particle-wave dual nature of electrons by hcj

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330 07’04: The DUAL NATURE OF THE ELECTRON CORRELATIONS
National Content Standards: Connecticut Standards: Ridgefield Themes:

ESSENTIAL CONCEPT
“Planck‟s work led to the explanation of the photoelectric effect by Einstein, who postulated that light consists of particles called photons, and the emission spectrum of the hydrogen atom by Bohr. Further advancements to quantum theory were made by de Broglie, who demonstrated that an electron possess both particle and wave properties, and Heisenberg, who derived an inherent limitation to measuring submicroscopic systems. These developments culminated in the Schrödinger equation, which describes the behavior and energy of electrons, atoms and molecules.”1

HOMEWORK
DUE: DO:

REVIEW:
1. Into (e.g., sci. method, classification of matter, SI units, significant figures, accuracy/precision, dimensional analysis, percent errors) 2. Atoms, Molecules & Ions (e.g., atomic structure, periodic table, names & formulas) 3. Stoichiometry (e.g., Avogadro‟s number, molecular mass, % composition, empirical & molecular formulas, mass/mole/number conversions, limiting reagents, percent yield) 4. Reactions in Aqueous Solutions (e.g., types of reactions, predicting products, acids/bases, M=g/L; m1v1=m2v2) 5. Gases (e.g., STP,

P1V1 P1V1 r  , V=kn, PV=nRT, Pi = XiPT, PT = P1 + P2 + ..., 1  n1T1 n1T1 r2

M2 ) M1

6. Thermochemistry (state functions; E=q+w; w= –PV; H=E+PV; E=H - PV; C=ms; q=mst; 0 q=CT; Hrnx  nH 0 ( prod )  nH 0 (react ) f f PREVIEW: 7. From Classical Physics to Quantum Theory (i.e., Modern Atomic Theory): c = ; E = h; Bohr model of the atom to explain the emission spectrum of the hydrogen atom; wave-particle dual nature of the electron; Heisenberg uncertainty principle; Schrödinger wave equation; quantum numbers (n, l, m, s); Aufbau/Pauli exclusion/Hund; orbitals (s, p, d, f); electron configuration/orbital notation/noble-gas configuration.

LESSON:
1

Yang, R. (2008). Chapter 7: “The Electronic Structure of Atoms.” In General Chemistry: The Essential Concepts (5th ed., p. 206). New York: McGraw Hill.

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§07‟04: THE DUAL NATURE OF THE ELECTRON
1. DUAL NATURE A. Background i. physicists questioned Bohr‟s Theory – Why is the electron in an atom

restricted to orbiting the nucleus at fixed distances? Why does it not lose energy and fall into the nucleus, obliterating the atom?
ii. Standing wave: like a musical note with harmonic resonance. source: http://id.mind.net/~zona/mstm/physics/waves/standingWaves/standingWaves1/StandingWaves1.html
(03/18/08)

iii. The following information is included for completeness. a. The relationship between the wavelength () of an orbit and the circumference (2r) of the orbit is: 2r = n (Eq. 7.6) where n is an integer (1, 2, 3, ...) b. Fig. 7.10 – standing waves for a guitar string (below). Each dot in the below figure represents a node (amplitude of the wave is zero). The length of the string (l) must equal an integer times one-half the wavelength (

   ; 2  etc). 2 2 2

Figure 7.11 – schematic of electron wave in an allowed orbit: The circumference of the orbit equals an integral number of wavelengths so this is an allowed orbit. The circumference does NOT equal an integral number of wavelengths so the wavelength is a nonallowed orbit.

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c. de Broglie argued that waves behave like particles and particles exhibit wave-like behavior: h (Eq. 7.7)  mu where = wavelength (m) 2 2 –34 h = Planck‟s constant (6.626 x 10 Js) (1 J = 1 kg * m /s ) m = mass of the particle (kg) u = velocity (m/s)

EXAMPLE 7.4
Problem. Calculate the wavelength of the „particle‟ (a) The fastest tennis serve is about 150 miles per hour. (68 m/s). What is the wavelength (m) associated with a 6.0 x 10–2 kg tennis ball? Solution.
= h m u

=

6.63E-34 J  s = 1.6E-34 m 6.0E-02 kg 68 m/s

Note: This is an exceedingly small wavelength considering that an atom, itself, is only on the order of 10–10 m. Practice Problems (not from textbook) A. Problem. Calculate the wavelength of an XM radio signal, e.g., 2340 MHz (See §7-1).
c =   = c  3.0E+08 m/s

=

2.340E+09 /s
= 1.3E-01 m

Which is about 10 cm (~ 4 inch) B. Problem. Calculate the wavelength of a VW Jetta (3,285 lbs or 1493.2 kg) traveling at 55 mph?
1. m/h = km/s 55 miles hr 2. wavelength 5280 ft 1 mile 12 inch 1 ft 2.54 cm 1 inch 1 m 100 cm 1 hr 1 hr 60 min 60 sec =

25 m s

=

h m u = 1.8E-38 m

=

6.63E-34 J  s 1493.2 kg 24.6 m/s

Note: Again, this is an exceedingly small wavelength considering that an atom, itself, is only on the order of 10–10 m.


								
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