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Question: A silicon specimen is made into a p-type semiconductor by doping on the average one indium atom per ...

A silicon specimen is made into a p-type semiconductor by doping on the average one indium atom per 5×1075\times {{10}^{7}} silicon atoms. If the number density of atoms in the silicon specimen is 5×1028atoms/m35\times {{10}^{28}}atoms/{{m}^{3}}. Then the number of acceptor atoms in the silicon will be:
A. 2.5×1030atoms/cm32.5\times {{10}^{30}}atoms/c{{m}^{3}}
B. 1.0×1015atoms/cm31.0\times {{10}^{15}}atoms/c{{m}^{3}}
C. 1.0×1013atoms/cm31.0\times {{10}^{13}}atoms/c{{m}^{3}}
D. 2.5×1034atoms/cm32.5\times {{10}^{34}}atoms/c{{m}^{3}}

Explanation

Solution

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 5×1028atoms/m35\times {{10}^{28}}atoms/{{m}^{3}}
As we know that 1m=100cm1m=100cm

Then, 1m3=(100)3cm3=106cm31{{m}^{3}}={{\left( 100 \right)}^{3}}c{{m}^{3}}={{10}^{6}}c{{m}^{3}}

Then the number density of atoms becomes 5×1022atoms/cm35\times {{10}^{22}}atoms/c{{m}^{3}}
Since, 1 atom of Indium is doped in 5×1075\times {{10}^{7}} Silicon atoms,
\therefore Total number of doped Indium atoms=5×10225×107=1×1015atoms/cm3=\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 1×1015atoms/cm31\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