Animated Solution for Physics - Semiconductors: If a semiconductor photodiode can detect a photon with a maximum wavelength of $400\text{ nm}$, then its band gap energy is (Take, Planck's constant, $h=6.63\times 10^{-34}\text{ J-s}$ and speed of light, $c=3\times 10^8\text{ m/s}$)

Its primary job is to detect light, but it doesn't just detect any light. For a photodiode to register a photon, that photon must pack enough punch to kick an electron from the valence band all the way up to the conduction band.

This minimum energy required is exactly what we call the band gap energy ($E_g$). The problem states that the photodiode can detect photons with a maximum wavelength of $400\text{ nm}$. Why maximum? Because wavelength and energy are inversely proportional. A maximum wavelength corresponds to the minimum energy required to bridge the band gap.

To find the energy of this photon, we turn to the famous Planck-Einstein relation:

Here, $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength. If we plug in the standard SI values ($h=6.63\times 10^{-34}\text{ J}\cdot \text{s}$ and $c=3\times 10^8\text{ m/s}$), we would get the energy in Joules. But look at our options—they are all in electron-volts (eV). Converting Joules to eV involves dividing by $1.6\times 10^{-19}$, which makes the calculation quite tedious.

Instead of dealing with those messy powers of 10, we can use the combined value of $hc$ directly in more convenient units:

This single substitution is a massive time-saver. Let's use it!

Now, we simply substitute our maximum wavelength $\lambda =400\text{ nm}$ into our modified equation:

Notice how beautifully the nanometer (nm) units cancel out, leaving us purely with electron-volts (eV).

And there we have it! The band gap energy of the semiconductor is $3.1\text{ eV}$, which perfectly matches option (c). Always remember this $1240$ trick when dealing with photons and electron-volts; it will save you precious minutes in the exam hall.