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Electronic Structure of Atoms Chapter 6

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Light is a Wave Electromagnetic wave Wavelength ( ), m Frequency ( ), Hz or s -1 Travels at c (3.00 X 10 8 m/s) in a vacuum

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Wavelength and frequency are inversely proportional c =

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How many complete waves are shown above? What is the wavelength of light shown above?

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Electromagnetic Spectrum

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Blu-Ray = 405 nanometers (blue light) DVD = 650 nanometers (red light)

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Visible light = 4 X 10 -7 m to 7X 10 -7 m (400 to 700 nm)

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Waves: Ex 1 Calculate the wavelength of a 60 Hz EM wave

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Waves: Ex 2 Calculate the wavelength of a 93.3 MHz FM radio station

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Waves: Ex 3 Calculate the frequency of 500 nm blue light.

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Planck’s Quantum Hypothesis: Light is a Particle Photon –light particle Photons emitted in “packets” (whole numbers) Light is both a wave and a particle E = h (for one photon) h = 6.63 X 10 -34 J s (Planck’s constant)

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Photons: Ex 1 Calculate the energy of a photon of wavelength 600 nm. 600 nm 1 X 10 -9 m = 6 X10 -7 m 1 nm

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Photons: Ex 2 Calculate the energy of a photon of wavelength 450 nm (blue light). Also, calculate the energy per mole. Ans: (4.42 X 10 -19 J, 266 kJ/mol)

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Photons: Ex 3 Calculate the energy of laser light with a frequency of 4.69 X 10 14 s -1. Also, calculate the energy per mole. Ans: (3.11 X 10 -19 J, 187 kJ/mol)

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Photons: Ex 3a If the laser emits 1.3 X 10 -2 J per pulse, how many photons (quanta) are released during this pulse? Ans: 4.18 X 10 16 photons

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Photoelectric Effect (Einstein) When light shines on a metal, electrons are emitted Can detect a current from the electrons Used in light meter, scanners, digital cameras

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Three Key Points 1.Below a certain frequency, no electrons are emitted 2.Greater intensity light produces more electrons 3.Greater Frequency light produces no more electrons, but they come off with greater speed

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Low FrequencyHigh Frequency Not enough energy to eject electron Can eject electron Energy of photon is greater than W (Work function)

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2.More intensity – More photons – More electrons ejected with same KE

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3.Greater Frequency – No more electrons ejected – Electrons ejected with greater speed (KE)

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Photon/Matter Interactions 1.Electron excitation (photon disappears) 2.Ionization/photoelectric effect (photon disappears) 3.Scattering by nucleus or electron 4.Pair production (photon disappears)

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Electron Excitation Photon is absorbed (disappears) Electron jumps to an excited state Ionization/Photoelectric Effect Photon is absorbed (disappears) Electron is propelled out of the atom

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Scattering Photon collides with a nucleus or electron Photon loses some energy Speed does not change, but the wavelength increases Pair Production Photon closely approaches a nucleus Photon disappears An electron and positron are created.

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Principle of Complimentarity Any experiment can only observe light’s wave or particle properties, not both Different “faces” that light shows

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Line Spectra Discharge tube – Low density gas (acts like isolated atoms) – high voltage Light emitted only at certain (discrete) wavelengths

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Hydrogen Helium Solar absorption spectrum

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Electrons orbit in ground state (without radiating energy) Jumps to excited state by absorbing a photon Returns to ground state by emitted a photon

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h = E e - E g

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4. Ways to make something glow Bohr Model Photon AbsorptionCollision -Glow in the dark-Heat -Electricity -Chemical Reaction

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Photon AbsorptionCollision

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Bohr’s Equation Works only for H and other 1- electron atoms (He +, Li 2+, Be 3+, etc…)

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E n = -R H n 2 E n = Energy of an orbital R H = Rydberg constant (2.18 X 10 -18 J) n = Principal quantum number of orbital

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Bohr’s Equation: Ex 1 Calculate the energy of the first three orbitals of hydrogen E n = - 2.18 X 10 -18 J n 2 E 1 = - 2.18 X 10 -18 J 1 2 E 1 = - 2.18 X 10 -18 J

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E 2 = - 2.18 X 10 -18 J 2 2 E 2 = - 5.45 X 10 -19 J E 3 = - 2.18 X 10 -18 J 3 2 E 3 = - 2.42 X 10 -19 J

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Bohr’s Equation: Ex 2 What wavelength of light is emitted if a hydrogen electron drops from the n=2 to the n=1 orbit? E 1 = - 2.18 X 10 -18 J E 2 = - 5.45 X 10 -19 J E = (- 2.18 X 10 -18 J - - 5.45 X 10 -19 J) E = -1.64 X 10 -18 J E = h = 2.48 X 10 15 s -1

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c = c/ = (3 X 10 8 m/s)(2.48 X 10 15 s -1 ) = 122 nm

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E = h c = E = hc/ = hc E = (6.626 X 10 -34 J s)(3.0 X 10 8 m/s) (1.63 X 10 -18 J) = 1.22 X 10 -7 m = 122 nm (UV)

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Bohr’s Equation: Ex 3 Calculate the wavelength of light emitted if a hydrogen electron drops from the n=6 to the n=2 orbit? (ANS: 410 nm (violet))

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Bohr’s Equation: Ex 4 Calculate the wavelength of light that must be absorbed to exite a hydrogen electron from the n=1 to the n=3 orbit? (ANS: 103 nm (UV))

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Bohr Model: Other Atoms E n = (Z 2 )(- 2.18 X 10 -18 J) n 2 Z= Atomic number of the element (H=1, He=2, etc)

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The Batcave Intruder Alert Laser uses a beam that has a frequency of 3.53 X 10 14 Hz. a)Calculate the wavelength in nanometers. (850) b)Calculate the energy per photon. (2.34X10 -19 J) c)Calculate the energy per mole of photons. (1.41 X 10 5 J/mol) d)Calculate the number of photons in a 50.0 mJ pulse of this laser. (2.13 X 10 17 photons) e)Calculate the moles of photons in that pulse. (3.54 X 10 -7 moles) f)Identify the range of the electromagnetic spectrum of this laser.

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Wave Nature of Matter Louis DeBroglie All matter has wave and particle properties “matter waves” = h mv momentum Everything has a wavelength

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Diffraction pattern of electrons scattered off aluminum foil Wavelike properties only matter for small objects Electrons, protons, light, etc…

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DeBroglie Wavelength: Ex 1 Calculate the wavelength of a baseball of mass 0.20 kg moving at 40.25 m/s (90 mph) = h mv = (6.626 X 10 -34 J s) (0.20 kg)(40.25 m/s) = 8.2 X 10 -35 m

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DeBroglie Wavelength: Ex 2 Calculate the wavelength of an electron (9.109 X 10 -31 kg) moving at 2.2 X 10 6 m/s = h mv = (6.626 X 10 -34 J s) (9.11 X 10 -31 kg)(2.2 X 10 6 m/s) = 3.3 X 10 -10 m or 0.33 nm

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DeBroglie Wavelength: Ex 3 What velocity must a neutron (1.67 X 10 -27 kg) move to have a wavelength of 500 pm (1 pm = 1X10 -12 m)? 500 pm1X10 -12 m = 5.0 X 10 -10 m 1 pm = h mv v = h=(6.626 X 10 -34 J s)= 794 m/s m (1.67 X 10 -27 kg)(5.0 X 10 -10 m/s)

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1.Electron is both wave and particle 2.Probabilistic view – Where does the electron “hang out” rather than where the electron “is Quantum Mechanical Model

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Electron as a particle Heisenberg Uncertainty Principle – can never know both the position and velocity of an electron at the same time Results a. Electron moves randomly (not like a planet) b. Electron cloud – 90% probability c. Important for transistors/chips Quantum Mechanical Model

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nucleus Random electron cloud

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Electron as Wave Schrodinger Wave Equation (1926) – treats electron solely as a wave Quantum Mechanical Model

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Result One Explains the forbidden zone (waves do not match) Quantum Mechanical Model

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Forbidden Zone

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Result Two Orbital are not circular a.Orbital – region of space where there is a significant chance of finding an electron b.Does not move like a planet Quantum Mechanical Model

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Three Major Discoveries Light is both a wave and a particle Electron Orbitals are Quantized The electron (and all matter) is both a wave and a particle Planck Einstein BohrDe Broglie (Later Schrodinger/ Heisenberg)

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1. First (principal) QN (n)– how far the electron is from the nucleus (larger the number, farther away) – Level or shell n = 2 n = 1 Quantum Numbers

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2. Second (azimuthal) QN (l) – the shape of the orbital Value of l0123 Letter usedspdf

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Quantum Mechanical Model

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3. Third (magnetic) QN (m l )– the suborbital Orbital # suborbitals Total e- s02 p3 (p x,p y,p z )6 d5 10 f714 Link to Periodic table

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In a magnetic field, you can “see” the three p suborbitals in a line spectrum

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4. Fourth (spin) QN (m s )– spin of the electron Pauli Exclusion Principle – No two electrons in an atom can have the same four quantum numbers – Two electrons in the same suborbital (ex: p y ) must have opposite spins – Can have values of +1/2 or -1/2

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Shows the two electrons, spin +½ and spin -½

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nPossible Values of l Possible Values of m l Subshell name 1001s 20101 0 -1, 0, +1 2s 2p 3012012 0 -1, 0, +1 -2, -1, 0, +1, +2 3s 3p 3d 401230123 0 -1, 0, +1 -2, -1, 0, +1, +2 -3, -2, -1, 0, +1, +2, +3 4s 4p 4d 4f

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Quantum Numbers: Ex 1 What is the designation for a subshell n=5 and l =1? How many suborbitals are in this subshell? What are the values of m l for each of the orbitals?

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Quantum Numbers: Ex 1 What is the designation for a subshell n=5 and l =1? 5p How many orbitals are in this subshell? 3 What are the values of ml for each of the orbitals? -1, 0, +1

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Quantum Numbers: Ex 2 Which of the following sets of quantum numbers are not allowed in the hydrogen atom? A. n=2, l=0, m l =0, m s =1/2 B. n=1, l=0, m l =0, m s = -1/2 C. n=3, l=1, m l = 2, m s =1/2 D. n=4, l=2, m l = -2, m s =1/2

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Quantum Numbers: Ex 2 Which of the following sets of quantum numbers are not allowed in the hydrogen atom? A. n=2, l=0, m l =0, m s =1/2 B. n=1, l=0, m l =0, m s = -1/2 C. n=3, l=1, m l = 2, m s =1/2 D. n=4, l=2, m l = -2, m s =1/2

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Quantum Numbers: Ex 3 Which of the following sets of quantum numbers are not allowed in the carbon atom? A. n=2, l=0, m l =0, m s =1/2 B. n=1, l=1, m l =0, m s = -1/2 C. n=2, l=1, m l = -1, m s =1/2 D. n=4, l=2, m l = -3, m s =1/2

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Quantum Numbers: Ex 3 Which of the following sets of quantum numbers are not allowed in the carbon atom? A. n=2, l=0, m l =0, m s =1/2 B. n=1, l=1, m l =0, m s = -1/2 C. n=2, l=1, m l = -1, m s =1/2 D. n=4, l=2, m l = -3, m s =1/2

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1.Electron Configuration – shorthand notation to tell you the locations of all the electrons in an atom or ion 2.Notation 2p 3 OrbitShape# e- Electron Configurations

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Electron Configuration

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Electron Configurations

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Why is “d” one less? Energy is slightly above the next s orbital Complete the orbital filling diagram for manganese

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1.Easy Examples HHeOFeS 2.Write e- configuration LiNSrPSe VFArMgKr Electron Configurations

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3.Which element is represented by the following electron configurations? 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 5 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 7 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 1s 2 2s 2 2p 6 3s 2 3p 3 1s 2 2s 2 2p 6 3s 1 1s 2 2s 2 2p 6 3s 2 3p 2 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 Electron Configurations

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1.Rule – Use the noble gas in the previous row 2.Examples Ne and P Ru Kr You try: Br Ar S Ca I Xe Noble Gas (Condensed) Shortcut

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Electron Configuration Exceptions Mostly with transition metal elements There is a special stability to filled and half-filled orbitals ElementActual configurationInstead of Cr[Ar]4s 1 3d 5 [Ar]4s 2 3d 4 Mo[Kr]5s 1 4d 5 [Kr]5s 2 4d 4 Cu[Ar]4s 1 3d 10 [Ar]4s 2 3d 9 Ag[Kr]5s 1 4d 10 [Kr]5s 2 4d 9

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pee - configuration Sr Sr + Sr 2+ Al 2+ Al 3+ Ions

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pee - configuration S S 1- S 2- Br 1- Ba Ba 2+ B 3+ Ions

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Transition Metal Ions Lose their “s” electrons first. Fe[Ar]4s 2 3d 6 Fe 1+ Fe 2+ Fe 3+ Cu Cu 2+ Ru Ru 3+

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Transition Metal Ions Zn 2+ Co 1+ Co 2+ V 3+ Sc 2+ Sc 3+

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4.a) Increaseb) Decrease c) line spectrum(forbidden zone) 14. a) Gamma < (d)yellow < (e)red < (b)93.1 radio < (c)680 AM 16. a) 3.00 X 10 17 Hz b) 3.94 X 10 -3 m c) (a) is X-Ray and (b) is microwave d) 7.64 X 10 -6 m 18. UV has a higher frequency and shorter than IR. UV produces more energy

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22.a) 5.09 X 10 14 Hzb) 20.3 kJ c) 3.37 X 10 -19 Jd) Na + 24. AM: 6.69 X 10 -28 JFM: 6.51 X 10 -26 J 26.1.56 X 10 -18 J/photon, 127 nm 28. a) microwaveb) 6.4 X 10 -11 J/hr 34.a) Absorbedb) Emittedc) Absorbed 36.a) 9.7 X10 -8 m, emittedb) 434 nm, emitted c) 1.06 X 10 -6 m, absorbed 38. a) n=1 larger Eb) 121 nm, 103 nm, 97.2 nm 40. a)2.626X10 -6 m (IR) b) 6 4 transition

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42. 3.97 X 10 -10 m (3.97 A) 44. 7.75 X 10 -11 m (0.775 A) 50. a) n = 3 (l=2,1,0) (9 m l values) b) n = 5 (l = 4,3,2,1,0) (25 m l values) 52. a) 2,1,1 2,1,0 2,1,-1 b) 5,2,2 5,2,1 5,2,0 5,2,-1 5,2,-2 54. 2pnot allowed 1s4d not allowed5f 3d not allowed

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Solution to 41c 6.941 g X 1kg=0.006941kg/mol 1mol 1000g 0.006941 kg X 1mol= 1.15 X 10 -26 kg/atom 1mol 6.022X10 23 atoms = h= 2.3 X 10 -13 m (1.15 X 10 -26 kg)(2.5 X 10 5 m/s)

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66 CPNe Config[He]2s 2 sp 2 [Ne]3s 2 3p 3 [He]2s 2 2p 6 Core2102 Valence458 Unpaired230

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68. a) [Ar]4s 2 3d 10 4p 1 (1 unpaired) b) [Ar]4s 2 (0 unpaired) c) [Ar]4s 2 3d 3 (3 unpaired) d) [Kr]5s 2 4d 10 5p 5 (1 unpaired) e) [Kr]5s 2 4d 1 (1 unpaired) f) [Xe]6s 1 4f 14 5d 9 (2 unpaired) g) [Xe]6s 2 4f 14 5d 1 (1 unpaired) 70. a) [Ar]3d 10 b) [Xe]4f 14 5d 8 c) [Ar]3d 3 d) [Ne]3s 2 3p 6 72.a) 7Ab) 4Bc) 3A d) Sm and Pm (f-block)

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74.a) [He]2s 2 2p 3 b) [Ar]4s 2 3d 10 4p 4 c) [Kr]5s 2 4d 7 76.a) Ba Ca K Nab) Au (shortest) Na(longest) c) 455 nm, Ba 78 a) 9.37X10 14 s-1b) 374 kJ/mol c) UV-B Shorter, higher energy d)UV-B

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100. O 3 O 2 + O H = 105.2 kJ/mol H = 1.052 X 10 5 J/mol 1.052 X 10 5 J 1 mol=1.75 X 10 -19 J 1 mol 6.022X10 23 molec. E = h = 2.63 X 10 14 Hz = c/ = 1140 nm

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Extra Problem A certain biomolecule requires 598 kJ/mol to break one of its bonds. What wavelength of light (nm)would a single photon need to break this bond? What region of the EM spectrum is this?

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A photon of wavelength 550 nm will break a bond in a synthetic dye. Calculate the energy per mole of that bond (kJ/mol) (218 kJ/mol)

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a) Calculate the wavelength of light emitted from a 7 2 transition in a hydrogen atom. (398 nm) b) Calculate the speed an electron would need to have the wavelength. (1830 m/s) c) Calculate the wavelength of a proton moving at that same speed. (2.17 X 10 -10 m or 0.217 nm) d) Why is the proton’s wavelength so much smaller? Which has more energy, e - or p + ? m e = 9.109 X 10 -31 kg m p = 1.673 X 10 -27 kg

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a) Calculate the wavelength of a proton that has been accelerated to 2 X 10 6 m/s. (1.98X10 -13 m) b) Calculate the frequency. (1.51 X 10 21 Hz) c) Calculate the energy. (1.00 X 10 -12 J/photon) m p = 1.673 X 10 -27 kg

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Quantum Theory and the Electronic Structure of Atoms Chapter 6.

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