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Survey of the Universe Tom Burbine tburbine@mtholyoke.edu
8/6/2018 Survey of the Universe Tom Burbine
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Prospective Astronomy Majors Lunch
Monday, March 4th 12:15-1:10pm Kendade 303 Meet the faculty, learn about the Mars Rover, and enjoy a delicious lunch.
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HW# 3 Due March 6
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Characteristics of waves
06/16/10 06/16/10 velocity = wavelength x frequency Wavelength = distance Frequency = cycles per second = hertz 4
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For light c = wavelength x frequency
06/16/10 06/16/10 For light c = wavelength x frequency In a vacuum, speed of light stays the same So if wavelength goes up Frequency does down f = frequency λ = wavelength c = λ x f 5
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Calculations c = λ x f So if the wavelength is 1 x 10-12 m
06/16/10 06/16/10 Calculations c = λ x f So if the wavelength is 1 x m 3 x 108 m/s = 1 x m * f f = 3 x 108 m/s/1 x m f = 3 x 1020 s-1 = 3 x 1020 Hz 6
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Calculations c = λ x f So if the frequency is 1 x 1015 Hz
06/16/10 06/16/10 Calculations c = λ x f So if the frequency is 1 x 1015 Hz 3 x 108 m/s = λ * 1 x 1015 Hz λ = 3 x 108 m/s/1 x 1015 Hz λ = 3 x 10-7 m 7
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Energy of light Energy is directly proportional to the frequency
06/16/10 06/16/10 Energy of light Energy is directly proportional to the frequency E = h * f h = Planck’s constant = x J*s since f = c/λ Energy is inversely proportional to the wavelength E = hc/λ 8
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06/16/10 06/16/10 Calculations What is the energy of a radio wave with a frequency of 1 x 107 Hz? E = h * f h = Planck’s constant = x J*s E = x J*s * 1 x 107 E = x J 9
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06/16/10 06/16/10 Calculations What is the energy of a gamma ray photon with wavelength of 1 x m E = hc/λ h = Planck’s constant = x J*s E = x J*s * 3 x 108 m/s / 1 x m E = 1.99 x J 10
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06/16/10 06/16/10 11
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Higher the frequency, Higher the energy of the photon
06/16/10 06/16/10 VIBGYOR violet red Higher the frequency, Higher the energy of the photon Higher the wavelength, Lower the energy of the photon 12
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ROYGBIV ROYGBIV Red – long wavelength Violet – short wavelength
06/16/10 ROYGBIV ROYGBIV Red – long wavelength Violet – short wavelength
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Any Questions?
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