Animated Solution for Physics - Semiconductors: Consider a situation in which reverse biased current of a particular p-n junction increases when it is exposed to a light of wavelength $\le 621$ nm. During this process, enhancement in carrier concentration takes place due to generation of hole-electron pairs. The value of band gap is nearly

Imagine you are looking at a tiny p-n junction diode. When it is reverse-biased, very little current flows through it. But suddenly, you shine a light on it, and the current spikes! What just happened? This is the magic of optoelectronic devices, specifically a photodiode. The light consists of tiny packets of energy called photons. When these photons hit the semiconductor material, they transfer their energy to the electrons sitting in the valence band.

If a photon has enough energy, it can kick an electron all the way across the forbidden energy gap (the band gap) into the conduction band. This creates a free electron and a hole, which are then swept away by the electric field, causing the reverse current to increase. The question tells us that this process happens for light with a wavelength of $\le 621\text{ nm}$. This maximum wavelength corresponds to the minimum energy required to just cross the band gap.

To find the band gap energy $E_g$, we need to calculate the energy of a photon with a wavelength of $\lambda =621\text{ nm}$. The fundamental equation connecting the energy of a photon to its wavelength is:

where $h$ is Planck's constant ($6.63\times 10^{-34}\text{ J s}$) and $c$ is the speed of light ($3\times 10^8\text{ m/s}$).

Let's plug in the raw numbers. Remember to convert the wavelength from nanometers to meters by multiplying by $10^{-9}$.

This gives us the energy in Joules. However, if you look at the options, they are all in electron-volts (eV). This is a classic trap! We must convert our answer.

To convert Joules to electron-volts, we divide by the elementary charge $e=1.6\times 10^{-19}\text{ C}$.

Now, it's just a matter of careful arithmetic. Let's group the numbers and the powers of 10:

The band gap is approximately $2\text{ eV}$, which perfectly matches option (a).

In competitive exams like JEE, doing this long calculation can eat up precious minutes. Here is a golden shortcut:

Using this formula:

You instantly see that the answer is $2\text{ eV}$ without wrestling with massive powers of 10!