A `p-n` photodiode is fabricated from a semiconductor with a band gap of `2.5 eV`. It can detect a signal of wavelength

Options

A. `6000 Å`
B. `4000 nm`
C. `6000nm`
D. `4000Å`

Explanation

To determine the wavelength that a p-n photodiode can detect, we can follow these steps:

### Step 1: Understand the relationship between energy and wavelength
The energy of a photon is related to its wavelength by the equation:
\[ E = \frac{hc}{\lambda} \]
where:
- \(E\) is the energy in electron volts (eV),
- \(h\) is Planck's constant (\(4.135667696 \times 10^{-15} \, \text{eV·s}\)),
- \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)),
- \(\lambda\) is the wavelength in meters.

### Step 2: Convert the energy from eV to Joules
The energy given is \(2.5 \, \text{eV}\). To convert this to Joules, we use the conversion factor:
\[ 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \]
Thus,
\[ E = 2.5 \, \text{eV} = 2.5 \times 1.602 \times 10^{-19} \, \text{J} = 4.005 \times 10^{-19} \, \text{J} \]

### Step 3: Rearrange the energy-wavelength equation to find wavelength
Rearranging the equation \(E = \frac{hc}{\lambda}\) gives:
\[ \lambda = \frac{hc}{E} \]

### Step 4: Substitute the values of \(h\), \(c\), and \(E\)
Using \(h = 4.135667696 \times 10^{-15} \, \text{eV·s}\) and \(c = 3 \times 10^8 \, \text{m/s}\):
\[ \lambda = \frac{(4.135667696 \times 10^{-15} \, \text{eV·s})(3 \times 10^8 \, \text{m/s})}{2.5 \, \text{eV}} \]

### Step 5: Calculate \(\lambda\)
Calculating the numerator:
\[ hc = (4.135667696 \times 10^{-15})(3 \times 10^8) = 1.240 \times 10^{-6} \, \text{eV·m} \]
Now substituting this into the equation for \(\lambda\):
\[ \lambda = \frac{1.240 \times 10^{-6}}{2.5} = 4.96 \times 10^{-7} \, \text{m} \]
To convert this to nanometers (1 nm = \(10^{-9}\) m):
\[ \lambda = 4.96 \times 10^{-7} \, \text{m} = 496 \, \text{nm} \]

### Step 6: Conclusion
The p-n photodiode can detect wavelengths shorter than \(496 \, \text{nm}\). Therefore, it can detect signals in the visible spectrum and some ultraviolet light.

### Final Answer
The p-n photodiode can detect signals with a wavelength of approximately \(496 \, \text{nm}\) or shorter.

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