A certain metallic surface is illuminated by monochromatic radiation of wavelength lambda. The stopp

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

Dividing the equations directly eliminates the $eV_0$ term because it appears as a multiple in both equations. Subtracting would require multiplying one equation by a factor first, which is slightly more work, though both methods are mathematically valid.

Q: What exactly is the threshold wavelength?

The threshold wavelength is the maximum wavelength of light that can eject an electron from a specific metal. If the incident light has a wavelength longer than this, its photons won't have enough energy to overcome the metal's work function.

Q: Why did the stopping potential decrease when the wavelength was doubled?

The energy of a photon is inversely proportional to its wavelength ($E = hc/\lambda$). Doubling the wavelength halves the photon's energy. With less incoming energy, the ejected electrons have less kinetic energy, so a smaller stopping potential is needed to halt them.

Q: Can the work function be negative?

No, the work function is the minimum energy required to remove an electron from the surface of a material. It is always a positive value, representing a binding energy barrier.

Q: What if the problem asked for threshold frequency instead?

You would use the relation $\phi = h\nu_t$ instead of $\phi = hc/\lambda_t$. Once you find $\phi = hc/4\lambda$, you would set it equal to $h\nu_t$ and solve for $\nu_t = c/4\lambda$.

Question

A certain metallic surface is illuminated by monochromatic radiation of wavelength $\lambda$. The stopping potential for photoelectric current for this radiation is $3V_0$. If the same surface is illuminated with a radiation of wavelength $2\lambda$, the stopping potential is $V_0$. The threshold wavelength of this surface for photoelectric effect is ...... $\lambda$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are conducting a photoelectric experiment. You have a metallic surface, and you shine light on it. When the light hits the metal, electrons are ejected. But they don't just gently float away; they are kicked out with some kinetic energy!

Einstein's photoelectric equation beautifully captures this energy conservation:
$$K_{\max} = \frac{hc}{\lambda} - \phi$$

Here, $K_{\max}$ is the maximum kinetic energy of the ejected electrons, $\lambda$ is the wavelength of the incident light, and $\phi$ is the work function of the metal. The work function is essentially the "toll fee" the electron must pay to escape the metal surface.

We also know that the maximum kinetic energy can be stopped by applying a reverse voltage, known as the stopping potential ($V_0$). So, we can rewrite the equation as:
$$eV_0 = \frac{hc}{\lambda} - \phi$$

### The Master Equations

The problem gives us two distinct scenarios. Let's translate them into math.

**Case 1:** The incident light has a wavelength of $\lambda$, and the stopping potential is $3V_0$. Plugging this into our master equation, we get:
$$e(3V_0) = \frac{hc}{\lambda} - \phi \quad \dots (1)$$

**Case 2:** The incident light is changed to a longer wavelength of $2\lambda$, and the stopping potential drops to $V_0$. This makes physical sense—a longer wavelength means less energetic photons, so the ejected electrons have less kinetic energy and are easier to stop. Our second equation becomes:
$$eV_0 = \frac{hc}{2\lambda} - \phi \quad \dots (2)$$

### Eliminating the Unknowns

We have a system of two equations. Our goal is to find the threshold wavelength, which is hidden inside the work function $\phi$. The stopping potential $V_0$ is an intermediate variable that we don't need. The most elegant way to eliminate it is to divide Equation (1) by Equation (2):

$$\frac{3eV_0}{eV_0} = \frac{\frac{hc}{\lambda} - \phi}{\frac{hc}{2\lambda} - \phi}$$

The $eV_0$ terms cancel out perfectly on the left side, leaving us with a simple algebraic equation:
$$3 = \frac{\frac{hc}{\lambda} - \phi}{\frac{hc}{2\lambda} - \phi}$$

Now, let's cross-multiply to isolate $\phi$:
$$3\left(\frac{hc}{2\lambda} - \phi\right) = \frac{hc}{\lambda} - \phi$$

Expanding the brackets, we get:
$$\frac{3hc}{2\lambda} - 3\phi = \frac{hc}{\lambda} - \phi$$

### Final Calculation

Let's group the $\phi$ terms on one side and the $\frac{hc}{\lambda}$ terms on the other:
$$3\phi - \phi = \frac{3hc}{2\lambda} - \frac{hc}{\lambda}$$

$$2\phi = \frac{hc}{2\lambda}$$

Dividing by 2, we find the work function:
$$\phi = \frac{hc}{4\lambda}$$

We are almost there! Recall the definition of the work function in terms of the threshold wavelength ($\lambda_t$):
$$\phi = \frac{hc}{\lambda_t}$$

Equating our two expressions for $\phi$:
$$\frac{hc}{\lambda_t} = \frac{hc}{4\lambda}$$

The $hc$ terms cancel out, revealing the final answer:
$$\lambda_t = 4\lambda$$

The threshold wavelength is exactly $4\lambda$. Therefore, the integer value we are looking for is **4**.

Enter Numerical Value:

Visualized Solution

$P h o t oe l ec t r i c E f f ec tS e t u p$

$C a se 1 : W a v e l e n g t hλ$

$C a se 2 : W a v e l e n g t h 2 λ$

$D i v i d in g t h e E q u a t i o n s$

$S o l v in g f or ϕ$

$F in d in g W or k F u n c t i o n$

$T h r es h o l d W a v e l e n g t h$

$C o n c l u s i o n$

The Sigma Insight: Photoelectric Effect

Analyzing the Setup

The Master Equations

Eliminating the Unknowns

Final Calculation

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The threshold wavelength for photoelectric emission from a material is $5200 \overset{˚}{A}$ . Photoelectrons will be emitted when this material is illuminated with monochromatic radiation from a

Enter Numerical Value:

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case1:Wavelengthλ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case2:Wavelength2λ

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

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

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

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

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