Before You Begin

Question #43610

Difficulty: 8/10 | 🟢 Free

The Question

An annular metal disk of inner radius $R_1 = R/e$ and outer radius $R_2 = R$ (where $e$ is the base of the natural logarithm) is placed in a vacuum chamber. The inner edge is maintained at a steady temperature $T_1 = 600 \text{ K}$ and the outer edge at $T_2 = 300 \text{ K}$ . The disk is in a steady state of radial heat conduction. The work function of the metal varies with local temperature $T$ as $\Phi(T) = \Phi_0 + \alpha T$ , where $\Phi_0 = 2.0 \text{ eV}$ and $\alpha = 10^{-3} \text{ eV/K}$ . The entire surface of the disk is illuminated uniformly with monochromatic light of photon energy $E = 2.6 \text{ eV}$ . A uniform retarding potential $V_0$ is applied between the disk and a parallel collector plate. It is observed that photoelectrons emitted from exactly half of the total surface area of the disk have sufficient kinetic energy to reach the collector. Find the value of $V_0$ .

Options

A.
$0.15 \ln\left( \frac{e^2 + 1}{2} \right) \text{ V}$
B.
$0.30 \ln\left( \frac{e + 1}{2} \right) \text{ V}$
C.
$0.15 \ln\left( \frac{e^2 - 1}{2} \right) \text{ V}$
D.
$0.30 - 0.15 \ln 2 \text{ V}$

Check Your Answer

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

Stopping Potential Independence|Steady State Temperature Gradient|Approximation

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of steady-state heat transfer and the photoelectric effect, specifically targeting the misconception that the stopping potential is a global constant for a given material.

Students often fail to recognize that a temperature gradient induces a spatial variation in the Fermi level and work function, making the cutoff voltage a local variable $V_s(r)$ .

Furthermore, the trap exploits the geometry of radial heat conduction: students who incorrectly assume a linear temperature gradient $T(r) \propto r$ or who calculate the 'half area' by simply halving the radius or ignoring the inner hole will arrive at incorrect logarithmic arguments.

The correct approach requires deriving the logarithmic temperature profile, mapping it to a spatially dependent kinetic energy, and applying an area integral constraint to find the critical emission boundary.

Quick Facts About This Question

Question Information

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