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. `10 xx 10^(14) Hz`
B. `5 xx 10^(14)Hz`
C. `1 xx 10^(14)Hz`
D. `20 xx 10^(14)Hz`

Answer

To find the minimum frequency of radiation that can be absorbed by a p-n photodiode made of a material with a band gap of 2.0 eV, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship Between Energy and Frequency**:
The energy of the absorbed radiation is related to its frequency by the equation:
\[ E = h \cdot f \]
where \( E \) is the energy, \( h \) is Planck's constant, and \( f \) is the frequency.

2. **Convert Band Gap Energy from eV to Joules**:
The band gap energy is given as 2.0 eV. We need to convert this energy into joules. The conversion factor is:
\[ 1 \text{ eV} = 1.6 \times 10^{-19} \text{ joules} \]
Therefore, the energy in joules is:
\[ E = 2.0 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 3.2 \times 10^{-19} \text{ J} \]

3. **Use Planck's Constant**:
Planck's constant \( h \) is given as:
\[ h = 6.634 \times 10^{-34} \text{ J s} \]

4. **Rearrange the Energy-Frequency Equation**:
We need to find the frequency \( f \), so we rearrange the equation:
\[ f = \frac{E}{h} \]

5. **Substitute the Values**:
Now substitute the values of \( E \) and \( h \) into the equation:
\[ f = \frac{3.2 \times 10^{-19} \text{ J}}{6.634 \times 10^{-34} \text{ J s}} \]

6. **Calculate the Frequency**:
Performing the calculation:
\[ f \approx \frac{3.2 \times 10^{-19}}{6.634 \times 10^{-34}} \approx 4.82 \times 10^{14} \text{ Hz} \]
Rounding this value gives approximately:
\[ f \approx 5.0 \times 10^{14} \text{ Hz} \]

### Final Answer:
The minimum frequency of the radiation that can be absorbed by the material is nearly \( 5.0 \times 10^{14} \text{ Hz} \).