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Electrons as Waves

This document explores the wave nature of electrons, detailing key characteristics such as wavelength, frequency, and amplitude. It explains how wavelength and frequency are inversely related and outlines the formula for wave speed in electromagnetic waves. Sample problems illustrate the calculation of wavelength from frequency for gamma rays, and the relationship between energy and frequency using Planck's constant. Additionally, it showcases how to find the energy of a photon based on its wavelength, providing practical insights into wave behavior in quantum physics.

beverly-leblanc
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Electrons as Waves

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  1. Electrons as Waves

  2. Characteristics of Waves • Wavelength • The shortest distance between equivalent points on a continuous wave; either crest to crest or trough to trough • Measured in m

  3. Characteristics of Waves • Frequency • the number of waves that pass a given point per second; one hertz (Hz) equals one wave per second

  4. Characteristics of Waves • Amplitude • The wave's height from the origin to crest or origin to trough • Measured in m

  5. The Wave

  6. As shown in the formula, if wavelength increases, frequency decreases. As wavelength decreases, frequency increases.

  7. Electromagnetic Wave Relationship • speed = λ υ • c = speed of light = 3.00 x 108 m/s

  8. Sample Problem • If a gamma ray traveling at the speed of light has a frequency of 2.88 x 1021Hz, what is its wavelength? • speed = λ υ • 3.00 x 108 m/s = λ (2.88 x 1021 Hz) Answer: λ = 1.04 x 10-13 m

  9. Energy • Energy will increase as frequency increases • E = h υ • h is a constant = 6.626 x 10-34Js

  10. Sample Problem • A photon has an energy of 2.93 x 10 -25J. What is its frequency? • E = h υ • 2.93 x 10-25 J = (6.626x10-34Js) υ Answer: υ = 4.42 x 105 Hz or 442,000 Hz

  11. What about in this one? • What is the energy of an ultraviolet photon that has a wavelength of 1.18 x 10-8m? • E = h υ • But wait? Do we have frequency? No? So…Find it! • Speed = c = λ υ • 3.00 x 108m/s = (1.18 x 10-8m) υ • υ = 2.54 x 1016 • Now solve for the energy • E = (6.626 x 10-34 Js) (2.54 x 1016) • E = 1.68 x 10-17 J

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