Three photodiodes `D_1 , D_2 ` and `D_3` are made of semiconductors having
band gaps of ` 2.5 eV , 2 eV` and 3 eV , respectively . Which one will be able to detect light of wavelength ` 6000 Å ` ?

Options

A. `D_1`
B. `D_2`
C. `D_3`
D. `D_1 and D_2` are both

Explanation

To determine which photodiode can detect light of wavelength 6000 Å, we need to follow these steps:

### Step 1: Convert Wavelength to Energy
We start by calculating the energy of the light corresponding to the wavelength of 6000 Å. The energy (E) of a photon can be calculated using the formula:

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

Where:
- \( h \) is Planck's constant, approximately \( 4.1357 \times 10^{-15} \, \text{eV s} \)
- \( c \) is the speed of light, approximately \( 3 \times 10^{10} \, \text{cm/s} \) or \( 3 \times 10^{18} \, \text{Å/s} \)
- \( \lambda \) is the wavelength in Ångströms (6000 Å)

### Step 2: Calculate Energy
Substituting the values into the equation:

\[ E = \frac{(4.1357 \times 10^{-15} \, \text{eV s}) \times (3 \times 10^{18} \, \text{Å/s})}{6000 \, \text{Å}} \]

Calculating this gives:

\[ E \approx \frac{1.240 \times 10^{4} \, \text{eV Å}}{6000 \, \text{Å}} \approx 2.0667 \, \text{eV} \]

### Step 3: Compare with Band Gaps
Now we compare this energy with the band gaps of the three photodiodes:

- For \( D_1 \): Band gap = 2.5 eV
- For \( D_2 \): Band gap = 2.0 eV
- For \( D_3 \): Band gap = 3.0 eV

### Step 4: Determine Which Photodiode Can Detect the Light
A photodiode can detect light if the energy of the incoming photon is greater than or equal to the band gap energy of the semiconductor.

- For \( D_1 \): \( 2.0667 \, \text{eV} < 2.5 \, \text{eV} \) (cannot detect)
- For \( D_2 \): \( 2.0667 \, \text{eV} > 2.0 \, \text{eV} \) (can detect)
- For \( D_3 \): \( 2.0667 \, \text{eV} < 3.0 \, \text{eV} \) (cannot detect)

### Conclusion
Only photodiode \( D_2 \) can detect light of wavelength 6000 Å.