A `p-n` photodiode is made of a material with a band gap of `2.0 eV`. The minimum frequency of the radiation that can be absorbed by the material is nearly

Options

A. `10xx10^(14) Hz`
B. `5xx10^(14) Hz`
C. `1xx10^(14) Hz`
D. `10xx10^(14) Hz`

Explanation

To find the minimum frequency of the radiation that can be absorbed by a p-n photodiode made of a material with a band gap of 2.0 eV, we can use the relationship between energy and frequency given 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).

### Step 1: Convert the band gap energy from eV to joules
The energy in electron volts (eV) can be converted to joules using the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \).

\[ E = 2.0 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 3.2 \times 10^{-19} \, \text{J} \]

### Step 2: Use the energy to find the frequency
Now that we have the energy in joules, we can rearrange the equation \( E = h \cdot f \) to solve for frequency \( f \):

\[ f = \frac{E}{h} \]

Substituting the values we have:

\[ f = \frac{3.2 \times 10^{-19} \, \text{J}}{6.626 \times 10^{-34} \, \text{J s}} \]

### Step 3: Calculate the frequency
Now we perform the division:

\[ f \approx \frac{3.2}{6.626} \times 10^{15} \, \text{Hz} \approx 0.482 \times 10^{15} \, \text{Hz} \approx 4.82 \times 10^{14} \, \text{Hz} \]

### Step 4: Round to significant figures
Rounding this to two significant figures, we find:

\[ f \approx 5.0 \times 10^{14} \, \text{Hz} \]

### Conclusion
The minimum frequency of the radiation that can be absorbed by the material is approximately \( 5.0 \times 10^{14} \, \text{Hz} \).

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