Animated Solution for Physics - Dual Nature of Matter and Radiation: When a monochromatic point source of light is at a distance of 0.2 m from a photoelectric cell, the cut-off voltage and the saturation current are respectively 0.6 V and 18.0 mA. If the same source is placed 0.6 m away from the photoelectric cell, then

Imagine you are standing in front of a sprinkler. The water droplets hitting you represent the photons of light. The energy of each individual droplet depends on how fast it was shot out, not on how many droplets are hitting you. In the photoelectric effect, the stopping potential is determined entirely by the maximum kinetic energy of the emitted photoelectrons.

According to Einstein's photoelectric equation:

Notice that the stopping potential $V_0$ depends only on the frequency $ u$ of the incident light and the work function $\varphi$ of the metal. When we move the light source from $0.2\text{ m}$ to $0.6\text{ m}$, we are not changing the color (frequency) of the light, nor are we changing the metal plate. Therefore, the energy of each individual photon remains exactly the same. Consequently, the stopping potential remains unchanged at $0.6\text{ V}$.

Now, let's think about the saturation current. This current is a measure of the total number of photoelectrons emitted per second, which is directly proportional to the total number of photons striking the metal plate per second—in other words, the intensity of the light.

For a point source, light spreads out spherically in all directions. As you move further away, the same amount of light energy is spread over a much larger area. This geometric spreading follows the famous inverse square law:

Since the saturation current $I_s$ is directly proportional to the intensity $I$, it must also follow the inverse square law:

Let's set up a ratio to find the new saturation current when the distance is increased. We know the initial distance $r_1=0.2\text{ m}$ and the final distance $r_2=0.6\text{ m}$. The initial saturation current is $I_{s1}=18.0\text{ mA}$.

Substituting the given values into our equation:

The ratio of the distances is $\frac{1}{3}$. Squaring this ratio gives us $\frac{1}{9}$. This means the new intensity is one-ninth of the original intensity!

By simply moving the source three times further away, the saturation current plummets to $2.0\text{ mA}$. This beautifully illustrates how spatial geometry governs the flow of photoelectrons in a circuit!