A silicon specimen is made into p - type semiconductor by doping, on an average, one indium atom per `5xx10^(7)` sillicon atoms. If the number density of atoms in the silicon per cubic centimetre is `5xx10^(28)` Then the number of acceptor atoms in silicon will be.

Options

A. `2.5xx10^(20)" atom cm"^(-3)`
B. `2.5xx10^(25)" atom cm"^(-3)`
C. `1xx10^(13)" atom cm"^(-3)`
D. `1xx10^(15)" atom cm"^(-3)`

Answer

To find the number of acceptor atoms (indium atoms) in a silicon specimen, we can follow these steps:

### Step 1: Understand the given data
- The average doping ratio is 1 indium atom per \(5 \times 10^7\) silicon atoms.
- The number density of silicon atoms is \(5 \times 10^{28}\) atoms per cubic centimeter.

### Step 2: Calculate the number of silicon atoms in 1 cm³
The number density of silicon atoms is already given as \(5 \times 10^{28}\) atoms/cm³. Therefore, in 1 cm³ of silicon, there are:
\[ N_{Si} = 5 \times 10^{28} \text{ atoms} \]

### Step 3: Calculate the number of indium (acceptor) atoms
Since there is 1 indium atom for every \(5 \times 10^7\) silicon atoms, we can find the number of indium atoms in 1 cm³ by using the following relationship:
\[ N_{In} = \frac{N_{Si}}{5 \times 10^7} \]
Substituting the value of \(N_{Si}\):
\[ N_{In} = \frac{5 \times 10^{28}}{5 \times 10^7} \]

### Step 4: Simplify the expression
\[ N_{In} = \frac{5}{5} \times \frac{10^{28}}{10^7} = 1 \times 10^{21} = 10^{21} \]

### Conclusion
The number of acceptor atoms (indium atoms) in the silicon specimen is:
\[ \boxed{10^{21}} \text{ atoms} \]

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