CHEM 101 1st Edition Lecture 17Outline of Last Lecture I. Electromagnetic RadiationII. Quantization of EnergyIII. Max PlanckIV. Atomic Line SpectraOutline of Current Lecture I. Bohr Model of the AtomII. Key TermsIII. Particle-Wave DualityCurrent Lecture Chapter 6: The Structure of Atoms- The Bohr Model of the Atomo Niels Bohr asserted that line spectra of elements indicated that the electrons were confined to specific energy states called orbits.o He proposed a planetary structure for the atom where the electrons circled the nucleus in these defined orbitso In this model, the attractive electrostatic forces of the electron and nucleus are balanced by the centripetal forces of the orbiting electrono The orbits or energy levels are quantized such that only certain levels are allowed (N= 1, 2,3,…,∞)o The total energy of a single electron in the nth orbit (energy level) of the hydrogen atom is described by En = -R x h x c n^2R (Rydberg constant) = 1.0974 x 10^(7)m^(-1)h (Planck’s constant) = 6.626 x 10^(-34) J x sc (Speed of light) = 2.9979 x 10^(8) m/sn= the energy levels of the electron (the principal quantum number)o Relation between n and rn:rn = n^(2)ao ao = Bohr radius (53 pm)Radius becomes larger with increasing “n”o The lines of the atomic spectra correspond to “jumps” or transitions between these levelso The difference between two energy levels can be calculate by ΔE = Efinal state – Einitial stateand the equation for the Bohr Model:En = -R x h x c n^2o As n increases, the energy levels get closer together- Key terms for spectroscopyo Ground state- the lowest energy level (n=1)o Excited state- a subsequently higher energy level, n=2 is the “first excited state” and so on.o Absorption- an electron moving from a lower energy level to a higher energy level via excitationo Emission- an electron moving from a higher to a lower energy level accompanied by the release of a photon- Ex: determine the photon wavelength of the transition from n=5 to n=2 (one blue line of the hydrogen spectrum)Step 1: Use Bohr model equation to calculate E for n=5 and n=2 ΔE = Efinal state – Einitial state = E(2) – E(5) = -R x h x c - (-R x h x c) = -Rhc ( 1 – 1 ) n(2)^2 n(5)^2 n^2(2) n^2(5)Step 2: Calculate energy ΔE = Efinal state – Einitial state = -1.097 x 10^(7) m^(-1) x 6.626x10^34) J x s x 2.998 x 10^(8) m/s = ( 1 – 1 ) 2^2 5^2 = -4.576 x 10^(-19) J Step 3: Use Plank’s equation to obtain λ (wavelength): E = h x c λ λ= h x c E = l ΔE l E = h x c l ΔE l = 6.626x10^(-34) J x s x 2.998x10^(8) m/s = 4.576 x 10^(-19) J = 4.341 x10 ^-7 m x 10^(9) nm = 4.341 nm 1 m- Clicker question: According to the Bohr model for the hydrogen atom, the energy necessary to excite an electron from n=1 to n=2 is _____ the energy necessary to excite an electron from n=2 to n=3A. Less thanB. Greater than correctC. Equal toD. Either less than or equal toE. Half of- The Particle-Wave Dualityo Louis de Broglie: Small particles such as photons should exhibit a characteristic wavelength Light waves have mass and photon (particles) have a wavelength Since E = m x c^2 and E = h x vthen h x v = m x c^2and thus h x v = m x c c ^= p or momentum Since c <--> v <--> 1 λ c λ then h = p <--> λ = h λ p- Ex: what is the wavelength associated with a tennis ball (m= 80 g, d = 8 cm) moving at 185 km/h? Compare that wavelength with an electron that travels at the same speed.Step 1: Calculate p from p = m x v80 g x 1 kg x 185 km x 10^3 m x 1 h = 4.1 kg x m 1000 g h 1 km 3600 s sStep 2: Calculate wavelength λ = h/p 1 J= (kg x m^2)/s^2 =6.626x10^(-36) J x s = 6.626x10^(-36) [(kg x m^2)/s^2] x s = 1.6 x 10^(-34) m 4.1 (kg x m)/s 4.1 (kg x m)/sStep 3: Determine λ of the electron using De Broglie’s equation λ = h/pλ = h/p = 6.626x10^(-36) J x s = 1.42 x 10^(-5) m (9.109 x 10^(-31) kg) x (51.4 (m/s))- The Uncertainty Principleo Werner Heisenberg: It’s impossible to determine simultaneously both the position and velocity of an electron or any other particle with any great degree of accuracy or certainty. Therefore an electron is represented by both a particle and a wave
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