Consider a silicon semiconductor doped with 5 x 10^16 cm − 3 phosphorus atoms at 300K. The intrinsic carrier concentration of silicon is 𝑛 𝑖 = 1.5 × 10^10 cm^ −3 . Calculate: The electron concentration 𝑛 is _____ cm^-3 . and the position of the Fermi level relative to the intrinsic Fermi level E Fi is _____ eV .

4.5 x 10^3 and 0.388 
 
5 x 10^16 and 0.388 
 
5 x 10^16 and 0.126
 
4.5 x 10^3 and 0.126

To solve this problem, we need to calculate two quantities:
1. The electron concentration \(n\) in the silicon semiconductor.
2. The position of the Fermi level relative to the intrinsic Fermi level \(E_{Fi}\).

### Given Data:
- Doping concentration \(N_D = 5 \times 10^{16} \, \text{cm}^{-3}\) (phosphorus atoms).
- Intrinsic carrier concentration \(n_i = 1.5 \times 10^{10} \, \text{cm}^{-3}\).
- Temperature \(T = 300 \, \text{K}\).
- Silicon is an n-type semiconductor (since it is doped with phosphorus, which donates electrons).

### Step 1: Calculate the electron concentration \(n\).

For an n-type semiconductor, the electron concentration \(n\) can be approximated as:

\[
n \approx N_D
\]

This is because in an n-type semiconductor, the majority carrier concentration (electrons) is approximately equal to the donor concentration when the doping is not too high and the temperature is not too low.

Thus:

\[
n = 5 \times 10^{16} \, \text{cm}^{-3}
\]

### Step 2: Calculate the position of the Fermi level relative to the intrinsic Fermi level \(E_{Fi}\).

The shift in the Fermi level relative to the intrinsic Fermi level \(E_{Fi}\) can be calculated using the formula:

\[
E_{F} - E_{Fi} = kT \ln\left(\frac{n}{n_i}\right)
\]

Where:
- \(k\) is Boltzmann's constant, \(8.617 \times 10^{-5} \, \text{eV/K}\).
- \(T = 300 \, \text{K}\).
- \(n = 5 \times 10^{16} \, \text{cm}^{-3}\).
- \(n_i = 1.5 \times 10^{10} \, \text{cm}^{-3}\).

Substitute the values:

\[
E_{F} - E_{Fi} = 8.617 \times 10^{-5} \times 300 \times \ln\left(\frac{5 \times 10^{16}}{1.5 \times 10^{10}}\right)
\]

Calculate the argument of the logarithm:

\[
\frac{5 \times 10^{16}}{1.5 \times 10^{10}} = 3.33 \times 10^{6}
\]

Now, take the natural logarithm:

\[
\ln(3.33 \times 10^{6}) \approx \ln(3.33) + \ln(10^{6}) = 1.203 + 13.816 = 15.019
\]

Now calculate \(E_F - E_{Fi}\):

\[
E_{F} - E_{Fi} = 8.617 \times 10^{-5} \times 300 \times 15.019
\]

\[
E_{F} - E_{Fi} \approx 0.388 \, \text{eV}
\]

### Final Answer:

- **Electron concentration \(n\)**: \(5 \times 10^{16} \, \text{cm}^{-3}\)
- **Position of the Fermi level \(E_{F} - E_{Fi}\)**: \(0.388 \, \text{eV}\)

The correct option from the choices given is:

**Option: \(5 \times 10^{16} \, \text{cm}^{-3}\) and \(0.388 \, \text{eV}\)**

Would you like any further explanations or details?

### Relative Questions:
1. How does doping concentration affect the position of the Fermi level in a semiconductor?
2. What is the significance of the intrinsic carrier concentration in semiconductors?
3. How does the Fermi level shift for p-type semiconductors?
4. Why is the Fermi level important in understanding the electrical properties of semiconductors?
5. How does temperature affect the Fermi level in semiconductors?
6. What assumptions are made when calculating the electron concentration in an n-type semiconductor?
7. How would the calculations change if the semiconductor were heavily doped?
8. What happens to the Fermi level in intrinsic semiconductors?

### Tip:
In semiconductor physics, always check whether the doping concentration significantly exceeds the intrinsic carrier concentration; this will help you determine if the majority carrier concentration is approximately equal to the doping level.

Learn how to calculate the electron concentration and the Fermi level shift in a silicon semiconductor doped with phosphorus atoms. Understand the impact of doping concentration on semiconductor properties. Suitable for undergraduate students studying semiconductor physics.

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