When a certain photosensitive surface is illuminated with monochromatic light of frequency v, the st

Q: Why is the stopping potential given as a negative value?

Stopping potential is inherently a retarding potential (negative voltage applied to the anode). Usually, problems state its magnitude, but here the negative sign is explicitly given to test your algebraic rigor. Just plug it into the equation as is!

Q: Can kinetic energy be negative?

No! Kinetic energy is always positive. The equation $K_{max} = eV_s$ uses the magnitude of the stopping potential. If the problem gives $V_s = -V_0$, it implies the actual potential difference is negative, but the energy term $e|V_s|$ must be positive. Here, we just treat $V_s$ as the coordinate on the graph.

Q: Why did we subtract the two equations?

Subtracting the equations is the fastest way to eliminate the work function $\phi_0$, allowing us to find the relationship between $eV_0$ and $h\nu$.

Q: What does the x-intercept of the $V_s$ vs $f$ graph represent?

The x-intercept is the point where the stopping potential is zero. This corresponds to the threshold frequency $\nu_0$, where the incident photon energy exactly equals the work function.

Q: Is the slope of the graph dependent on the metal?

No, the slope of the $V_s$ vs $f$ graph is always $h/e$, which is a universal constant. Only the y-intercept (and consequently the x-intercept) depends on the specific metal's work function.

Question

When a certain photosensitive surface is illuminated with monochromatic light of frequency $
u$, the stopping potential for the photocurrent is $-V_0 / 2$. When the surface is illuminated by monochromatic light of frequency $
u / 2$, the stopping potential is $-V_0$. The threshold frequency for photoelectric emission is

Options:

(A) $\frac{4}{3}
u$

(B) $2
u$

(C) $\frac{3}{2}
u$

(D) $\frac{5}{3}
u$

Sigma Percentile · Accepted Answer

Have you ever looked at a physics problem and felt like the universe was playing a trick on you? This question is a perfect example of that. It presents us with a scenario where the stopping potential is given as a negative value.

Usually, we talk about the **magnitude** of the stopping potential, but here, the negative sign is explicitly included. I know this might look terrifying, but let's take a breath. Mathematics is the language of the universe, and it never lies. If we trust our fundamental equations and follow the logic, the answer will reveal itself beautifully.

### The Master Equation

Let's start by recalling the bedrock of the photoelectric effect: **Einstein's Photoelectric Equation**.

This elegant equation relates the maximum kinetic energy of emitted photoelectrons to the frequency of the incident light. We can write it in terms of the stopping potential $V_s$:

$$eV_s = h
u - \phi_0$$

Here, $e$ is the elementary charge, $h$ is Planck's constant, $
u$ is the frequency of the incident light, and $\phi_0$ is the work function of the metal.

This equation is essentially a straight line! If we plot $V_s$ on the y-axis and $
u$ on the x-axis, we get a linear graph where the slope is $h/e$ and the y-intercept is $-\phi_0/e$.

### Setting Up the Scenarios

The problem gives us two distinct experimental scenarios. Let's translate the English into pristine mathematics.

**Scenario 1:**
When the incident frequency is $
u$, the stopping potential is given as $-V_0/2$. Plugging this into our master equation, we get:

$$e\left(-\frac{V_0}{2}ight) = h
u - \phi_0 \quad \dots (1)$$

**Scenario 2:**
When the incident frequency is halved to $
u/2$, the stopping potential drops to $-V_0$. Substituting these values yields:

$$e(-V_0) = h\left(\frac{
u}{2}ight) - \phi_0 \quad \dots (2)$$

Take a moment to appreciate what we have here. We have successfully translated a physical phenomenon into a system of two linear equations. The physics is done; now, it's time for the algebraic dance!

### The Algebraic Dance

Our goal is to find the threshold frequency, which is directly tied to the work function $\phi_0$. Therefore, we need to solve for $\phi_0$. The easiest way to do this is to first eliminate the $eV_0$ term.

Let's subtract Equation (1) from Equation (2):

$$-eV_0 - \left(-\frac{eV_0}{2}ight) = \frac{h
u}{2} - h
u$$

Simplifying the left side:

$$-\frac{eV_0}{2} = -\frac{h
u}{2}$$

The negative signs and the denominators cancel out perfectly, leaving us with a beautifully simple relation:

$$eV_0 = h
u$$

### The Grand Finale

Now that we have the value of $eV_0$, we can substitute it back into Equation (1) to find our elusive work function $\phi_0$.

$$-\frac{h
u}{2} = h
u - \phi_0$$

Rearranging the terms to isolate $\phi_0$:

$$\phi_0 = h
u + \frac{h
u}{2}$$

$$\phi_0 = \frac{3}{2}h
u$$

We are almost at the finish line! Remember the physical meaning of the work function. It is the minimum energy required to eject an electron, and it is defined as $\phi_0 = h
u_0$, where $
u_0$ is the **threshold frequency**.

Equating the two expressions for the work function:

$$h
u_0 = \frac{3}{2}h
u$$

Canceling Planck's constant $h$ from both sides, we arrive at our final, glorious answer:

$$
u_0 = \frac{3}{2}
u$$

And there it is! By trusting the linear relationship and carefully handling our signs, we navigated through the trap of negative potentials and arrived at the exact threshold frequency. Never let a strange sign convention intimidate you; your fundamental equations are your ultimate guide!

Select Answer:

Visualized Solution

Visualizing the Data

Einstein's Photoelectric Equation

Substituting the Coordinates

Solving the System of Equations

Finding the Work Function

Calculating Threshold Frequency

The Way Forward

The Sigma Insight: Photoelectric Effect

The Master Equation

Setting Up the Scenarios

The Algebraic Dance

The Grand Finale

Similar Questions

When radiation of wavelength $λ$ is incident on a metallic surface, the stopping potential of ejected photoelectrons is $4.8$ V. If the same surface is illuminated by radiation of double the previous wavelength, then the stopping potential becomes $1.6$ V. The threshold wavelength of the metal is

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 $(\frac{h c}{e} = 1240 nm V)$

The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength $491 nm$ is $0.710 V$ . When the incident wavelength is changed to a new value, the stopping potential is $1.43 V$ . The new wavelength is

For photo-electric effect with incident photon wavelength $λ$ , the stopping potential is $V_{0}$ . Identify the correct variation(s) of $V_{0}$ with $λ$ and $\frac{1}{λ}$ .

The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 4 eV. The stopping potential in volt is

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