When radiation of wavelength lambda is used to illuminate a metallic surface, the stopping potential

Q: Why does the stopping potential decrease when we increase the wavelength?

The energy of a photon is inversely proportional to its wavelength ($E = \frac{hc}{\lambda}$). So, a longer wavelength means less incident energy, which results in ejected electrons having less kinetic energy, thus requiring a smaller stopping potential to halt them.

Q: Can I solve this by dividing the two equations instead of substituting?

Yes, you can divide the two equations to eliminate $eV$. However, cross-multiplying and solving the resulting fractional equation often leads to more algebraic mistakes under exam pressure. Eliminating $eV$ by substitution is usually cleaner and faster.

Q: What exactly is the threshold wavelength?

It is the maximum wavelength of incident light that can just barely eject an electron from the metal surface with zero kinetic energy. Any wavelength longer than this will not have enough energy to overcome the work function, and no photoelectric emission will occur.

Q: Why do we use $K_{\max}$ in the equation and not just average kinetic energy?

Electrons are bound at different depths inside the metal. The ones at the very surface require the least energy (exactly the work function) to escape, so they come out with the maximum possible kinetic energy. Electrons deeper inside lose some energy in collisions before escaping, so they emerge with less energy.

Q: What would happen if we used a wavelength of $10\lambda$ in this experiment?

Since we calculated the threshold wavelength to be $9\lambda$, any wavelength greater than $9\lambda$ (like $10\lambda$) has less energy than the work function. Therefore, absolutely no photoelectric emission would occur!

Question

When radiation of wavelength $\lambda$ is used to illuminate a metallic surface, the stopping potential is $V$. When the same surface is illuminated with radiation of wavelength $3\lambda$, the stopping potential is $\frac{V}{4}$. If the threshold wavelength for the metallic surface is $n\lambda$, then value of $n$ will be ......... .

Sigma Percentile · Accepted Answer

### The Magic of the Photoelectric Effect

Imagine you are shining a beam of light onto a pristine metallic surface. If the light has enough energy, it acts like a barrage of tiny billiard balls—photons—knocking electrons right out of the metal! This beautiful phenomenon is the **Photoelectric Effect**, and Albert Einstein won a Nobel Prize for explaining it with a remarkably elegant equation.

The core idea is simple: The energy of the incoming photon ($E = \frac{hc}{\lambda}$) is used for two things. First, it pays the "exit toll" required to free the electron from the metal, known as the **work function** ($\phi_0$). Whatever energy is left over becomes the kinetic energy of the escaping electron.

Mathematically, the maximum kinetic energy is given by:

$$K_{\max} = \frac{hc}{\lambda} - \phi_0$$

We also know that to stop these energetic electrons, we need to apply a reverse voltage called the **stopping potential** ($V$). The work done to stop them is $eV$, so $K_{\max} = eV$.

### Analyzing the Two Scenarios

The problem presents us with two distinct experimental setups using the same metal. Because it's the same metal, the work function $\phi_0$ remains absolutely constant. Let's translate the physical situations into pure algebra.

**Case 1:** We use light of wavelength $\lambda$, and the stopping potential is $V$. Plugging this into our master equation yields:

$$\frac{hc}{\lambda} - \phi_0 = eV \quad \dots 	ext{(Equation 1)}$$

**Case 2:** We switch to a less energetic light with a longer wavelength, $3\lambda$. Consequently, the stopping potential drops to $\frac{V}{4}$. Our new equation becomes:

$$\frac{hc}{3\lambda} - \phi_0 = \frac{eV}{4} \quad \dots 	ext{(Equation 2)}$$

### The Master Equation and Elimination

We now have a system of two linear equations. Our ultimate goal is to find the **threshold wavelength** ($\lambda_0$), which is deeply connected to the work function $\phi_0$. Therefore, the stopping potential $V$ is just an intermediate variable that we need to eliminate.

Let's isolate $eV$ in the second equation by multiplying the entire equation by 4:

$$4 \left( \frac{hc}{3\lambda} - \phi_0 ight) = eV$$

$$\frac{4hc}{3\lambda} - 4\phi_0 = eV \quad \dots 	ext{(Equation 3)}$$

Now, we have two different expressions that both equal $eV$. Let's equate Equation 1 and Equation 3:

$$\frac{hc}{\lambda} - \phi_0 = \frac{4hc}{3\lambda} - 4\phi_0$$

### Final Calculation and The Threshold

It's time for some careful algebraic maneuvering. Let's group all the $\phi_0$ terms on the left side and the $\frac{hc}{\lambda}$ terms on the right side:

$$4\phi_0 - \phi_0 = \frac{4hc}{3\lambda} - \frac{hc}{\lambda}$$

$$3\phi_0 = \frac{4hc - 3hc}{3\lambda}$$

$$3\phi_0 = \frac{hc}{3\lambda}$$

Dividing both sides by 3, we isolate the work function:

$$\phi_0 = \frac{hc}{9\lambda}$$

We are at the final stretch! The physical definition of the work function is the energy of a photon exactly at the threshold wavelength $\lambda_0$. So, $\phi_0 = \frac{hc}{\lambda_0}$.

By equating our two expressions for $\phi_0$, we get:

$$\frac{hc}{\lambda_0} = \frac{hc}{9\lambda}$$

The $hc$ terms gracefully cancel out, leaving us with:

$$\lambda_0 = 9\lambda$$

The problem states that the threshold wavelength is $n\lambda$. By direct comparison, we can confidently conclude that **$n = 9$**.

Enter Numerical Value:

Visualized Solution

Einstein's Photoelectric Equation

Case 1: Wavelength $λ$

Case 2: Wavelength $3 λ$

Eliminating Stopping Potential $V$

Solving for Work Function $ϕ_{0}$

Finding Threshold Wavelength $λ_{0}$

The Way Forward

The Sigma Insight: Photoelectric Effect

The Magic of the Photoelectric Effect

Analyzing the Two Scenarios

The Master Equation and Elimination

Final Calculation and The Threshold

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