When light falls on a photo diode, its conductivity increases. Experimetally it is found that the conductivity changes only when the wavelength of the incident light is less than 620 nm. The band gap of the diode is

Options

A. 0.75 eV
B. 1.1 eV
C. 1.5 eV
D. 2.0 eV

Answer

To determine the band gap of the photodiode based on the given wavelength of light, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Wavelength**:
The problem states that the conductivity of the photodiode increases when the wavelength of the incident light is less than 620 nm. We will use this wavelength for our calculations.

\[ \text{Wavelength} (\lambda) = 620 \, \text{nm} = 620 \times 10^{-9} \, \text{m} \]

2. **Use the Band Gap Energy Formula**:
The energy of the band gap (\(E_g\)) can be calculated using the formula:

\[ E_g = \frac{hc}{\lambda} \]

where:
- \(h\) is the Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\))
- \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\))
- \(\lambda\) is the wavelength in meters.

3. **Substitute the Values**:
Plugging in the values into the formula:

\[ E_g = \frac{(6.626 \times 10^{-34} \, \text{Js}) \times (3 \times 10^8 \, \text{m/s})}{620 \times 10^{-9} \, \text{m}} \]

4. **Calculate the Energy**:
Performing the calculation step-by-step:

\[ E_g = \frac{1.9878 \times 10^{-25}}{620 \times 10^{-9}} = 3.20484 \times 10^{-19} \, \text{J} \]

5. **Convert Joules to Electron Volts**:
To convert the energy from joules to electron volts, we use the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\):

\[ E_g \, (\text{in eV}) = \frac{3.20484 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 2.003 \, \text{eV} \]

6. **Final Result**:
Rounding the result gives us:

\[ E_g \approx 2.0 \, \text{eV} \]

### Conclusion:
The band gap of the photodiode is approximately **2.0 eV**.