1. Atomic Theory and Spectroscopy Clues to the Nature of Atoms

2. The Nature of Light and Radiation  Electrons exist in discrete, unique energy levels.  If exposed to energy which matches those levels, they leap to unstable higher energy levels.  As they fall back they emit that energy at a wavelength and frequency which can be detected.

3. Electromagnetic Radiation  Energy emitted by electrons can be detected at any part of the electromagnetic spectrum  Ultraviolet light is emitted by many elements  Many atoms emit energy at many different wavelengths at once  Electromagnetic SpectrumElectromagnetic Spectrum  Most elements emit visible radiation as different colors of light  A spectroscope can view the unique, bright-line spectrum of light emitted by each element

4. Infrared Detection Pit vipers can detect heat emitted by animals.

5. Examples of nonvisible detection  http://home.centurytel.net/Arkcite/uv.htm http://home.centurytel.net/Arkcite/uv.htm  Ultraviolet photography  http://www.youtube.com/watch?v=kaPIh2XlaVM& feature=related http://www.youtube.com/watch?v=kaPIh2XlaVM& feature=related  Infrared ‘thermal’ photography

6. Measuring Electromagnetic Radiation  Radiation can take many forms across the electromagnetic spectrum  All radiation constantly travels through space at the same velocity (speed) = 3.0 x 10 8 m/s  The speed of electromagnetic radiation AKA  “The speed of light”

7. Measuring radiation II  Two properties can be measured:  Wavelength ( )  The distance from the same point on successive waves (measured in meters)  Frequency ( ν )  The number of times a wave passes a fixed point per second, or the number of times it travels up or down per second  Measured in cycles per second or hertz (Hz)

8. Measuring radiation III  The relationship can be described mathematically as follows: c= ν  Or inversely proportional  If one knows the frequency or wavelength, the other can be calculated  ν = c/  = c/ ν

9. Examples  What is the wavelength of a radio signal that is broadcast at 99.7MHz?  If =c/ ν, then the wavelength of this signal is the constant (the velocity) divided by the frequency  3.0 x 10 8 meters/second / 9.97 x 10 7 Hz (1/second) =  3.0 m

10. Example problem 2  What is the frequency of a radar beam that has a wavelength of 4.52cm?  If f= c/, then  3.0 x 10 8 meters/second / 4.52 x 10 -2 m =  6.63 x 10 9 Hz  What is this in MHz?  6.63 x 10 9 Hz / 1 x 10 6 Hz/MHz =  6.63 x 10 3

11. Electromagnetic spectrum and Energy  Energy and frequency are directly proportional  The higher the energy, the higher the frequency  Energy and wavelength are inversely proportional  The higher the energy, the shorter the wavelength  Planck discovered the energy of a wave or photon of light is constant (6.626x 10 -34 J/Hz)  Planck’s constant (h)

12. Energy and Radiation II  If one knows the frequency of a wave, then the energy of the wave can be calculated as well in the same manner  E=h ν,  Sample: What is the energy of a wave which has a frequency of 103.5MHz? E= 6.626 x 10 -34 J/Hz x 1.035 x 10 8 Hz 6.86 x 10 -26 J