Animated Solution for Physics - Dual Nature of Matter and Radiation: In a photoelectric experiment ultraviolet light of wavelength $280\text{ nm}$ is used with lithium cathode having work-function $\varphi =2.5\text{ eV}$. If the wavelength of incident light is switched to $400\text{ nm}$, find out the change in the stopping potential. ($h=6.63\times 10^{-34}\text{ Js}$, and $c=3\times 10^8{\text{ ms}}^{-1}$)

Imagine shining a beam of light onto a piece of metal and watching electrons pop out like tiny sparks. This isn't magic; it's the photoelectric effect, a phenomenon that fundamentally changed our understanding of light and matter. In this problem, we are dealing with a lithium cathode and exploring how changing the color (or wavelength) of the incident light affects the energy of these ejected electrons.

To solve this, we turn to Albert Einstein's elegant photoelectric equation:

Here, $E$ is the energy of the incoming photon, $\varphi$ is the work function (the minimum energy required to break an electron free from the metal's surface), and $(KE{)}_{\max }$ is the maximum kinetic energy of the escaping electron.

We can also express the kinetic energy in terms of the stopping potential ($V_0$), which is the voltage needed to stop even the fastest electrons:

So, our equation becomes:

Calculating $\frac{hc}{\lambda }$ with standard SI units can be a nightmare of scientific notation. But there's a brilliant shortcut! If we express the wavelength $\lambda$ in nanometers (nm) and the energy in electron-volts (eV), the product $hc$ is approximately $1240\text{ eV}\cdot \text{nm}$.

This little trick will save us a tremendous amount of time.

First, we shine ultraviolet light with a wavelength of $280\text{ nm}$. Let's find the energy of these photons:

Now, we subtract the work function of lithium ($\varphi =2.5\text{ eV}$) to find the stopping potential:

So, the first stopping potential is $V_{01}=1.93\text{ V}$.

Next, we switch to a longer wavelength of $400\text{ nm}$. Longer wavelength means lower energy. Let's calculate it:

Again, we subtract the work function:

The new stopping potential is $V_{02}=0.6\text{ V}$.

The question asks for the change in the stopping potential. We simply find the difference between the two values we calculated:

Rounding to one decimal place, we get $1.3\text{ V}$.

Notice how a relatively small change in the wavelength of light resulted in a significant drop in the stopping potential. This perfectly illustrates the particle nature of light: individual photons interact with individual electrons, and their energy is strictly dictated by their wavelength!