A silicon wafer is doped with 10^16 cm−3 arsenic atoms. The donor level 𝐸𝑑 is located at 54 meV below 𝐸c. What percentage of the donor atoms are ionized at T = 77K?  [Assume, 𝑁c(𝑇 = 77𝐾) = 4.2 × 10^18 𝑐𝑚−3 and the degeneracy factor of donor level =2

a) 100 %

b) 65.5 %

c) 22.3 %

d) 10.9 %

To determine the percentage of donor atoms that are ionized, we can use the Fermi-Dirac distribution function. The fraction of donor atoms that are ionized can be calculated using:

\[
f(E_d) = \frac{1}{1 + g \exp\left(\frac{E_d - E_f}{kT}\right)}
\]

Where:
- \( f(E_d) \) is the probability that a donor atom is not ionized (i.e., occupied by an electron).
- \( g \) is the degeneracy factor, given as 2.
- \( E_d \) is the donor energy level below the conduction band \( E_c \).
- \( E_f \) is the Fermi level.
- \( k \) is the Boltzmann constant, \( k \approx 8.617 \times 10^{-5} \text{ eV/K} \).
- \( T \) is the temperature in Kelvin.

The ionization percentage is \( 1 - f(E_d) \).

### Steps to solve:

1. **Calculate the Fermi level \( E_f \)**:
   
   The Fermi level \( E_f \) for a non-degenerate semiconductor can be approximated by:

   \[
   N_d = N_c \exp\left(\frac{E_f - E_c}{kT}\right)
   \]
   
   Rearranging the equation:
   
   \[
   E_f = E_c + kT \ln\left(\frac{N_d}{N_c}\right)
   \]
   
   Substituting values:
   
   - \( N_d = 10^{16} \, \text{cm}^{-3} \)
   - \( N_c = 4.2 \times 10^{18} \, \text{cm}^{-3} \)
   - \( T = 77 \, \text{K} \)
   - \( k = 8.617 \times 10^{-5} \, \text{eV/K} \)
   
   \[
   E_f = E_c + (8.617 \times 10^{-5} \times 77) \ln\left(\frac{10^{16}}{4.2 \times 10^{18}}\right)
   \]
   
2. **Calculate the ionization probability \( f(E_d) \)**:

   \[
   f(E_d) = \frac{1}{1 + 2 \exp\left(\frac{E_d - E_f}{kT}\right)}
   \]
   
3. **Calculate the ionized fraction**:

   The ionized fraction is:

   \[
   1 - f(E_d)
   \]

Finally, compare the calculated value with the given options to determine the correct answer.

Let's solve it step by step.The calculated percentage of ionized donor atoms at \( T = 77 \, \text{K} \) is approximately **19.7%**. This value does not exactly match any of the given options, but the closest provided answer is:

- **d) 10.9%**

This suggests that there may be some approximation or rounding in the provided options, or the exact calculation falls between the provided choices.

Do you have any questions or would you like more details?

---

**Related Questions:**

1. How does the ionization energy of donor atoms affect the percentage of ionization?
2. What would happen to the percentage of ionized donors if the temperature increased?
3. How is the Fermi level influenced by the concentration of donor atoms?
4. Why is the degeneracy factor important in determining the ionization percentage?
5. How does the density of states in the conduction band influence the calculation of the Fermi level?
6. What is the significance of the donor level being 54 meV below the conduction band?
7. How would the ionization percentage change if the donor concentration increased?
8. Why is the Boltzmann constant used in the calculation of the Fermi level?

**Tip:** Remember that as temperature increases, more donor atoms will ionize due to increased thermal energy.

Learn how to calculate the ionization percentage of donor atoms doped with arsenic in a silicon wafer at T = 77K. This problem involves semiconductor physics and the Fermi-Dirac distribution function. Explore the detailed steps and the approximation involved in determining the ionized fraction.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation