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. 6000 nm
C. 4000 nm
D. 4000 Å

Explanation

To solve the problem of determining the detectable signal wavelength for a p-n photodiode made from a semiconductor with a band gap of 2.5 eV, we will 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 of the photon,
- \( h \) is 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.

### Step 2: Convert the band gap energy from eV to Joules
The energy given is in electron volts (eV). To convert this to Joules, use the conversion factor:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
Thus, the energy in Joules is:
\[ E = 2.5 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 4.0 \times 10^{-19} \, \text{J} \]

### Step 3: Rearrange the energy-wavelength equation to find the threshold wavelength
From the equation \( E = \frac{hc}{\lambda} \), we can rearrange to find \( \lambda \):
\[ \lambda = \frac{hc}{E} \]

### Step 4: Substitute the values into the equation
Now, substituting the known values:
- \( h = 6.626 \times 10^{-34} \, \text{Js} \)
- \( c = 3 \times 10^8 \, \text{m/s} \)
- \( E = 4.0 \times 10^{-19} \, \text{J} \)

We get:
\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{Js}) \times (3 \times 10^8 \, \text{m/s})}{4.0 \times 10^{-19} \, \text{J}} \]

### Step 5: Calculate the threshold wavelength
Calculating this gives:
\[ \lambda = \frac{1.9878 \times 10^{-25}}{4.0 \times 10^{-19}} \]
\[ \lambda = 4.9695 \times 10^{-7} \, \text{m} \]
To convert this to angstroms (1 m = \( 10^{10} \) angstroms):
\[ \lambda = 4.9695 \times 10^{-7} \, \text{m} \times 10^{10} \, \text{Å/m} = 4969.5 \, \text{Å} \]

### Step 6: Determine the detectable wavelengths
The photodiode can detect signals with wavelengths less than the threshold wavelength of approximately 5000 Å.

### Step 7: Analyze the options
If we have options for wavelengths, we can now check which ones are less than 5000 Å. For example:
- If the options are 4000 Å, 6000 Å, 7000 Å, and 5000 Å, only 4000 Å is less than 5000 Å and can be detected.

### Final Answer
The p-n photodiode can detect signals with wavelengths less than approximately 5000 Å.

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