To detect light of wavelength 500 nm, the photodiode must be fabricated from a semiconductor of minimum bandwidth of

Options

A. 1.24 eV
B. 0.62 eV
C. 2.48 eV
D. 32. eV

Answer

To find the minimum bandwidth of a semiconductor photodiode required to detect light of wavelength 500 nm, we will follow these steps:

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

The frequency \( \nu \) can also be related to the wavelength \( \lambda \) by the equation:
\[ \nu = \frac{c}{\lambda} \]
where:
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength of the light.

### Step 2: Convert the wavelength from nanometers to meters
Given the wavelength \( \lambda = 500 \, \text{nm} \):
\[ \lambda = 500 \times 10^{-9} \, \text{m} \]

### Step 3: Calculate the frequency of the light
Using the formula for frequency:
\[ \nu = \frac{c}{\lambda} = \frac{3 \times 10^8 \, \text{m/s}}{500 \times 10^{-9} \, \text{m}} \]
Calculating this gives:
\[ \nu = 6 \times 10^{14} \, \text{Hz} \]

### Step 4: Calculate the energy of the photon
Now, substituting the frequency into the energy formula:
\[ E = h \nu = (6.626 \times 10^{-34} \, \text{J s}) \times (6 \times 10^{14} \, \text{Hz}) \]
Calculating this gives:
\[ E = 3.976 \times 10^{-19} \, \text{J} \]

### Step 5: Convert energy from joules to electron volts
To convert joules to electron volts, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ E \, (\text{in eV}) = \frac{3.976 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \]
Calculating this gives:
\[ E \approx 2.485 \, \text{eV} \]

### Conclusion
The minimum bandwidth of the semiconductor required to detect light of wavelength 500 nm is approximately **2.48 eV**.