A

104m−310^4 m^{-3}

B

102m−310^2 m^{-3}

C

1017m−310^{17} m^{-3}

D

1015m−310^{15} m^{-3}

Answer

1017m−310^{17} m^{-3}

Explanation

Solution

Given that, the density of electron hole pair in an intrinsic
semiconductor (ni)=1019m3(n_i) =\frac{10^{19}}{m^3}
density of holes (nh)=1021m3(n_h) =\frac{10^{21}}{m^3}
We know that nh⋅ne=ni2n_h \cdot n_e =n_i^2
∴      1021×ne=(1019)2\therefore \, \, \, \, \, \, 10^{21} \times n_e =(10^{19})^2
⇒       ne=1019×10191021\Rightarrow \, \, \, \, \, \, \, n_e =\frac{10^{19} \times 10^{19}}{10^{21}}
\hspace30mm =10^{17}m^{-3}