In a photoelectric experiment set-up, photons of energy 5 eV falls on the cathode having work functi

Q: Why does the stopping potential remain the same for both curves?

The stopping potential depends on the maximum kinetic energy of the photoelectrons, which in turn depends only on the energy of the incident photons and the work function of the material. Since both of these remain constant, the stopping potential does not change with intensity.

Q: Why does the saturation current double when the intensity is doubled?

Intensity is the energy incident per unit area per unit time. For a given frequency, doubling the intensity means doubling the number of incident photons per second. This results in twice as many photoelectrons being emitted per second, thereby doubling the saturation current.

Q: What does the negative sign in the stopping potential indicate?

The negative sign indicates that the anode is at a lower potential compared to the cathode. This creates a retarding electric field that opposes the motion of the emitted photoelectrons.

Q: Why does the current not increase indefinitely with anode potential?

Once the anode potential is sufficiently positive, all the emitted photoelectrons are collected by the anode. Further increasing the potential cannot increase the current because the rate of emission of photoelectrons is limited by the intensity of the incident light. This maximum current is called the saturation current.

Q: Can the kinetic energy of photoelectrons be negative?

No, kinetic energy is a scalar quantity and is always non-negative. If the energy of the incident photon is less than the work function, no photoelectrons are emitted.

Question

In a photoelectric experiment set-up, photons of energy $5 	ext{ eV}$ falls on the cathode having work function $3 	ext{ eV}$. (a) If the saturation current is $i_A = 4 \mu	ext{A}$ for intensity $10^{-5} 	ext{ W}/	ext{m}^2$, then plot a graph between anode potential and current. (b) Also, draw a graph for intensity of incident radiation $2 	imes 10^{-5} 	ext{ W}/	ext{m}^2$.

Sigma Percentile · Accepted Answer

The photoelectric effect is one of the most beautiful phenomena in modern physics, bridging the gap between the wave and particle theories of light. In this problem, we are asked to visualize how the photoelectric current varies with the anode potential under different intensities of incident light. Let's break down the physics step by step!

### Analyzing the Setup

We are given a cathode with a work function $W = 3 	ext{ eV}$. Photons of energy $E = 5 	ext{ eV}$ are incident on it.

According to **Einstein's Photoelectric Equation**, the maximum kinetic energy $K_{\max}$ of the emitted photoelectrons is the difference between the energy of the incident photons and the work function of the material:

$$K_{\max} = E - W$$

Substituting the given values, we get:

$$K_{\max} = 5 	ext{ eV} - 3 	ext{ eV} = 2 	ext{ eV}$$

This means the fastest electrons leave the cathode with a kinetic energy of $2 	ext{ eV}$.

### The Stopping Potential

To stop these fastest electrons from reaching the anode, we must apply a negative potential (retarding potential) to the anode. The stopping potential $V_0$ is related to the maximum kinetic energy by:

$$K_{\max} = e V_0$$

Since $K_{\max} = 2 	ext{ eV}$, the stopping potential is:

$$V_0 = 2 	ext{ V}$$

On our graph, this means the photocurrent will drop to zero when the anode potential is **$-2 	ext{ V}$**. Because the energy of the incident photons and the work function remain constant, this stopping potential will be the same for both cases (a) and (b).

### The Role of Intensity

Next, let's look at the saturation current. The saturation current is the maximum current that flows when all emitted photoelectrons are collected by the anode. This current is directly proportional to the number of photoelectrons emitted per second, which in turn is directly proportional to the **intensity** of the incident light.

For case (a), the intensity is $I_1 = 10^{-5} 	ext{ W}/	ext{m}^2$, and the saturation current is $i_1 = 4 \mu	ext{A}$.

For case (b), the intensity is $I_2 = 2 	imes 10^{-5} 	ext{ W}/	ext{m}^2$. Notice that the intensity has exactly **doubled**.

Because $i \propto I$, the new saturation current will also double:

$$i_2 = 2 	imes 4 \mu	ext{A} = 8 \mu	ext{A}$$

### Plotting the Final Graph

Now we have all the pieces to draw the graph:

1. Both curves must originate from the same point on the voltage axis: **$-2 	ext{ V}$**.
2. As the anode potential increases, the current rises and eventually levels off (saturates).
3. Curve (a) will level off at a height corresponding to **$4 \mu	ext{A}$**.
4. Curve (b) will level off at a height corresponding to **$8 \mu	ext{A}$**.

This elegant graph perfectly encapsulates the core principles of the photoelectric effect: the stopping potential depends only on the photon energy, while the saturation current depends only on the light intensity!

Visualized Solution

$K_{m a x} = E - W$

$K_{m a x} = 2 eV$

$V_{0} = 2 V$

$i \propto I$

\text{Plotting the Graphs}

The Sigma Insight: Photoelectric Effect

Analyzing the Setup

The Stopping Potential

The Role of Intensity

Plotting the Final Graph

Similar Questions

The figure shows the variation of photocurrent with anode potential for a photosensitive surface for three different radiations. Let $I_{a}, I_{b}$ and $I_{c}$ be the intensities and $f_{a}, f_{b}$ and $f_{c}$ be the frequencies for the curves $a, b$ and $c$ respectively. Then :

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

In a photoelectric effect experiment, the graph of stopping potential $V$ versus reciprocal of wavelength $(1/ λ)$ obtained is shown in the figure. As the intensity of incident radiation is increased,

The anode voltage of a photocells kept fixed. The wavelength $λ$ of the light falling on the cathode is gradually changed. The plate current $I$ of photocell varies as follows

In a photoelectric experiment, the wavelength of the light incident on a metal is changed from $300 nm$ to $400 nm$ . The decrease in the stopping potential is close to $(\frac{h c}{e} = 1240 nm V)$