
A silicon specimen is made into a p-type semiconductor by doping on the average one indium atom per $5\times {{10}^{7}}$ silicon atoms. If the number density of atoms in the silicon specimen is $5\times {{10}^{28}}atoms/{{m}^{3}}$. Then the number of acceptor atoms in the silicon will be:
A. $2.5\times {{10}^{30}}atoms/c{{m}^{3}}$
B. $1.0\times {{10}^{15}}atoms/c{{m}^{3}}$
C. $1.0\times {{10}^{13}}atoms/c{{m}^{3}}$
D. $2.5\times {{10}^{34}}atoms/c{{m}^{3}}$
Answer
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Hint: In production of semiconductors, the doping is the purposeful introduction of external impurities to the intrinsic semiconductor which is called pure conductor. After doping the intrinsic semiconductor becomes a doped semiconductor. Based on the relative number and the nature of external impurities the semiconductor is called either p-type or n-type semiconductor.
Complete step by step solution:
When a pure semiconductor is doped with an external impurity, then there are two types of impurities. The impurities are called a dopant.
An electron donor is a dopant atom (impurity) that, when added to a pure semiconductor provides free electrons to the semiconductor. The resultant semiconductor is called an n-type semiconductor.
An electron acceptor is a dopant atom (impurity) that, when added to a semiconductor provides the hole to the semiconductor. The resulting semiconductor is called a p-type semiconductor.
It is given that the number density of the atoms in Silicon is $5\times {{10}^{28}}atoms/{{m}^{3}}$
As we know that $1m=100cm$
Then, $1{{m}^{3}}={{\left( 100 \right)}^{3}}c{{m}^{3}}={{10}^{6}}c{{m}^{3}}$
Then the number density of atoms becomes $5\times {{10}^{22}}atoms/c{{m}^{3}}$
Since, 1 atom of Indium is doped in $5\times {{10}^{7}}$ Silicon atoms,
$\therefore $Total number of doped Indium atoms$=\dfrac{5\times {{10}^{22}}}{5\times {{10}^{7}}}=1\times {{10}^{15}}atoms/c{{m}^{3}}$
As Indium is an acceptor atom, therefore the total number of acceptors atoms in silicon is $1\times {{10}^{15}}atoms/c{{m}^{3}}$.
Thus, option B is correct.
Note: - When a pure semiconductor is doped with a donor dopant then the semiconductor formed is a n-type semiconductor.
- When a pure semiconductor is doped with a acceptor dopant then the semiconductor formed is a p-type semiconductor
Complete step by step solution:
When a pure semiconductor is doped with an external impurity, then there are two types of impurities. The impurities are called a dopant.
An electron donor is a dopant atom (impurity) that, when added to a pure semiconductor provides free electrons to the semiconductor. The resultant semiconductor is called an n-type semiconductor.
An electron acceptor is a dopant atom (impurity) that, when added to a semiconductor provides the hole to the semiconductor. The resulting semiconductor is called a p-type semiconductor.
It is given that the number density of the atoms in Silicon is $5\times {{10}^{28}}atoms/{{m}^{3}}$
As we know that $1m=100cm$
Then, $1{{m}^{3}}={{\left( 100 \right)}^{3}}c{{m}^{3}}={{10}^{6}}c{{m}^{3}}$
Then the number density of atoms becomes $5\times {{10}^{22}}atoms/c{{m}^{3}}$
Since, 1 atom of Indium is doped in $5\times {{10}^{7}}$ Silicon atoms,
$\therefore $Total number of doped Indium atoms$=\dfrac{5\times {{10}^{22}}}{5\times {{10}^{7}}}=1\times {{10}^{15}}atoms/c{{m}^{3}}$
As Indium is an acceptor atom, therefore the total number of acceptors atoms in silicon is $1\times {{10}^{15}}atoms/c{{m}^{3}}$.
Thus, option B is correct.
Note: - When a pure semiconductor is doped with a donor dopant then the semiconductor formed is a n-type semiconductor.
- When a pure semiconductor is doped with a acceptor dopant then the semiconductor formed is a p-type semiconductor
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