Animated Solution for Physics - Dual Nature of Matter and Radiation: The threshold frequency for a metallic surface corresponds to an energy of 6.2 eV and the stopping potential for a radiation incident on this surface is 5 V. The incident radiation lies in

Imagine you are a detective trying to identify a mysterious beam of light. You can't see it directly, but you can observe its effects on a piece of metal. This is exactly what we are doing in this classic photoelectric effect problem!

We are given two crucial clues. First, the 'threshold frequency' corresponds to an energy of $6.2\text{ eV}$. This is just a fancy way of stating the work function ($\varphi$) of the metal. Think of the work function as the toll fee an electron must pay to escape the metal's surface.

Our second clue is the stopping potential, which is $5\text{ V}$. The stopping potential is the electrical 'brakes' applied to stop the fastest-moving electrons. Because the charge of an electron is $e$, a stopping potential of $5\text{ V}$ means the maximum kinetic energy ($K_{\max }$) of the ejected electrons is exactly $5\text{ eV}$.

Now, we bring in the heavy artillery: Einstein's Photoelectric Equation. Albert Einstein brilliantly proposed that light consists of packets of energy called photons. When a photon hits an electron, it transfers all its energy ($E$) to it.

The equation is beautifully simple:

The total energy of the incoming photon is split into two parts: paying the toll fee ($\varphi$) and giving the electron some kinetic energy to fly away ($K_{\max }$).

Let's plug in our clues:

So, our mysterious incident radiation consists of photons, each carrying $11.2\text{ eV}$ of energy. But what kind of light is this? To find out, we need its wavelength.

In the quantum world, energy and wavelength are inversely related. The formula connecting them is:

Here, $h$ is Planck's constant and $c$ is the speed of light. Now, you could plug in the messy SI values, but as elite JEE problem solvers, we use the ultimate shortcut: $hc\approx 1240\text{ eV}\cdot \text{nm}$.

Let's substitute our energy into the shortcut formula:

Doing the quick math, we find:

We have our wavelength! Now, we just need to check the electromagnetic spectrum.

Remember the golden rule for the visible spectrum: it stretches from about $400\text{ nm}$ (violet) to $700\text{ nm}$ (red). Our calculated wavelength is $110.7\text{ nm}$, which is significantly shorter than $400\text{ nm}$.

Light with a wavelength shorter than violet light is known as Ultraviolet (UV) radiation. Therefore, our mysterious incident beam lies squarely in the ultra-violet region. Mystery solved!