MCQPhysics

CBSE · Class 12 · Physics · Dual Nature of Radiation and MatterThe work function of a metal is $4.5 \text{ eV}$. What is the threshold wavelength for this metal?

Step-by-Step Solution

The threshold wavelength $\lambda_0$ is related to the work function $\phi_0$ by the formula $\lambda_0 = \frac{hc}{\phi_0}$. Substituting $hc \approx 12400 \text{ eV}\cdot\text{Å}$ and $\phi_0 = 4.5 \text{ eV}$, we get $\lambda_0 = \frac{12400}{4.5} \approx 2755 \text{ Å}$ or approximately $276 \text{ nm}$.

Detailed Options Breakdown
Option : 276 nm (Correct Answer)

Correct choice. Refer to the step-by-step verified solution guidelines above for details.

Option 1: 550 nm

Incorrect choice. This distractor represents a common misunderstanding of the core principles of Dual Nature of Radiation and Matter.

Option 2: 138 nm

Incorrect choice. This distractor represents a common misunderstanding of the core principles of Dual Nature of Radiation and Matter.

Option 3: 690 nm

Incorrect choice. This distractor represents a common misunderstanding of the core principles of Dual Nature of Radiation and Matter.

💡 Study Guide: This question tests core syllabus concepts from Dual Nature of Radiation and Matter. For formulas, key summaries, and mock exam reference guides, read the full Dual Nature of Radiation and Matter Revision Notes.
✓
Verified by Academic Committee
← All Chapter QuestionsPhysics Chapters