VIEWS: 133 PAGES: 39 CATEGORY: Education POSTED ON: 12/31/2009
Fig. 4-9, p. 125 HYDROGEN ATOM VISIBLE SPECTRUM This grouping of lines has a name. It is called the Balmer Series. (1885, Johann Balmer ) Notice how the spacing increases as the wavelength gets longer This will become important later. • the energy levels are found to follow the empirical formulas given by Balmer and Rydberg. ⎛ Z 2 ⎞ E = − 2 . 178 x 10 − 18 J ⎜ 2 ⎟ ⎝ n ⎠ Z = atomic number (1 for H) n = integer (2, 3, ….., ….) • R = -2.178 x 10-18 J is the Rydberg Constant • We can use the equation to predict what ΔE is for any two energy levels ΔE = E final − E initial ⎛ 1 ⎞ ⎛ 1 ⎞ −18 −18 ΔE = −2.178x10 J⎜ 2 ⎟ − (−2.178x10 J)⎜ 2 ⎟ ⎜n ⎟ ⎝ final ⎠ ⎝ n initial ⎠ ⎛ 1 1 ⎞ ΔE = −2.178x10−18 J⎜ 2 − 2 ⎟ ⎜n ⎝ final n initial ⎟ ⎠ • H Atom Energy level diagram Visible Spectrum Energy 0 Note: As n goes to n=∞ -R/25 5 Infinity energy levels -R/16 4 get closer and closer -R/9 3 together.Looks more Continuous -R/4 2 Balmer R En i = − 2 ni R En f = − 2 nf ⎛ 1 1 ⎞ ΔE = E n f − En i = − R 2 − 2 ⎟ ⎜ ⎜n ni ⎟ ⎝ f ⎠ -R 1 Fig. 4-13, p. 129 ⎛ Z2 ⎞ E = −2.178x10−18 J⎜ 2 ⎟ ⎝n ⎠ • Energy levels get closer together as n increases • at n = infinity, E = 0 Bohr Model of the Atom • Postulates: 1. The coulomb force on a planetary electron provides the centripetal acceleration required for a dynamically stable circular orbit. 2. The only permissible orbits are those in the discrete set for which the angular momentum of the electron equals an integer n times h, where h = h / 2π Note Quantization of Angular Momentum Postulates Part II 3. An electron moving in one of these stable orbits does not radiate. 4. Emission or absorption of radiation occurs only when an electron makes a transition from one orbit to another. The Bohr Hydrogen Atom • We can use Bohr’s postulates to derive the energy levels in Hydrogen • Postulate 1 implies: Z=1 2 2 mv ke = 2 r r From the second postulate we can write mvr = nh Radius of the Hydrogen Atom • We get the equation 2 2 n h 2 r= 2 = n a0 mke Which if evaluated comes out as: a0 = 5.29 10-11 m ≈ 0.53Å Energy Levels in the Bohr Atom • The kinetic energy of an electron in circular motion is: 2 2 1 2 ke ke K = mv = r 2 = 2 2r 2r The potential energy in a Coulomb Potential 2 ke V =− r The total energy is then 2 2 2 ke ke ke E = K +V = − + =− r 2r 2r Energy Levels Part II • The energy of a given level depends on the radius of the electron: Actually “Reduced Mass m m /(m + m ) e p e p 2 2 4 ke mk e w0 En = 2 =− 2 2 =− 2 2n a0 2h n n 2.178 eV w0 = 13.60 x 10 -18 Joules Fig. 4-13, p. 129 • Example: At what wavelength will emission from n = 4 to n = 1 for the H atom be observed? ⎛ 1 1 ⎞ ΔE = −2.178x10−18 J⎜ 2 − 2 ⎟ ⎜n ⎝ final n initial ⎟ ⎠ 1 4 −18 ⎛ 1⎞ ΔE = −2.178x10 J⎜1− ⎟ = −2.04 x10−18 J ⎝ 16 ⎠ hc ΔE = 2.04 x10 −18 J= λ = 9.74 x10−8 m = 97.4nm λ Alternative Units • 1 electron volt = 1.602 x 10-19 joules • 2.18 x 10 – 18 joules/ 1.602 x 10-19 joules/eV = 13.6 eV • We can use the equation to predict the Photon Energy Connecting any two energy levels ΔE = E final − E initial ⎛ 1 ⎞ ⎛ 1 ⎞ −18 −18 ΔE = −2.178x10 J⎜ 2 ⎟ − (−2.178x10 J)⎜ 2 ⎟ ⎜n ⎟ ⎝ final ⎠ ⎝ n initial ⎠ ⎛ 1 1 ⎞ Ephoton = ΔE = −2.178x10−18 J⎜ 2 − 2 ⎟ ⎜n ⎝ final n initial ⎟ ⎠ ⎛ 1 1 ⎞ eV Ephoton = 13.6⎜ 2 − 2 ⎟ ⎝n m ⎠ HYDROGEN ATOM IONIZATION Energy = R/n2 = 2.179874113 x 10 -18 J 0.00 Fig. 4-12, p. 129 Extension to Higher Z • The Bohr model can be extended to any single electron system….must keep track of Z (atomic number). ⎛ Z2 ⎞ Z = atomic number E = −2.178x10−18 J⎜ 2 ⎟ ⎝n ⎠ n = integer (1, 2, ….) • Examples: He+ (Z = 2), Li+2 (Z = 3), etc. Extension to Higher Z (cont.) • Example: At what wavelength will emission from n = 4 to n = 1 for the He+ atom be observed? ⎛ 1 1 ⎞ ΔE = −2.178x10−18 J (Z 2 ) 2 − 2 ⎟ ⎜ ⎜n ⎝ final n initial ⎟ ⎠ 2 1 4 ⎛ 1⎞ ΔE = −2.178x10 −18 J (4)⎜1− ⎟ = −8.16x10−18 J ⎝ 16 ⎠ hc −18 ΔE = 8.16x10 J= λ = 2.43x10−8 m = 24.3nm λ λ H > λ He + Albert Einstein explained the photoelectric effect Explanation involved light having particle-like behavior. – Fig. 4-14, p. 132 EM Radiation Has Wave Properties λν = c EM Waves Have Particle Properties hc E = hν or E= λ = h = Planck’ constant 6.626x 10-34 J ⋅ s s Momentum P = h/λ Next Advance in Understanding Atomic Structure Came From Asking “If EM Waves Have Particle Properties-Do Other particles than the photon Have Wave Properties” ? • yes- Particles also exhibit duality De Broglie Associated a Wave Length With Particles Planck’s Constant Mass of particle λ = h/(mV) Velocity of particle Wavelength of an electron traveling 1.24 x 107 meters/sec 0.041 x speed of light λ = h/(mV) = (6.626 x 10-34 Joule sec ) (9.11 x 10-31 kilograms x 1.24 x 107 meter/sec) = 5.87 x 10-11 meters comparable to interatomic distances Wavelength of a baseball traveling 92.5 miles/hour 1.4 x 10-7 speed of light λ = h/(mV) = (6.626 x 10-34 Joule sec ) (0.149 kilograms x 41.3 meter/sec) = 1.08 x 10-34 meters Much smaller than baseball or distance to home plate Bohr’s Model was improved upon in the 1920’s “WAVE MECHANICAL” or “QUANTUM MECHANICAL” Model. An “Operator Equation” – performing the operation indicated by H on the wave function leads to a simple answer , the energy times the wave function They govern the motion of the electrons IN PARTICULAR IF CORRECT THEN OUR JOB IS TO DEFINE H AND THE WAVE FUNCTION H Must Contain the Interaction terms of the electrons • Schrödinger proposed a Wave Equation to describe how electrons move in atoms • It has a form HΨ=ΕΨ • Where H is the Hamiltonian “Operator” * Ψ is the Wave Function E is the energy of the system *Simple Example of Operator equation: In the equation 2a = b, the “operation” is multiply by 2. Operating on a gives b. Operations can be more complicated STANDING WAVES Fig. 4-17, p. 135 STANDING WAVES Fig. 4-18, p. 135 Particle in a One Dimensional Box Fig. 4-24, p. 146 H GENERAL SOLUTION OF THIS DIFFERENTIAL EQUATION General Solution AT ENDS OF BOX DEMAND Ψ = 0 WHY? TRUE WHEN INTEGER OR 0 L SO If Wave Function squared is probability then integrated over all space must equal 1 SO We Can Now Plot These Solutions For n =, 1.2, 3 etc. 1 2 3 Fig. 4-24, p. 146 Fig. 4-24, p. 146