Chapter 23 The Early Atom & Quantum Physics

1. Chapter 23 The Early Atom & Quantum Physics When we consider the motion of objects on the atomic level, we find that our classical approach does not work very well. For understanding motion on the microscopic scale we must use Quantum Mechanics.

2. ATOMIC STRUCTURE Historical Development: • Greek Concepts of Matter • Aristotle - Matter is continuous, infinitely • divisible, and is composed of only 4 elements: • Earth, Air, Fire, and Water • Won the philosophical/political battle. • Dominated Western Thought for Centuries. • Seemed very “logical”. • Was totally WRONG!!

3. ATOMIC STRUCTURE The “Atomists” (Democritus, Lucippus, Epicurus, et. al.) - Matter consists ultimately of “indivisible” particles called “atomos” that canNOT be further subdivided or simplified. If these “atoms” had space between them, nothing was in that space - the “void”. • Lost the philosophical/political battle. • Lost to Western Thought until 1417. • Incapable of being tested or verified. • Believed the “four elements” consisted of “transmutable” atoms. • Was a far more accurate, though quite imperfect “picture” of reality.

4. ATOMIC STRUCTURE Modern Concepts of Matter John Dalton (1803)- An atomist who formalized the idea of the atom into a viable scientific theory in order to explain a large amount of empirical data that could not be explained otherwise. • Matter is composed of small “indivisible”particles called “atoms”. • The atoms of each element are identical to each other in mass but different from the atoms of other elements. • A compound contains atoms of two or more elements bound together in fixed proportions by mass.

5. ATOMIC STRUCTURE Present Concepts - An atom is an electrically neutral entity consisting of negatively charged electrons (e-) situated outside of a dense, posi- tively charged nucleus consisting of positively charged protons (p+) and neutral neutrons (n0). ParticleChargeMass Electron - 1 9.109 x 10 -28 g Proton +1 1.673 x 10 -24 g Neutron 0 1.675 x 10 -24 g

6. ATOMIC STRUCTURE Nucleus Model of a Helium-4 (4He) atom p+no e- e- no p+ Electron Cloud How did we get this concept? - This portion of our program is brought to you by: Democritus, Dalton, Thompson, Planck, Einstein, Millikan, Rutherford, Bohr, de Broglie, Heisenberg, Schrödinger, Chadwick, and many others.

7. ATOMIC STRUCTURE • Democritus - First atomic ideas • Dalton - 1803 - First Atomic Theory • J. J. Thompson - 1890s - Measured the charge/mass • ratio of the electron (Cathode Rays) Fluorescent Material _ Cathode + Anode Electric Field Source (Off) With the electric field off, the cathode ray is not deflected.

8. ATOMIC STRUCTURE - Fluorescent Material - Cathode + + Anode Electric Field Source (On) With the electric field on, the cathode ray is deflected away from the negative plate. The stronger the electric field, the greater the amount of deflection. - Cathode + Anode Magnet

9. ATOMIC STRUCTURE With the magnetic field present, the cathode ray is deflected out of the magnetic field. The stronger the magnetic field, the greater the amount of deflection. e/m = E/H2r e = the charge on the electron m = the mass of the electron E = the electric field strength H = the magnetic field strength r = the radius of curvature of the electron beam Thompson, thus, measured the charge/mass ratio of the electron - 1.759 x 108 C/g

10. ATOMIC STRUCTURE • Summary of Thompson’s Findings: • Cathode rays had the same properties no matter what metal was being used. • Cathode rays appeared to be a constituent of all matter and, thus, appeared to be a “sub-atomic” particle. • Cathode rays had a negative charge. • Cathode rays have a charge-to-mass ratio of 1.7588 x 108 C/g.

11. ATOMIC STRUCTURE R. A. Millikan - Measured the charge of the electron. In his famous “oil-drop” experiment, Millikan was able to determine the charge on the electron independently of its mass. Then using Thompson’s charge-to-mass ratio, he was able to calculate the mass of the electron. e = 1.602 10 x 10-19 coulomb e/m = 1.7588 x 108 coulomb/gram m = 9.1091 x 10-28 gram Goldstein - Conducted “positive” ray experiments that lead to the identification of the proton. The charge was found to be identical to that of the electron and the mass was found to be 1.6726 x 10-24 g.

12. ATOMIC STRUCTURE Ernest Rutherford - Developed the “nuclear” model of the atom. The Plum Pudding Model of the atom: A smeared out “pudding” of positive charge with negative electron “plums” imbedded in it. + + + + + + Electrons The Metal Foil Experiments: Fluorescent Screen a-particles Radioactive Material in Pb box. Metal Foil

13. ATOMIC STRUCTURE If the plum pudding model is correct, then all of the massive a-particles should pass right through without being deflected. In fact, most of the a - particles DID pass right through. However, a few of them were deflected at high angles, disproving the “plum pudding” model. Rutherford concluded from this that the atom con- sisted of a very dense nucleus containing all of the positive charge and most of the mass surrounded electrons that orbited around the nucleus much as the planets orbit around the sun.

14. Problems with the Rutherford Model: It was known from experiment and electromagnetic theory that when charges are accelerated, they continuously emit radiation, i.e., they lose energy continuously. The “orbiting” electrons in the atom were, obviously, not doing this. ATOMIC STRUCTURE Planck • Atomic spectra and blackbody radiation were known to be DIScontinuous. Bohr • The atoms were NOT collapsing.

15. ATOMIC STRUCTURE Atomic Spectra - Since the 19th century, it had been known that when elements are heated until they emit light (glow) they emit that light only at discrete frequencies, giving a line spectrum. - + Hydrogen Gas Line Spectrum

16. ATOMIC STRUCTURE When white light is passed through a sample of the vapor of an element, only discrete frequencies are absorbed, giving a absorption ban spectrum. These frequencies are identical to those of the line spectrum of the same element. For hydrogen, the spectroscopists of the 19th Century found that the lines were related by the Rydberg equation: n/c = R[(1/m2) - (1/n2)] n = frequency R = Rydberg Constant c = speed of light m = 1, 2, 3, …. n = (m+1), (m+2), (m+3), ….

17. Blackbody Radiation One of the earliest indications that classical physics was incomplete came from attempts to describe blackbody radiation. A blackbody is an ideal surface that absorbs all incident radiation. Blackbody radiation is the emission of electromagnetic waves from the surface of an object. The distribution of blackbody radiation depends only the temperature of the object.

18. The Blackbody Distribution The intensity spectrum emitted from a blackbody has a characteristic shape. The maximum of the intensity is found to occur at a wavelength given by Wien’s Displacement Law: fpeak = (5.88  1010 s-1·K-1)T T = temperature of blackbody (K)

19. The Ultraviolet Catastrophe Classical physics can describe the shape of the blackbody spectrum only at long wavelengths. At short wavelengths there is complete disagreement. This disagreement between observations and the classical theory is known as the ultraviolet catastrophe.

20. Planck’s Solution In 1900, Max Planck was able to explain the observed blackbody spectrum by assuming that it originated from oscillators on the surface of the object and that the energies associated with the oscillators were discrete or quantized: En = nhf n = 0, 1, 2, 3… n is an integer called the quantum number h is Planck’s constant: 6.62  10-34 J·s f is the frequency

21. Quantization of Light Einstein proposed that light itself comes in chunks of energy, called photons. Light is a wave, but also a particle. The energy of one photon is E = hf where f is the frequency of the light and h is Planck’s constant. Useful energy unit: 1 eV = 1.6  10-19 J

22. Quantum Mechanics The essence of quantum mechanics is that certain physical properties of a system (like the energy) are not allowed to be just any value, but instead must be only certain discrete values.

23. Example (a) Find the energy of 1 (red) 650 nm photon. (b) Find the energy of 2 (red) 650 nm photons.

24. The PhotoElectric Effect When light is incident on a surface (usually a metal), electrons can be ejected. This is known as the photoelectric effect. Around the turn of the century, observations of the photoelectric effect were in disagreement with the predictions of classical wave theory.

25. Observations of the Photoelectric Effect • No electrons are emitted if the frequency of the incident photons is below some cutoff value, independent of intensity. • The maximum kinetic energy of the emitted electrons does not depend on the light intensity. • The maximum kinetic energy of the emitted electrons does depend on the photon frequency. • Electrons are emitted almost instantaneously from the surface.

26. The Photoelectric Effect Explained(Einstein 1905, Nobel Prize 1921) The photoelectric effect can be understood as follows: • Electrons are emitted by absorbing a single photon. • A certain amount of energy, called the work function, W0, is required to remove the electron from the material. • The maximum observed kinetic energy is the difference between the photon energy and the work function. Kmax = E – W0 E = photon energy

27. Walker Problem 25, pg. 1008 Zinc and cadmium have photoelectric work functions given by WZn = 4.33 eV and WCd = 4.22 eV, respectively. (a) If both metals are illuminated by UV radiation of the same wavelength, which one gives off photoelectrons with the greater maximum kinetic energy? Explain. (b) Calculate the maximum kinetic energy of photoelectrons from each surface if l = 275 nm.

28. The Mass and Momentum of a Photon Photons have momentum, but no mass. We cannot use the formula p = mv to find the momentum of the photon. Instead:

29. The Wave Nature of Particles We have seen that light is described sometimes as a wave and sometimes as a particle. In 1924, Louis deBroglie proposed that particles also display this dual nature and can be described by waves too! The deBroglie wavelength of a particle is related to its momentum: • = h/p (Use p = gmv if the velocity is large.)

30. Example If a baseball, with a mass of 0.200 kg has a speed of 45 m/s, what is its wavelength?

31. Example If an electron has a speed of 1.00  106 m/s, what is its wavelength?

32. Example • The maximum momentum of electrons at the Jefferson Lab accelerator in Newport News is 6 GeV/c. • What is the wavelength of those electrons? • Why is the wavelength well suited to the study of nuclear physics?