10/17/11. Chapter 7. Quantum Theory and Atomic Structure. Amplitude (intensity) of a wave. Quantum Theory and Atomic Structure

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1 Quantum Theory and Atomic Structure Chapter 7 7. The Nature of Light Quantum Theory and Atomic Structure 7. Atomic Spectra 7. The Wave-Particle Duality of Matter and Energy 7.4 The Quantum-Mechanical Model of the Atom Chap 7- Chap 7- Amplitude (intensity) of a wave. Frequency and wavelength. c = ν c =.998 x 0 8 m/sec Chap 7- Chap 7-4 Regions of the electromagnetic spectrum. Interconverting Wavelength and Frequency PROBLEM: A dental hygienist uses x-rays ( =.00 Ǻ) to take a series of dental radiographs while the patient listens to a radio station ( = 5 cm) and looks out the window at the blue sky ( = 47 nm). What is the frequency (in s - ) of the electromagnetic radiation from each source? (Assume that the radiation travels at the speed of light,.00 x 0 8 m/s.) PLAN: Use c = ν Chap 7-5 Chap 7-6

2 Interconverting Wavelength and Frequency: Solution c = ν Ǻ x 0 m =.00 x 0 m Ǻ 5 cm x 47 nm x Chap m = 5 x 0- m cm 0-9 m = 47 x 0-9 m nm x 08 m/s = x 08 s- ν =.00 x 0-0 m ν= Different behaviors of waves and particles. x 08 m/s = 9. x 07 s- 5 x 0- m ν= x 08 m/s 47 x 0-9 m = 6.4 x 04 s- The diffraction pattern caused by light passing through two adjacent slits. Chap 7-8 Particle Nature of Light: Blackbody Radiation Three phenomena could not be explained by classical physics: blackbody radiation, photoelectric effect, and atomic spectra. A new picture of energy was required. Blackbody Radiation: Heat a solid object: at 000 K it begins to emit soft red light; at 500 K it begins to glow orange and is brighter; at 000 K it is still brighter and white. How to explain? Planck: proposed that only certain quantities of energy could be emitted or absorbed: E = nhν = nhc These packets of energy are called quanta, and n is a quantum number. Since it was (correctly) assumed that most transitions occur between adjacent states, Δn = and the change in energy is: ΔE = hν = Chap 7-9 hc Chap 7-0 Calculating the Energy of Radiation from Its Wavelength Demonstration of the photoelectric effect. NOT a function of light intensity; IS a function of light wavelength (and therefore frequency) Threshold frequency ; NO time lag. PROBLEM: A cook uses a microwave oven to heat a meal. The wavelength of the radiation is.0 cm. What is the energy of one photon of this microwave radiation? PLAN: After converting cm to m, we can use the energy equation, E = hν combined with ν = c/ to find the energy. SOLUTION: E = hν = E= Chap 7- Chap 7- hc ( J sec)( m / sec) =.66 0 J.0 0 m

3 Atomic Spectra: A Problem to Solve! Continuum : Classical Physics Chap 7- Rydberg equation for emission = R where n > n n n The line spectra of several elements. Chap 7-4 Quantum staircase. R is the Rydberg constant = x 07 m- Three series of spectral lines of atomic hydrogen. for the visible series, n = and n =, 4, 5,... Chap 7-5 The Bohr explanation of the three series of spectral lines. Chap 7-7 Chap 7-6 Wave motion in restricted systems. Chap 7-8

4 Wave-Particle Duality = Energy (electromagnetic radiation) behaves like waves (diffraction, wavelength, frequency) but also possesses particle behavior (photons in photoelectric effect). Invoke wave-particle duality of EMR to explain. If energy can exhibit both particle and wave natures, what about matter? debroglie: Matter Waves h h = mv p Large items have very small wavelengths Electrons are small enough however to exhibit measurable wave properties Chap 7-9 Chap 7-0 Determining E and of an Electron Transition Comparing diffraction patterns of x-rays and electrons. PROBLEM: A hydrogen atom absorbs a photon of visible light and its electron enters the n = 4 energy level. Calculate (a) the change in energy of the atom and (b) the wavelength (in nm) of the photon. PLAN: (a) The H atom absorbs visible light, so the electron is going from n = to n = 4. Calculate ΔE. (b) ΔE = hν = hc/ SOLUTION: Z Z = J 4 n final ninitial (a) ΔE = J ΔE = J (b) = Chap 7- Electron diffraction of aluminum X-ray diffraction of aluminum hc ( J s)( m/s) = = m = 486 nm ΔE J Chap 7- Summary of the major observations and theories leading from classical theory to quantum theory. Wave-Particle Duality: Problems Classical mechanics: particles have trajectories and can precisely be described according to location (x) and linear momentum (p) Quantum mechanics: can t think like this: waves are spread out as they oscillate Duality denies the ability of knowing the trajectory of particles (complementarity) Heisenberg Uncertainty principle: Important for wave equations Chap 7- Chap 7-4 4

5 Schrödinger Equation The SWE relates the second derivative of Ψ to the value of Ψ at each point; impossible to solve exactly (except simple cases) One simple solution: particle in a box The Heisenberg Uncertainty Principle Δx m Δu h Λ = ενγτη οφ βοξ ν =,, ξ = διστανχε βετωεεν 0 ανδ Λ 4π Solution of SWE using this yields: Since n is an integer, E is quantized with discrete values of E and, since ΔE = hν and ΔE = En+ - En Chap 7-5 Chap 7-6 SWE: Particle in a Rectangular Box Electron probability density in the ground-state H atom. What if the box is -dimensional? Solution of SWE using this yields: Chap 7-7 Chap 7-8 Hydrogen Wavefunctions Quantum Numbers and Atomic Orbitals Angle independent An atomic orbital is specified by three quantum numbers. n the principal quantum number - a positive integer (,,, ) l the angular momentum quantum number - an integer from 0 to (n - ) ml the magnetic moment quantum number - an integer from l to +l R decreases exponentially as r increases a0 is the Bohr radius Chap 7-9 Chap 7-0 5

6 Determining Quantum Numbers for an Energy Level PROBLEM: What values of the angular momentum (l) and magnetic (ml) quantum numbers are allowed for a principal quantum number (n) of? How many orbitals exist for n =? PLAN: Follow the rules for allowable quantum numbers found in the text. l values can be integers from 0 to (n ); ml can be integers from -l through 0 to +l. SOLUTION:For n =, l = 0,, For l = 0, ml = 0 For l =, ml = -, 0, + For l =, ml = -, -, 0, +, + There are 9 ml values and therefore, 9 orbitals with n =. Chap 7- Chap 7- Determining Sublevel Names and Orbital Quantum Numbers PROBLEM: Give the name, magnetic quantum numbers, and number of orbitals for each sublevel with the following quantum numbers: (a) n =, l = (b) n =, l = 0 (c) n = 5, l = The s, s, and s orbitals. (d) n = 4, l = PLAN: Combine the n value and l designation to name the sublevel. Knowing l, we can find ml and the number of orbitals. SOLUTION: sublevel name possible ml values n l (a) d -, -, 0, +, + 5 (b) 0 s 0 (c) 5 5p -, 0, + (d) 4 4f -, -, -, 0, +, +, + 7 Chap 7- # of orbitals Chap 7-4 The d orbitals. The p orbitals. Chap 7-5 Chap 7-6 6

7 One of the seven possible 4f orbitals. Chap 7-7 7