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Announcements First project is due in two weeks. In addition to a short (~10 minute) presentation you must turn in a written report on your project Homework Set 4: Chapter 4 # 41 & 42 + Supplemental Problems

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Light! Frequency = Wavelength = Speed = c = Energy E = h c = 299792458 m / s ≈ 3.00x10 8 m / s h = 6.626x10 -34 J-s

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Some simple examples Determine the frequency of the red emission line of hydrogen whose wavelength is 656.3 nm Determine the energy of a sodium D 1 photon whose wavelength is 589.592 nm What is the wavelength of an gamma-ray photon whose energy is 5.0 MeV? 1 eV = 1.602x10 -19 Joules?

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Example Solution Given a wavelength of 656.3 nm, find the frequency.

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Example Solution 2 Given a wavelength of 589.592 nm, find the energy.

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Example 3 Solution Given an energy of 5.0 MeV, find the wavelength. First convert energy from MeV to Joules since h has units of J-s Now use energy formula to find wavelength

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Light comes from electron transitions within the atom For the hydrogen atom these transitions are named after the scientist who studied them. The Lyman lines are in the UV range, the Balmer are visible and the Paschen and Brackett are in the IR

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The energy of an emitted photon is just the difference in energy of two energy levels n is the quantum number for the initial or final energy level. Note that there is no n = 0 energy level. The lowest energy level, n = 1, is referred to as the ground state. For the hydrogen atom

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Examples The Balmer emission lines of hydrogen are transitions whose final energy level is the n = 2 level. Determine the wavelength, in nm, of the first four Balmer lines.

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Example Solution The Balmer lines all have n f = 2 so the first four lines will have n i = 3, 4, 5 and 6. First do some algebra to get a formula for wavelength in terms of quantum numbers. Convert eV to J while we are at it.

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Example Solution 2 Lump all the constants together since we have to do the same calculation several times

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Example Solution 3 Now just plug in values for n i

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