If a semiconductor photo diode can detect a photon with a maximum wavelength of 400 nm, then its band gap energy is: Planck's constant h = 6.63 $$\times$$ 10$$^{-34}$$ J.s, Speed of light c = 3 $$\times$$ 10$$^8$$ m s$$^{-1}$$

We have a photodiode that just manages to detect photons of wavelength $$\lambda_{\max}=400\ \text{nm}$$. Detection is possible only when the photon energy is at least equal to the semiconductor’s band-gap energy $$E_g$$. Therefore the maximum detectable wavelength corresponds exactly to the condition

$$E_g = \dfrac{h\,c}{\lambda_{\max}}.$$

Here, $$h=6.63\times10^{-34}\ \text{J·s}$$ (Planck’s constant) and $$c=3\times10^{8}\ \text{m·s}^{-1}$$ (speed of light). First we convert the wavelength into metres:

$$\lambda_{\max}=400\ \text{nm}=400\times10^{-9}\ \text{m}=4.00\times10^{-7}\ \text{m}.$$

Now we substitute the numerical values into the energy formula:

$$E_g =\dfrac{(6.63\times10^{-34}\ \text{J·s})\,(3\times10^{8}\ \text{m·s}^{-1})} {4.00\times10^{-7}\ \text{m}}.$$

Multiplying the constants in the numerator, we have

$$6.63\times10^{-34}\times3\times10^{8} =19.89\times10^{-26}\ \text{J·m} =1.989\times10^{-25}\ \text{J·m}.$$

Dividing this result by the denominator gives

$$E_g =\dfrac{1.989\times10^{-25}\ \text{J·m}} {4.00\times10^{-7}\ \text{m}} =1.989\times10^{-25}\times\frac{1}{4.00}\times10^{7}\ \text{J}.$$

Because $$\dfrac{1}{4.00}=0.25$$ and $$10^{-25}\times10^{7}=10^{-18}$$, we get

$$E_g=0.49725\times10^{-18}\ \text{J}=4.9725\times10^{-19}\ \text{J}.$$

The band-gap energy is usually expressed in electron-volts. The conversion factor is

$$1\ \text{eV}=1.6\times10^{-19}\ \text{J}.$$

So we convert:

$$E_g=\dfrac{4.9725\times10^{-19}\ \text{J}} {1.6\times10^{-19}\ \text{J/eV}} =\dfrac{4.9725}{1.6}\ \text{eV} =3.1078\ \text{eV}\approx3.1\ \text{eV}.$$

Hence, the correct answer is Option D.