Question

A germanium specimen is doped with aluminum. The concentration of acceptor atoms is ${10}^{21}atoms/m^3$. Given that the intrinsic concentration of electron-hole pairs is ${10}^{19}/m^3$, the concentration of electrons in the specimen is

• A. ${10}^{17}/m^3$
• B. ${10}^{15}/m^3$
• C. ${10}^4/m^3$
• D. ${10}^3/m^3$

Answer 1

The correct option is A ${10}^{17}/m^3$

For intrinsic semiconductors:
$n_i=n=p\Rightarrow (n_i{)}^2=np$................(1)
where, $n_i$ is intrinsic carrier concentration, $n$ and $p$ are the concentration of electrons and holes respectively.
Here, a Ge specimen is doped with Al,
Let, $p^{\prime }$ be the concentration of acceptor atoms in the specimen and $n^{\prime }$ be the concentration of electrons in the specimen.

Now, the equation (1) becomes
$(n_i{)}^2=n^{\prime }{p}^{\prime }$
$n^{\prime }=\frac{(n_i{)}^2}{p^{\prime }}$

Using, $n_i={10}^{19}{m}^{-3}$ and $p^{\prime }={10}^{21}{m}^{-3}$
$n^{\prime }=\frac{({10}^{19}{)}^2}{{10}^{21}}$
$n^{\prime }={10}^{(38-21)}={10}^{17}{m}^{-3}$