Pure Si at 500K has equal number of electrons (${{n}_{e}}$) and hole (${{n}_{h}}$ ) concentration of $1.5\times {{10}^{16}}{{m}^{-3}}$ . Doping by indium increases ${{n}_{h}}$ to $4.5\times {{10}^{22}}{{m}^{-3}}$ . The doped semiconductor is of
a)n-type with electron concentration ${{n}_{e}}=5\times {{10}^{22}}{{m}^{-3}}$
b) p-type with electron concentration ${{n}_{e}}=2.5\times {{10}^{10}}{{m}^{-3}}$
c) n-type with electron concentration ${{n}_{e}}=2.5\times {{10}^{23}}{{m}^{-3}}$
d) p-type having electron concentration ${{n}_{e}}=5\times {{10}^{9}}{{m}^{-3}}$
Answer
576.9k+ views
Hint: Pure semiconductors i.e. intrinsic semiconductor has equal number of hole and electron concentration. The holes carry a net positive charge and electrons carry negative charge. Depending on the excess of the above two the semiconductors are named as p-type and n-type respectively.
Formula used:
${{n}_{e}}{{n}_{p}}=n_{i}^{2}$
Complete answer:
When pure silicon semiconductor is doped with indium, the hole concentration in the semiconductor increases. The holes are characterized by a net positive charge. Hence the impure semiconductor i.e. extrinsic semiconductor obtained by doping is p-type.
Let us say the intrinsic charge of a carrier concentration is ${{n}_{i}}$ . After doping the semiconductor, the electron charge carrier concentration is ${{n}_{e}}$ and that of the holes is ${{n}_{h}}$ . Hence thermodynamically the relation between ${{n}_{i}},{{n}_{e}}$ and ${{n}_{h}}$ is given by,
${{n}_{e}}{{n}_{p}}=n_{i}^{2}$
In a intrinsic semiconductor the hole concentration,
${{n}_{e}}={{n}_{p}}={{n}_{i}}$
Using the concentration of the charge carries given in the question we get,
$\begin{align}
& {{n}_{e}}{{n}_{p}}=n_{i}^{2} \\
& \Rightarrow 4.5\times {{10}^{22}}{{m}^{-3}}\times {{n}_{e}}={{(1.5\times {{10}^{16}})}^{2}}{{m}^{-3}} \\
& \Rightarrow {{n}_{e}}=\dfrac{2.25\times {{10}^{3}}^{2}}{4.5\times {{10}^{22}}}{{m}^{-3}}=5\times {{10}^{9}}{{m}^{-3}} \\
\end{align}$
Therefore the correct answer of the above question is option d.
Note:
The thermodynamic relation between the intrinsic charge carriers and the extrinsic charge carriers only holds valid as long as the lattice structure of the crystal solid remains the same. If after doping the structure changes the above equation does not hold valid. The above equation also holds valid as the number density of electrons and holes in extrinsic semiconductor is very large and hence the rate of recombination is the same.
Formula used:
${{n}_{e}}{{n}_{p}}=n_{i}^{2}$
Complete answer:
When pure silicon semiconductor is doped with indium, the hole concentration in the semiconductor increases. The holes are characterized by a net positive charge. Hence the impure semiconductor i.e. extrinsic semiconductor obtained by doping is p-type.
Let us say the intrinsic charge of a carrier concentration is ${{n}_{i}}$ . After doping the semiconductor, the electron charge carrier concentration is ${{n}_{e}}$ and that of the holes is ${{n}_{h}}$ . Hence thermodynamically the relation between ${{n}_{i}},{{n}_{e}}$ and ${{n}_{h}}$ is given by,
${{n}_{e}}{{n}_{p}}=n_{i}^{2}$
In a intrinsic semiconductor the hole concentration,
${{n}_{e}}={{n}_{p}}={{n}_{i}}$
Using the concentration of the charge carries given in the question we get,
$\begin{align}
& {{n}_{e}}{{n}_{p}}=n_{i}^{2} \\
& \Rightarrow 4.5\times {{10}^{22}}{{m}^{-3}}\times {{n}_{e}}={{(1.5\times {{10}^{16}})}^{2}}{{m}^{-3}} \\
& \Rightarrow {{n}_{e}}=\dfrac{2.25\times {{10}^{3}}^{2}}{4.5\times {{10}^{22}}}{{m}^{-3}}=5\times {{10}^{9}}{{m}^{-3}} \\
\end{align}$
Therefore the correct answer of the above question is option d.
Note:
The thermodynamic relation between the intrinsic charge carriers and the extrinsic charge carriers only holds valid as long as the lattice structure of the crystal solid remains the same. If after doping the structure changes the above equation does not hold valid. The above equation also holds valid as the number density of electrons and holes in extrinsic semiconductor is very large and hence the rate of recombination is the same.
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