A p type semiconductor has acceptor level 114 meV above the valance band .What is maximum wavelength of light required to create a hole ?

Answer

To solve the problem of finding the maximum wavelength of light required to create a hole in a p-type semiconductor with an acceptor level of 114 meV above the valence band, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Energy Relation**:
The energy required to create a hole is given as 114 meV. We need to convert this energy into joules for our calculations.

2. **Convert MeV to Joules**:
- 1 eV = \(1.6 \times 10^{-19}\) Joules
- Therefore, \(114 \text{ meV} = 114 \times 10^{-3} \text{ eV} = 114 \times 10^{-3} \times 1.6 \times 10^{-19} \text{ Joules}\).

\[ E = 114 \times 10^{-3} \times 1.6 \times 10^{-19} = 1.824 \times 10^{-20} \text{ Joules} \]

3. **Use the Energy-Wavelength Relation**:
The relationship between energy (E), wavelength (λ), and Planck's constant (h) is given by:

\[ E = \frac{hc}{\lambda} \]

Rearranging this equation to solve for wavelength (λ):

\[ \lambda = \frac{hc}{E} \]

4. **Substitute Constants**:
- Planck's constant \(h = 6.626 \times 10^{-34} \text{ J s}\)
- Speed of light \(c = 3 \times 10^8 \text{ m/s}\)

Now substituting these values into the equation:

\[ \lambda = \frac{(6.626 \times 10^{-34} \text{ J s}) \times (3 \times 10^8 \text{ m/s})}{1.824 \times 10^{-20} \text{ J}} \]

5. **Calculate the Wavelength**:
Performing the calculation:

\[ \lambda = \frac{1.9878 \times 10^{-25}}{1.824 \times 10^{-20}} \approx 1.090 \times 10^{-5} \text{ m} \]

To convert this to angstroms (1 m = \(10^{10}\) angstroms):

\[ \lambda \approx 1.090 \times 10^{-5} \text{ m} \times 10^{10} \text{ angstrom/m} \approx 1090 \text{ angstroms} \]

### Final Answer:
The maximum wavelength of light required to create a hole in the p-type semiconductor is approximately **1090 angstroms**.