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 1: Understand the Energy Levels
In an n-type semiconductor, donor levels are introduced due to the presence of impurities (donors) that donate extra electrons. The donor level is located above the valence band, and in this case, it is 500 meV above the valence band.

### Step 2: Convert Energy to Joules
The energy required to create a hole (which is equivalent to promoting an electron from the valence band to the conduction band) can be expressed in joules. We know that:
1 eV = 1.6 × 10^(-19) Joules.

Given that the energy level is 500 meV, we convert this to joules:
\[ 500 \text{ meV} = 500 \times 10^{-3} \text{ eV} = 0.5 \text{ eV} \]
Now converting to joules:
\[ 0.5 \text{ eV} = 0.5 \times 1.6 \times 10^{-19} \text{ J} = 8.0 \times 10^{-20} \text{ J} \]

### Step 3: Use the Energy-Frequency Relation
The energy of a photon is related to its frequency by the equation:
\[ E = h \cdot f \]
where:
- \(E\) is the energy in joules,
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \text{ J s}\)),
- \(f\) is the frequency in hertz.

We can rearrange this equation to solve for frequency:
\[ f = \frac{E}{h} \]

### Step 4: Calculate the Frequency
Substituting the values we have:
\[ f = \frac{8.0 \times 10^{-20} \text{ J}}{6.626 \times 10^{-34} \text{ J s}} \approx 1.21 \times 10^{14} \text{ Hz} \]

### Conclusion
The frequency of light required to create a hole in the n-type semiconductor is approximately \(1.21 \times 10^{14} \text{ Hz}\).

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