1. Concept used

For a photodiode to detect a photon, the photon energy must be at least equal to the band gap energy:

$E_g = h\nu = \frac{hc}{\lambda_{\max}}$

Here, the maximum wavelength corresponds to the minimum photon energy that can still be detected.

2. Given data

$h = 6.63 \times 10^{-34}\ \text{J·s}$ $c = 3 \times 10^8\ \text{m/s}$ $\lambda_{\max} = 400\ \text{nm} = 400 \times 10^{-9}\ \text{m} = 4 \times 10^{-7}\ \text{m}$

3. Calculate band gap energy in joules

$E_g = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{4 \times 10^{-7}}$

First compute the numerator:

$6.63 \times 10^{-34} \times 3 \times 10^8 = 19.89 \times 10^{-26} = 1.989 \times 10^{-25}$

Now divide by $4 \times 10^{-7}$:

$E_g = \frac{1.989 \times 10^{-25}}{4 \times 10^{-7}} = 0.49725 \times 10^{-18} = 4.9725 \times 10^{-19}\ \text{J}$

4. Convert joules to electron volts

Using:

$1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$

So,

$E_g = \frac{4.9725 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.11\ \text{eV}$

5. Match with options

$E_g \approx 3.1\ \text{eV}$

So the correct option is:

C: 3.1 eV

6. Comparison with stored answer

Stored correct answer: C

My derived answer: C

They agree.