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 ` ?

Explanation

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

### Step 1: Convert the wavelength from Ångströms to meters.
1 Å = \(10^{-10}\) m, so:
\[ 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} = 6 \times 10^{-7} \, \text{m} \]

### Step 2: Calculate the energy of the incident light using the formula \(E = \frac{hc}{\lambda}\).
Where:
- \(h\) (Planck's constant) = \(6.626 \times 10^{-34} \, \text{Js}\)
- \(c\) (speed of light) = \(3 \times 10^8 \, \text{m/s}\)
- \(\lambda\) = \(6 \times 10^{-7} \, \text{m}\)

Now, substituting the values:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{Js}) \times (3 \times 10^8 \, \text{m/s})}{6 \times 10^{-7} \, \text{m}} \]
Calculating this gives:
\[ E \approx \frac{1.9878 \times 10^{-25}}{6 \times 10^{-7}} \approx 3.313 \times 10^{-19} \, \text{J} \]

### Step 3: Convert the energy from Joules to electronvolts (eV).
1 eV = \(1.6 \times 10^{-19} \, \text{J}\), so:
\[ E \approx \frac{3.313 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 2.07 \, \text{eV} \]

### Step 4: Compare the energy of the incident light with the band gaps of the photodiodes.
- For photodiode \(D_1\): Band gap = 2.5 eV
- For photodiode \(D_2\): Band gap = 2.0 eV
- For photodiode \(D_3\): Band gap = 3.0 eV

To detect the light, the energy of the incident light must be greater than or equal to the band gap energy of the photodiode.

### Step 5: Determine which photodiode can detect the light.
- \(D_1\) (2.5 eV): Cannot detect (2.07 eV < 2.5 eV)
- \(D_2\) (2.0 eV): Can detect (2.07 eV > 2.0 eV)
- \(D_3\) (3.0 eV): Cannot detect (2.07 eV < 3.0 eV)

### Conclusion:
The photodiode \(D_2\) with a band gap of 2.0 eV will be able to detect light of wavelength 6000 Å.

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