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Question #43605

Difficulty: 8/10 | 🟢 Free

The Question

An infinitely long thin straight wire with a uniform positive linear charge density $\lambda$ is placed parallel to a large, grounded, flat metal plate at a distance $d$ . The entire space above the plate is filled with a transparent liquid dielectric of relative permittivity $K$ . The unperturbed work function of the metal is $\Phi_0$ . Due to the applied electric field, the work function is locally reduced (Schottky effect). Assuming the escaping photoelectrons are subjected to the macroscopic electric field and the modified image charge force in the dielectric, what is the maximum vacuum threshold wavelength $\lambda_{th}$ of incident light that can cause photoelectric emission from the region of the plate directly below the wire?

(Here $e$ is the elementary charge, $h$ is Planck's constant, $c$ is the speed of light in vacuum, and $\epsilon_0$ is the vacuum permittivity)

Options

A.
$\frac{hc}{\Phi_0 - \frac{1}{2\pi K \epsilon_0} \sqrt{\frac{e^3 \lambda}{d}}}$
B.
$\frac{hc}{\Phi_0 - \frac{1}{2\pi K \epsilon_0} \sqrt{\frac{e^3 \lambda}{2d}}}$
C.
$\frac{hc}{\Phi_0 - \frac{1}{2\pi \epsilon_0} \sqrt{\frac{e^3 \lambda}{K d}}}$
D.
$\frac{hc}{\Phi_0 - \frac{1}{2\pi \epsilon_0} \sqrt{\frac{e^3 \lambda}{d}}}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Work Function Dependence|Continuous Charge Distribution|Dielectric Polarization Logic

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of the Schottky effect, the method of images for continuous charge distributions, and the role of dielectric polarization.

The primary trap lies in failing to account for the dielectric constant $K$ in both the macroscopic field calculation and the microscopic image force on the escaping electron.

Students often blindly apply the standard vacuum Schottky formula $\Delta \Phi = \sqrt{e^3 E / 4\pi\epsilon_0}$ (leading to Option C) or forget the image charge of the wire itself, missing a factor of 2 in the macroscopic field (leading to Option B).

It rigorously evaluates whether a student understands the physical origin of the work function modification rather than just memorizing the final formula.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Dual Nature of Radiation and Matter - Class 12
Concept	Work Function Dependence
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free