The number density n, of conduction electrons in pure silicon at room temperature is about `10^(16) m^(-3)`. Assume that, by doping the silicon lattice with phosphorus, we want to increase this number by a factor of a million `(10^6)`. What fraction of `((n_p)/(n_(Si)))` silicon atoms must we replace with phosphorus atoms? (Recall that at room temperature, thermal agitation is so effective that essentially every phosphorus atom donates its "extra" electron to the conduction band) given density of silicon `= 2.33x10^3`.

Options

A. `1/(2 xx 10^5)`
B. `2/(3 xx 10^5)`
C. `8/(6 xx 10^4)`
D. `1/(5 xx 10^6)`

Explanation

To solve the problem, we need to find the fraction of silicon atoms that must be replaced by phosphorus atoms to achieve a desired increase in the number density of conduction electrons. Here’s a step-by-step solution:

### Step 1: Determine the target number density of conduction electrons
The initial number density of conduction electrons in pure silicon, \( n_{Si} \), is given as \( 10^{16} \, m^{-3} \). We want to increase this number by a factor of \( 10^6 \).

\[ n_p = 10^6 \times n_{Si} = 10^6 \times 10^{16} = 10^{22} \, m^{-3} \]

### Step 2: Calculate the number of phosphorus atoms needed
Since each phosphorus atom donates one extra electron, the number density of phosphorus atoms, \( n_p \), must also equal \( 10^{22} \, m^{-3} \).

### Step 3: Calculate the number density of silicon atoms
To find the number density of silicon atoms, \( n_{Si} \), we can use the formula:

\[ n_{Si} = \frac{\rho \cdot N_A}{M_{Si}} \]

Where:
- \( \rho = 2.33 \times 10^3 \, kg/m^3 \) (density of silicon)
- \( N_A = 6.022 \times 10^{23} \, atoms/mol \) (Avogadro's number)
- \( M_{Si} = 28 \times 10^{-3} \, kg/mol \) (molar mass of silicon)

Substituting the values:

\[ n_{Si} = \frac{(2.33 \times 10^3) \cdot (6.022 \times 10^{23})}{28 \times 10^{-3}} \]

Calculating this gives:

\[ n_{Si} \approx 5.02 \times 10^{28} \, atoms/m^3 \]

### Step 4: Calculate the fraction of silicon atoms replaced by phosphorus
Now, we need to find the fraction of silicon atoms that must be replaced by phosphorus atoms:

\[ \text{Fraction} = \frac{n_p}{n_{Si}} = \frac{10^{22}}{5.02 \times 10^{28}} \]

Calculating this gives:

\[ \text{Fraction} \approx 1.99 \times 10^{-7} \]

### Step 5: Conclusion
Thus, the fraction of silicon atoms that must be replaced with phosphorus atoms is approximately \( 1.99 \times 10^{-7} \).

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