1 / 43

CLASSICAL DESCRIPTION OF AN ATOM

CLASSICAL DESCRIPTION OF AN ATOM. Instructor: Dr. Mehr Nigar. Classical Description of an atom. Coulomb’s Force Law F(r) = (-e)(e)/4 πε 0 r 2 = -e 2 /4 πε 0 r 2. e = absolute value of electron charge r = distance between two charges

bonniereeder
Download Presentation

CLASSICAL DESCRIPTION OF AN ATOM

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. CLASSICAL DESCRIPTION OF AN ATOM Instructor: Dr. Mehr Nigar

  2. Classical Description of an atom Coulomb’s Force Law F(r) = (-e)(e)/4πε0r2 = -e2/4πε0r2 e = absolute value of electron charge r = distance between two charges ε0= permittivity constant of a vacuum 8.854 x 10-12 C2J-1m-1

  3. Topics: • Failure of the Classical description of Atom • Introduction to Quantum Mechanics: Wave-particle duality • Light as a wave; characteristics of waves • Light as a particle; photoelectric effect

  4. Classical Description of an Atom Coulomb’s Force Law: Describes force as a function of r (distance) Does NOT tell how r changes with time, t When r → ∞ F(r) → 0 F(r) → ∞ When r → 0 There is however a CLASSICAL EQUATION OF MOTION that can describe how the electron and the nucleus change position with time.

  5. Newton’s Second Law F = ma Force = mass x acceleration F = m (dv/dt) = m(d2r/dt2) Put in Coulomb’s Force law for F For rinitial = 10 Å r = 0 at t = 10-10 s!!! Qualitatively the electron should plummet into the nucleus in 0.1 nano seconds!!! Laws of classical mechanicsno longer work at this size scale! Quantum Mechanics explains the behavior of atom in a better way.

  6. Failure of Classical Mechanics • What is wrong in this mathematical calculation??? • Coulomb’s Force Law? • Newtonian Mechanics? F = ma The laws of classical mechanics no longer work at this size scale!

  7. The Quantum Revolution! • “Will we ever understand the atom?” (question posed to Niels Bohr) • Bohr: “Yes, we will, but only if we change what we mean by the word understand.” • “Things on small scale behave like nothing that you have any direct experience about.” Richard Feynman (1963)

  8. Assumptions made by Quantum Mechanics • Matter and radiation display both wave-like and particle-like properties. • Light consists of discrete packets of energy called photons.

  9. Properties of Waves • Waves tend to have a periodic variation of some quantity

  10. Examples of Waves Water Waves Sound Waves

  11. Amplitude, Wavelength and Frequency: Amplitude is defined as the deviation from the average level, it can be positive or negative. Wavelength is the distance between the successive maxima or minima. Frequency is the number of cycles per unit time.

  12. What is the amplitude of the water wave below? • 10 m • 12 m • 5 m • 6 m x 10 m 12 m

  13. Light Waves Image courtesy of MIT open coursware Light (electromagneticradiation) is a periodic variation of an electric field.

  14. Speed =  1/ f How fast is a wave of light travelling? At t = 0 At t = 1/f (one period) Speed = distance traveled / time elapsed = fm/sec

  15. Speed of EM Radiation • Electromagnetic radiation has a constant speed, c (the speed of light) • f = c = 2.9979 x 108 ms-1 • (c  670,000,000 mph, or 186,000 miles/sec!

  16. Compare light wave A to light wave B as illustrated below: • Light wave A has a shorter  and a lower f. • Light wave A has a shorter  and a higher f. • Light wave A has a longer  and a higher f. • Light wave A has a longer  and a lower f. light wave A light wave B

  17. Electromagnetic Spectrum Long Wavelengths Low Frequencies Short Wavelengths High Frequencies

  18. Applications of Flourescence Fluorescent Paint Quantum Dots Frequency  of the emitted light is proportional to the size of the quantum dot.

  19. Superposition of Waves

  20. Applications of Destructive Interference • Noise Cancellation Headphones: Take in the ambient noise waves and inbuilt batteries Produce waves to cancel out the noise waves

  21. Lasers • A laser is source of light that emits the purest, brightest color of any device known. The word laser is an acronym standing for light amplification by stimulated emission of radiation. A gas laser is just a discharge tube with mirrors on each end. One of the mirrors is only partially reflecting, to allow some light to escape. • The earliest gas laser (invented in 1961) is the helium-neon laser, which emits red light of wavelength  = 6.328 x 10-5 cm (632.8 nm); it came into widespread use in barcode scanning. Calculate the frequency and wave number of this laser light. • Solution: f = c/ = (2.998 x 1010 cm s-1) / (6.328 x 10-5 cm) = 4.738 x 1014 cycles s-1 = 4.738 x 1014 Hz wave number = 1/  = 15,800 cm-1

  22. Light as as Particle: The Photoelectric Effect • Established in 1800’s that light was a wave. • Diffraction of light was observable • Constructive and destructive interference of light was observable

  23. Lenard’s Photoelectric Apparatus:

  24. The Experiment: By varying the voltage on a negatively charged grid between the ejecting surface and the collector plate, Lenard was able to: • Determine that the particles had a negative charge. • Determine the kinetic energy of the ejected particles.

  25. Perplexing Problems: • The intensity of light had no effect on the kinetic energy of the electrons. • There was a threshold frequency f0 for electron ejection. • Classical Physics failed to explain these observations. Lenard won the Nobel Prize in Physics in 1905!

  26. Graph of number of electrons vs. frequency

  27. Observations: • No electrons were ejected below a certain threshold frequency, which was different for each metal. • The number of electrons ejected was not related to the frequency of the incident light!

  28. Kinetic Energy of Electrons

  29. Intensity of Light vs. number of Electrons

  30. Number of electrons as a function of Intensity:

  31. Einstein’s Observations y = mx + b Slope (m) = 6.626 x 10-34 Js Intercept (b) = (6.626 x 10-34 Js)0 Planck had observed this number much earlier as a fitting constant to explain Black body radiation.

  32. Einstein’s Relations: Einstein predicted that a graph of the maximum kinetic energy versus frequency would be a straight line, given by the linear relation: KE = hf - Φ …Therefore light energy comes in multiples of hv This was published in his famous 1905 paper: “On a Heuristic Point of View About the Creation and Conversion of Light”

  33. Einstein’s Interpretation A new theory of light: • Electromagnetic waves carry discrete energy packets, called photons. • Energy of a photon is proportional to its frequency. • More intense light corresponds to a greater number of photons, not higher energy photons. This was published in his famous 1905 paper: “On a Heuristic Point of View About the Creation and Conversion of Light”

  34. New Model for Photoelectric Effect: K.E +  Ei = K.E = Ei - 

  35. If a beam of light with energy = 7.0 eV (1 eV = 1.602 x 10-19 J) strikes a gold surface, what is the maximum kinetic energy of the ejected electrons. • K.E. = 12.1 eV • K.E. = 5.1 eV • K.E. = 1.9 eV • K.E. = 7.0 eV • No electrons will be ejected. light (Ei = 4.0 eV) gold ( = 5.1 eV)

  36. Ei = Incident Energy f0 = threshold frequency  = Work Function If f  f0 No electrons are ejected! If f  f0 Electrons are ejected!

  37. Energy of a photon: • Energy of a photon must be equal to or greater than the work function of a metal, for an electron to be ejected. e-is ejected! e- is NOT ejected! e- is NOT ejected! In photoelectric effect, light is acting as particles and NOT waves, therefore the energies cannot add up; like they do in case of waves.

  38. Summary: • If the incident beam has insufficient energy i.e Ei< , no matter how intense the beam, electrons will not be ejected. • If Ei>  then does the intensity of the incident beam has any effect on the ejected electrons? Each photon ejects a single electron, therefore the greater the number of incident photons the higher the number of ejected electrons. Intensity of light (energy/sec) is proportional to the # of photons emitted/sec. High intensity means more photons/sec, Unit of Intensity = W = Js-1 NOT more E / photon

  39. Summary of Terminology: • Photons: also called light, electromagnetic radiation, may be described in terms of energy,  or f. • Electrons: also referred to as “photelectrons”, may be described in terms of K.E., velocity or .

  40. Problem 1 • Consider two light sources; a UV lamp ( = 254 nm) and a laser pointer ( = 700 nm), are these two light sources capable of ejecting electrons from a Zinc plate? ( = 6.9 x 10-19J) 1)What is the energy/photon emitted by the UV lamp? E = 7.82 x 10-19J, the UV lamp does have enough energy to eject an electron from Zn surface. 2) What is the energy/photon emitted by the Laser pointer? E = 2.84 x 10-19J, the laser pointer does NOT have enough energy to eject an electron from Zn surface. 3) What is the total number of photons emitted by the laser pointer in 60 seconds if intensity (I) = 1.00mW # of photons emitted if the laser pointer is used for 60 s = 2.1 x 1017 photons. The intensity of light is NOT related to the energy of its photons!

  41. Problem 2: • The “electric eyes” used to open doors automatically are based on the photoelectric effect. The metal surface inside the “eye” which is basically an evacuated glass envelope containing the surface (cathode) and a photocurrent collector (anode), continuously provide current until the light beam that actuates it is interrupted. What is the maximum kinetic energy of photoelectrons produced by a mercury vapor lamp, emitting 436 nm violet light, actuating an electric eye of work function 2.1eV? • K.E = Ei -  • K.E = hc/ -  • K.E = 6.626 x 10-34 Js x 2.998 x 108 ms-1 – (2.1 x 1.602 x 10-19Js-1) 436 x 10-9 m K.E = 1.19 x 10-19 Js-1 = 0.74 eV

  42. Problem 3: • If a beam of light with energy = 5.0 eV (1 eV = 1.602 x 10-19 J) strikes a gold surface, what is the maximum kinetic energy of the ejected electrons. Light Ei = 5.0 eV gold ( = 5.1 eV) No electrons will be ejected!

  43. Practical Applications of Photoelectric Effect: • Photocell: A vacuum tube with a photosensitive cathode, electrons flow from cathode to anode to complete the circuit in presence of light. Can be made responsive to light orto removal of light. • Photoconductive devices: lead to an increase in the conductivity of a non-metallic material when exposed to light. • Solar Cells: act like a battery when exposed to light. Individual solar cells produce voltages of about 0.6 volts but higher voltages and large currents can be obtained by appropriately connecting many solar cells togetherRead more: • <a href="http://science.jrank.org/pages/5169/Photoelectric-Effect-Applications.html">Photoelectric Effect - Applications</a>

More Related