Animated Solution for Physics - Dual Nature of Matter and Radiation: The stopping potential in the context of photoelectric effect depends on the following property of incident electromagnetic radiation

Imagine you are standing in a dark room, and suddenly, a beam of light hits a metal plate. Almost instantly, tiny invisible particles—electrons—are kicked out of the metal surface. This beautiful phenomenon is known as the Photoelectric Effect. But these electrons aren't just lazily falling out; some of them are ejected with tremendous kinetic energy, zooming away from the metal plate.

To study these energetic electrons, physicists set up an experiment where they apply a reverse voltage to push the electrons back. The exact minimum negative voltage required to stop even the fastest, most energetic electron from reaching the opposite plate is called the Stopping Potential, denoted by $V_0$.

Albert Einstein won his Nobel Prize for explaining this very effect. He proposed that light is made of tiny packets of energy called photons. The energy of a single photon is given by $h u$, where $h$ is Planck's constant and $ u$ is the frequency of the light.

When a photon hits an electron, it transfers all its energy. However, the electron has to pay a 'toll tax' to escape the metal's surface. This toll tax is called the Work Function ($\varphi$). Whatever energy is left over becomes the maximum kinetic energy ($K_{\max }$) of the escaping electron.

This gives us Einstein's famous photoelectric equation:

Now, how does the stopping potential $V_0$ fit into this? The work done by the electric field to stop the fastest electron is equal to the electron's charge ($e$) multiplied by the stopping potential ($V_0$). By the work-energy theorem, this work must exactly equal the maximum kinetic energy of the electron:

Let's substitute this back into Einstein's equation:

To see exactly what the stopping potential depends on, let's isolate $V_0$ by dividing the entire equation by the charge of an electron ($e$):

Look closely at this elegant equation. It is in the form of a straight line, $y=mx+c$. For any given metal, Planck's constant ($h$), the charge of an electron ($e$), and the work function ($\varphi$) are all strict constants.

Therefore, the stopping potential $V_0$ depends on only one variable: the frequency ($ u$) of the incident electromagnetic radiation.

A very common trap that students fall into is thinking that a brighter, more intense light will increase the stopping potential. Don't make this silly mistake!

Increasing the intensity of light simply means you are firing more photons per second. More photons will eject more electrons, which increases the photoelectric current. However, the energy of each individual photon remains exactly the same ($h u$). Since the individual photon energy hasn't changed, the maximum kinetic energy of the ejected electrons doesn't change, and thus, the stopping potential remains completely unaffected.

Always remember: Frequency dictates energy and stopping potential, while intensity dictates the number of electrons and photocurrent.