When radiation of wavelength lambda is incident on a metallic surface, the stopping potential of eje

Q: Why does the stopping potential decrease when the wavelength is doubled?

The energy of a photon is inversely proportional to its wavelength ($E = \frac{hc}{\lambda}$). When you double the wavelength, you halve the energy of the incoming photons. With less incident energy, the ejected electrons have less kinetic energy, so a smaller stopping potential is required to halt them.

Q: Why do we divide the two equations instead of subtracting them?

While subtracting is a valid mathematical method, dividing the equations is much faster here. Division immediately cancels out the constants $hc$ and $e$, transforming a physics equation into a straightforward algebraic ratio.

Q: Does the work function of the metal change in the second case?

No, the work function is an intrinsic property of the specific metal surface being used. Since the problem states 'the same surface is illuminated', the work function (and therefore the threshold wavelength $\lambda_0$) remains perfectly constant.

Q: What exactly is the threshold wavelength?

The threshold wavelength ($\lambda_0$) is the absolute maximum wavelength of light that possesses just enough energy to eject an electron from the metal surface with zero kinetic energy. Any wavelength longer than $\lambda_0$ will not cause photoelectric emission.

Q: What would happen if we increased the intensity of the light instead of changing the wavelength?

Increasing the intensity means more photons are hitting the surface per second, which would eject more electrons per second (increasing the photocurrent). However, the energy of each individual photon remains the same, so the maximum kinetic energy and the stopping potential would not change at all.

Question

When radiation of wavelength $\lambda$ 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

Options:

(A) $2\lambda$

(B) $4\lambda$

(C) $8\lambda$

(D) $6\lambda$

Sigma Percentile · Accepted Answer

### The Magic of the Photoelectric Effect

Imagine a pristine metallic surface resting in a vacuum. When a beam of light strikes this surface, something magical happens: electrons are violently ejected from the metal. This phenomenon, known as the photoelectric effect, was a monumental discovery that proved light behaves as a particle—a photon.

Every metal holds onto its electrons with a certain amount of binding energy, known as the **work function** ($\phi$). To kick an electron out, the incoming photon must have an energy greater than this work function. The minimum energy required corresponds to a maximum wavelength, which we call the **threshold wavelength** ($\lambda_0$).

### Setting Up Einstein's Master Equation

Albert Einstein beautifully summarized this energy exchange with his famous photoelectric equation. He stated that the maximum kinetic energy ($K_{\max}$) of the ejected electron is simply the energy of the incoming photon minus the work function of the metal.

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

We can express the maximum kinetic energy in terms of the stopping potential ($V_0$), which is the voltage required to stop even the fastest ejected electron. So, $K_{\max} = eV_0$. The energy of the photon is $\frac{hc}{\lambda}$, and the work function is $\frac{hc}{\lambda_0}$. Substituting these, our master equation becomes:

$$eV_0 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$$

### Analyzing the Two Scenarios

The problem presents us with two distinct cases. Let's translate the physical situations into mathematical equations.

**Case 1:** The incident light has a wavelength of $\lambda$, and the stopping potential is $4.8 	ext{ V}$. Plugging this into our master equation gives us:

$$4.8 e = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} \quad \dots (i)$$

**Case 2:** The wavelength of the incident light is doubled to $2\lambda$. Because the photons now have less energy, the ejected electrons are slower, and the stopping potential drops to $1.6 	ext{ V}$. Our second equation is:

$$1.6 e = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0} \quad \dots (ii)$$

### The Mathematical Execution

We now have a system of two equations. Our goal is to find the threshold wavelength $\lambda_0$ in terms of $\lambda$. The most elegant way to solve this is to divide equation (i) by equation (ii). This brilliant move instantly eliminates the constants $h$, $c$, and $e$, leaving us with a pure algebraic relationship.

$$\frac{4.8 e}{1.6 e} = \frac{\frac{hc}{\lambda} - \frac{hc}{\lambda_0}}{\frac{hc}{2\lambda} - \frac{hc}{\lambda_0}}$$

Simplifying the left side gives us $3$. On the right side, we can factor out and cancel $hc$:

$$3 = \frac{\frac{1}{\lambda} - \frac{1}{\lambda_0}}{\frac{1}{2\lambda} - \frac{1}{\lambda_0}}$$

Now, we cross-multiply to flatten the equation:

$$3 \left( \frac{1}{2\lambda} - \frac{1}{\lambda_0} ight) = \frac{1}{\lambda} - \frac{1}{\lambda_0}$$

Expanding the brackets:

$$\frac{3}{2\lambda} - \frac{3}{\lambda_0} = \frac{1}{\lambda} - \frac{1}{\lambda_0}$$

Next, we group the terms containing $\lambda$ on one side and the terms containing $\lambda_0$ on the other side. This is where you must be careful not to make a silly sign mistake!

$$\frac{3}{2\lambda} - \frac{1}{\lambda} = \frac{3}{\lambda_0} - \frac{1}{\lambda_0}$$

Finding a common denominator for the left side:

$$\frac{3 - 2}{2\lambda} = \frac{2}{\lambda_0}$$

$$\frac{1}{2\lambda} = \frac{2}{\lambda_0}$$

Finally, cross-multiplying one last time reveals our answer:

$$\lambda_0 = 4\lambda$$

### The Final Takeaway

The threshold wavelength of the metal is exactly four times the initial incident wavelength. This means that if you were to shine light with a wavelength greater than $4\lambda$, no electrons would be ejected at all, regardless of how bright the light is! This beautifully highlights the quantum nature of light.

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Visualized Solution

$Photoelectric Effect Setup$

$Einstein’s Photoelectric Equation$

$Case 1: Incident Wavelength λ$

$Case 2: Incident Wavelength 2 λ$

$Dividing the Equations$

$Algebraic Simplification$

$Final Calculation$

$Conceptual Insight$

The Sigma Insight: Photoelectric Effect

The Magic of the Photoelectric Effect

Setting Up Einstein's Master Equation

Analyzing the Two Scenarios

The Mathematical Execution

The Final Takeaway

Similar Questions

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Photoelectric Effect Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Einstein’s Photoelectric Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case 1: Incident Wavelength λ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case 2: Incident Wavelength 2λ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Dividing the Equations

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Algebraic Simplification

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conceptual Insight