A p-n photodiode is made of a material with a band gap of 2 e V. The minimum frequency of the radiation that can be absorbed by the material is nearly
(hc= 1240 eV nm)

Options

A. `1xx10^14` Hz
B. `20xx10^14` Hz
C. `10xx10^14` Hz
D. `5xx10^14` Hz

Explanation

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 eV, we can follow these steps:

### Step 1: Understand the relationship between energy and frequency
The energy of a photon can be expressed using the equation:
\[ E = h \cdot f \]
where:
- \( E \) is the energy of the photon,
- \( h \) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J s}\)),
- \( f \) is the frequency of the photon.

### Step 2: Convert the band gap energy from eV to Joules
The band gap energy given is 2 eV. To convert this energy into Joules, we use the conversion factor:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
Thus, the band gap energy in Joules is:
\[ E = 2 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 3.2 \times 10^{-19} \, \text{J} \]

### Step 3: Rearrange the equation to solve for frequency
From the equation \( E = h \cdot f \), we can rearrange it to find the frequency:
\[ f = \frac{E}{h} \]

### Step 4: Substitute the values into the equation
Now we substitute the values we have:
\[ f = \frac{3.2 \times 10^{-19} \, \text{J}}{6.626 \times 10^{-34} \, \text{J s}} \]

### Step 5: Calculate the frequency
Perform the calculation:
\[ f \approx \frac{3.2 \times 10^{-19}}{6.626 \times 10^{-34}} \approx 4.82 \times 10^{14} \, \text{Hz} \]

### Step 6: Round the result
Rounding \( 4.82 \times 10^{14} \, \text{Hz} \) gives approximately \( 5 \times 10^{14} \, \text{Hz} \).

### Conclusion
The minimum frequency of the radiation that can be absorbed by the material is nearly:
\[ \text{Answer: } 5 \times 10^{14} \, \text{Hz} \]