E must be negative positive nucleus attracts the

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E - must be negative Positive nucleus attracts the negative electron So, energy has to be supplied to move the electron to a higher energy level Excited state- more energy than in its ground state The difference in energy, E = E f – E i (orbit!) = h c/ λ E = 0 for electron and proton completely separated ( n = ∞)
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Bohr’s H-atom model Bohr’s H-atom model For transitions between H-atom levels: Δ E = −2.179 x 10 -18 J Bohr Model of Hydrogen Example Example Calculate the energy and wavelength (in nm) for an H-atom n = 4 → n = 2 transition. 1 n f 2 1 n i 2
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Bohr Model Example Calculate the energy and wavelength (in nm) for an H-atom n = 4 → n = 2 transition. Δ E = −2.179 x 10 -18 [( ½ ) 2 −( ¼ ) 2 ] J = −4.086 x 10 -19 J = 4.086 x 10 -19 J (negative sign omitted. Losing energy = emission) ν = Δ E/h = 4.086 x 10 -19 J /6.626 x 10 -34 Js = 6.166 x 10 14 s -1 = 6.166 x 10 14 Hz = 616.6 THz λ= c/ ν = 2.9979 x 10 8 ms -1 /6.166 x 10 14 s -1 λ= 4.862 x 10 -7 m = 486.2 nm
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Electron Transitions in an excited H-atom B a l m e r S e ri e s Energies are quantized Orbit, principal quantum number Ground State, lowest energy level
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20 E photon = E = E f - E i E f = - R H ( ) 1 n 2 f E i = - R H ( ) 1 n 2 i i f E = R H ( ) 1 n 2 1 n 2 n f = 1 n i = 2 n f = 1 n i = 3 n f = 2 n i = 3
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21
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22 E photon = 2.18 x 10 -18 J x (1/25 - 1/9) E photon = E = -1.55 x 10 -19 J λ = 6.63 x 10 -34 (J•s) x 3.00 x 10 8 (m/s)/1.55 x 10 -19 J λ = 1280 nm Calculate the wavelength (in nm) of a photon emitted by a hydrogen atom when its electron drops from the n = 5 state to the n = 3 state. E photon = h x c / λ λ = h x c / E photon i f E = R H ( ) 1 n 2 1 n 2 E photon =
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23 De Broglie (1924) reasoned that e - is both particle and wave . Why is e - energy quantized? u = velocity of e- m = mass of e- 2 π r = n λ λ = h mu Left- wave
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24 λ = h / mu λ = 6.63 x 10 -34 / (2.5 x 10 -3 x 15.6) λ = 1.7 x 10 -32 m = 1.7 x 10 -23 nm What is the de Broglie wavelength (in nm) associated with a 2.5 g Ping-Pong ball traveling at 15.6 m/s? m in kg h in J s u in (m/s)
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25 Chemistry in Action: Laser – The Splendid Light Laser light is (1) intense, (2) monoenergetic, and (3) coherent Light Amplification by Stimulated Emission of Radiation
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26 Chemistry in Action: Electron Microscopy STM image of iron atoms on copper surface λ e = 0.004 nm Electron micrograph of a normal red blood cell and a sickled red blood cell from the same person
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The Uncertainty Principle Heisenberg showed that the more precisely the momentum of a particle is known, the less precisely is its position known: In many cases, our uncertainty of the whereabouts of an electron is greater than the size of the atom itself! ( x ) ( mv ) h 4 π
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De Broglie Equation Compare the wavelength of an e - moving at 2.0 x 10 5 ms -1 to that of a baseball moving at 98 mph (44 ms -1 ). Masses: e - = 9.11 x 10 -31 kg; baseball = 0.143 kg. λ(ball) = 6.626 x 10 -34 kg m 2 s -1 (0.143 kg)(44 m s -1 ) = 1.1 x 10 -34 m = 3.6 nm = 3.6 x 10 -9 m λ(e - ) = h m v 6.626 x 10 -34 kg m 2 s -1 (9.11 x 10 -31 kg)(2.0 x 10 5 m s -1 ) = The wavelength of baseball is far too short to observe.
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Quantum Model of the Atom In 1926, Schrödinger wrote an equation which revolutionized the study of atoms and molecules.
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