An n-type semiconductor has donor levels at 500 meV above the valence band, the frequnecy of light required to create a hole is nearly

Options

A. `8xx10^(13) Hz`
B. `12xx10^(13)Hz`
C. `22xx10^(13)Hz`
D. `15xx10^(13)Hz`

Answer

To find the frequency of light required to create a hole in an n-type semiconductor with donor levels at 500 meV above the valence band, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Energy Requirement**:
The energy required to create a hole corresponds to the energy difference between the valence band and the donor level. In this case, the donor level is at 500 meV above the valence band.

2. **Convert Energy from MeV to Joules**:
The energy in electron volts (eV) can be converted to joules using the conversion factor:
\[ 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \]
Therefore, we convert 500 meV to joules:
\[ E = 500 \text{ meV} = 500 \times 10^{-3} \text{ eV} = 0.5 \text{ eV} \]
\[ E = 0.5 \times 1.6 \times 10^{-19} \text{ J} = 8.0 \times 10^{-20} \text{ J} \]

3. **Use Planck's Equation**:
The energy of a photon can be expressed using Planck's equation:
\[ E = h \nu \]
where \( E \) is the energy of the photon, \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \text{ J s} \)), and \( \nu \) is the frequency.

4. **Rearrange to Find Frequency**:
Rearranging the equation gives:
\[ \nu = \frac{E}{h} \]
Substituting the values we have:
\[ \nu = \frac{8.0 \times 10^{-20} \text{ J}}{6.626 \times 10^{-34} \text{ J s}} \]

5. **Calculate the Frequency**:
Performing the calculation:
\[ \nu \approx 1.206 \times 10^{14} \text{ Hz} \]
This can be expressed as:
\[ \nu \approx 12.06 \times 10^{13} \text{ Hz} \]

6. **Select the Closest Option**:
Comparing with the given options, the closest frequency is:
\[ \nu \approx 12 \times 10^{13} \text{ Hz} \]
Therefore, the answer is option 2.

### Final Answer:
The frequency of light required to create a hole is nearly \( 12 \times 10^{13} \text{ Hz} \).

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