A p - n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Which of the following wavelengths it can detect?

Options

A. 950 nm
B. 850 nm
C. 580 nm
D. 440 nm

Explanation

To determine the wavelength that a p-n photodiode can detect, we need to relate the energy of the photon to the bandgap energy of the semiconductor. Here’s the step-by-step solution:

### Step 1: Understand the relationship between energy and wavelength
The energy (E) of a photon is related to its wavelength (λ) by the equation:
\[ E = \frac{hc}{\lambda} \]
where:
- \( E \) is the energy in electron volts (eV),
- \( h \) is Planck's constant (\( 4.1357 \times 10^{-15} \) eV·s),
- \( c \) is the speed of light (\( 3 \times 10^8 \) m/s),
- \( \lambda \) is the wavelength in meters.

### Step 2: Convert the constants to appropriate units
We can use the value of \( hc \) in eV·nm for convenience:
\[ hc \approx 1242 \text{ eV·nm} \]

### Step 3: Set up the equation with the given bandgap
Given that the bandgap energy \( E \) is 2.8 eV, we can set up the equation:
\[ 2.8 = \frac{1242}{\lambda} \]

### Step 4: Rearrange the equation to solve for wavelength
Rearranging the equation to find \( \lambda \):
\[ \lambda = \frac{1242}{2.8} \]

### Step 5: Calculate the wavelength
Now, we can perform the calculation:
\[ \lambda = \frac{1242}{2.8} \approx 443.57 \text{ nm} \]

### Step 6: Round the answer and match with options
Rounding the value gives us approximately 444 nm. We then check the options provided in the question to find the closest match, which is option 4 (440 nm).

### Final Answer:
The p-n photodiode can detect wavelengths around **440 nm**.

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