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# Chapter 4: Atomic Structure

Chapter 4: Atomic Structure. The Nature of Light Atomic Spectra Atomic Structure. Suggested problems for Ch 4: 34, 38, 45, 49, 52, 56, 61, 66, 69, 72, 73, 80, 85, 86, 88, 97, 99, 105, 113, 114, 121, 125, 133, 137, 153, 162, 167, 171. The Spectrum of Electromagnetic Radiation.

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## Chapter 4: Atomic Structure

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1. Chapter 4: Atomic Structure The Nature of Light Atomic Spectra Atomic Structure

2. Suggested problems for Ch 4: 34, 38, 45, 49, 52, 56, 61, 66, 69, 72, 73, 80, 85, 86, 88, 97, 99, 105, 113, 114, 121, 125, 133, 137, 153, 162, 167, 171

3. The Spectrum of Electromagnetic Radiation • The wavelength of visible light is between 400 and 700 nanometers • Radio, TV , microwave and infrared radiation have longer wavelengths (shorter frequencies), and lower energies than visible light. • Gamma rays and X-rays have shorter wavelengths (larger frequencies), and higher energies than visible light!

4. Electromagnetic Radiation c   • WAVELENGTH - The distance between identical points on successive waves. (  ) • FREQUENCY - The number of waves that pass through a particular point per second. () • AMPLITUDE - The vertical distance from the midline to a peak, or trough in the wave.

5. Calculation of Frequency from Wavelength Problem: The wavelength of an x-ray is 1.00 x10 -9 m or 1 nm, what is the frequency of this x-ray? If the wavelength of long-wavelength electromagnetic radiation is 7.65 x 104 m, what is the frequency of this long-wavelength radiation used to contact submerged nuclear submarines at sea? Plan: Use the relationship between wavelength and frequency to obtain the answer. wavelength x frequency = speed of light! Solution: speed of light wavelength(m) frequency(cycles/sec) = 3.00 x 108 m/s 1.00 x 10 - 9 m a) frequency = = 3.00 x 1017 cycles/sec 3.00 x 108 m/s 7.65 x 104 m b) frequency = = 3.92 x 103 cycles/s

6. Photons and the Photoelectric Effect • Einstein also stated that the change in the photon’s energy was equal to the ejected electron’s energy.

7. Calculation of Energy from Frequency Problem: What is the energy of a photon of electromagnetic radiation being emitted by radio station KBSG 97.3 FM ( 97.3 x 108 cycles/sec)? What is the energy of a gamma ray emitted by Cs137 if it has a frequency of 1.60 x 1020/s? Plan: Use the relationship between energy and frequency to obtain the energy of the electromagnetic radiation (E = hv). Solution: EKBSG =hv = (6.626 x 10 -34Js)(9.73 x 109/s) = 6.447098 x 10 -24J EKBSG = 6.45 x 10 - 24 J Egamma ray =hv = ( 6.626 x 10-34Js )( 1.60 x 1020/s ) = 1.06 x 10 -13J Egamma ray = 1.06 x 10 - 13J

8. Calculation of Energy from Wavelength Problem: What is the energy of a photon of electromagnetic radiation that is used in microwave ovens to cook things by rotating water molecules, the wavelength of the radiation is 122 mm. Plan: Convert the wavelength into meters, then the frequency can be calculated using the relationship;wavelength x frequency = C (where C is the speed of light), then using E=hv to calculate the energy. Solution: wavelength = 122 mm = 1.22 x 10 -1m C wavelength 3.00 x108 m/s 1.22 x 10 -1m frequency = = = 2.46 x 1010 cycles/s Energy = E = hv = (6.626 x 10 -34Js)(2.46 x 1010/s) = 1.63 x 10 - 23 J

9. Light and Atoms • When an atom gains a photon, it enters an excited state. • This state has too much energy - the atom must lose it and return back down to its ground state, the most stable state for the atom. • An energy level diagram is used to represent these changes.

10. The Line Spectra of Several Elements

11. Energy Level Diagram • Energy Excited States photon’s path Ground State Light Emission Light Emission Light Emission

12. Quantum Numbers - I • 1) Principal Quantum Number = n • Also called the “energy “ quantum number, indicates the approximate distance from the nucleus . • Denotes the electron energy shells around the atom, and is derived directly from the Schrodinger equation. • The higher the value of “n” , the greater the energy of the orbital, and hence the energy of electrons in that orbital. • Positive integer values of n = 1 , 2 , 3 , etc.

13. Energy-level diagram for the electron in the hydrogen atom.

14. Transitions of the electron in the hydrogen atom.

15. 1 n12 1 n22 1 2 2 1 4 2 Using the Rydberg Equation Problem: Find the energy change when an electron changes from the n=4 level to the n=2 level in the hydrogen atom? What is the wavelength of this photon? Plan: Use the Rydberg equation to calculate the energy change, then calculate the wavelength using the relationship of the speed of light. Solution: Ephoton = -2.18 x10 -18J - = Ephoton = -2.18 x 10 -18J - = - 4.09 x 10 -19J h x c E (6.626 x 10 -34Js)( 3.00 x 108 m/s) wavelength = = = 4.09 x 10 -19J wavelength = 4.87 x 10 -7 m = 487 nm

16. Back to Ch 3

17. Modern Reassessment of the Atomic Theory 1. All matter is composed of atoms. Although atoms are composed of smaller particles (electrons, protons, and neutrons), the atom is the smallest body thatretains the unique identity of the element. 2. Atoms of one element cannot be converted into atoms of another element ina chemical reaction. Elements can only be converted into other elements in Nuclear reactions in which protons are changed. 3. All atoms of an element have the same number of protons and electrons, which determines the chemical behavior of the element. Isotopes of an element differ in the number of neutrons, and thus in mass number, but a sample of the element is treated as though its atoms have an average mass. 4. Compounds are formed by the chemical combination of two or more elements in specific ratios, as originally stated by Dalton.

18. Predicted Properties Observed properties (average of Si and Sn) Ge Atomic mass 72 72.61 Density 5.5 g/cm3 5.32 g/cm3 Melting Point 82.5C 93.8C Oxide formula XO2 GeO2 Density of oxide 4.7 g/cm3 4.70 g/cm3 Chloride formula XCl4 GeCl4 Boiling point of chloride 100C 86C

19. Orbital energies of the hydrogen atom.

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