Chapter 25: Problem 14
Given that a photon of light of wavelength \(10,000 \AA\) has an energy equal to \(1.23 \mathrm{eV}\). When light of wavelength \(5000 \AA\) and intensity \(I_{0}\) falls on a photoelectric cell, the surface current is \(0.40 \times 10^{-6} \mathrm{~A}\) and the stopping potential is \(1.36 \mathrm{~V}\), then the work function is (a) \(0.43 \mathrm{eV}\) (b) \(0.55 \mathrm{eV}\) (c) \(1.10 \mathrm{eV}\) (d) \(1.53 \mathrm{eV}\)
Short Answer
Step by step solution
Understand Energy-Wavelength Relationship
Calculate Energy of 5000 Ã… Photon
Use Photoelectric Equation to Find Work Function
Choose the Correct Option
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Wavelength-Energy Relationship
- \( E \) is the energy of the photon,
- \( h \) is Planck's constant \((6.626 \times 10^{-34} \text{ J s})\),
- \( c \) is the speed of light in vacuum \((3.0 \times 10^8 \text{ m/s})\),
- \( \lambda \) is the wavelength of the photon.
Work Function
- \( E_{\text{photon}} \) is the energy of the incoming photon,
- \( \phi \) is the work function of the material,
- \( E_{\text{kinetic}} \) is the maximum kinetic energy of the emitted electrons.
Stopping Potential
- \( e \) is the charge of the electron, \((1.602 \times 10^{-19} \text{ C})\),
- \( V_{0} \) is the stopping potential in volts.
Kinetic Energy of Electrons
- \( E_{\text{photon}} \) - energy of the incident photon,
- \( \phi \) - the work function of the material.