5.111 Lecture Summary #2

5.111 Lecture Summary #2 Readings for today: A.2-A.3 (pgs F10-F13), B.1-B.2 (pgs F15-F18), and Section 1.1. Read for Lecture #3: Section 1.2 (1.1 in 3rd ed) – The Characteristics of Electromagnetic Radiation, Section 1.4 (1.2 in 3rd ed)- Radiation, Quanta, and Photons (focus on pgs 10-12 in the 4th ed or pgs 5-7 in the 3rd ed). Topics: I. Discovery of the electron and the nucleus II. (Failure of) the classical description of the atom Heading into the 20th century there was a feeling by many in the chemistry and physics communities that our scientific knowledge was nearly complete. It was universally accepted that atoms were the most basic constituent of matter and that the behavior of all matter could be explained through Newtonian mechanics. BUT…several discoveries and observations contradicted these theories: I. A) DISCOVERY OF THE ELECTRON J.J. Thomson (English physicist, 1856-1940) in 1897 discovers the electron and determines the charge to mass ratio. In 1906 he wins the Nobel Prize. The Experiment: deflection plate (+) cathode(-) anode(+) H2(g) !V >> 0 !x deflection cathode rays !V = 0 deflection plate (-) high voltage difference cathode rays = _____________________-charged particles = electrons If ΔV (the voltage difference between the two deflection plates) = 0 there is no deflection, but when ΔV is large there is Δx deflection towards the positive plate. From classical electromagnetism, Thompson knew that Δx(-) ∝ _______ e(-)= ___________________ of the negative particle m(-) = __________________ of the negative particle 1 BUT when a very large ΔV is applied a slight deflection towards the negative plate is also observed. Therefore there is also a positive particle! Δx(+) ∝ _______ e(+)= charge of the positive particle m(+) = mass of the positive particle BUT Δx(-) >>> Δx(+) + Thompson suspected that the positive particle was H and H was the source of the negative particle. Since H is electronically neutral, Then  e(-) =  e(+) So:  Δx(-)  =  Δx(+) Since the ratio of Δx is large the ratio of masses will be large : m(-) ___________ m(+) This negative particle from the cathode ray tubes was named the electron (m = 9.11 x 10 31 kg). Atoms are NOT indivisible! Thomson proposed a model of the atom in which electrons are embedded in a cloud of positive charge. His (inaccurate) model is often referred to as the "Plum pudding model". I. B) DISCOVERY OF THE NUCLEUS Ernest Rutherford (Cambridge University, England, 1871-1937) studied α emission from newly-discovered radioactive elements. • α particles were known to be heavy, charged particles • α particles are ________ atoms- this was unknown in 1911 In 1911 Rutherford tested the plum pudding model. (Experiments were done with a postdoc, Hans Geiger, and an undergraduate, E. Marsden.) 2 The Experiment: RaBr2 ! particles detector Count rate = ____________________ α particles / min Au foil (~10-9 m thick) RaBr2 ! particles detector Count rate = ____________________ α particles / min It appeared that all of the α particles passed through the Au foil. A detector was built that could swing around to the front side and measure any potential back scattered particles. Au foil (~10-9 m thick) RaBr2 ! particles detector detector for backscattering Count rate = ____________________ α particles / min Backscattering was detected!! 3 Probability (P) of backscattering: P= count rate backscattered count rate of incident particles = ___________ = Interpretation: Plum pudding model disproved! New atomic model • Since the vast majority of α particles pass through the Au foil undeflected, the Au atoms are mostly ________________. • A very tiny percentage of α particles hit something massive in the atom and backscatter (bounce back). This indicates that most of the mass of the atom is concentrated in a very small volume relative to the volume of the entire atom. We now call this the NUCLEUS. • Rutherford calculated the diameter of the nucleus to be _________ m. His calculation (see below) related the probability of backscattering to the diameter of the nucleus based on the size of the atom (10-10 m) and the thickness of the foil. • Rutherford proposed the charge on the nucleus to be positive, since electrons are negatively charged and atoms are neutral. Charge on the nucleus = _________ Z = atomic number e = absolute value of electron’s charge NOW FOR AN IN-CLASS RUTHERFORD BACKSCATTERING EXPERIMENT! Please imagine: white Styrofoam balls = 1 monolayer of Au nuclei ping pong balls = α particles _________ = radioactive material emitting α particles 4 Probability of backscattering (P) = ________________________________________ (We will experimentally determine this value.) P is related to r, the radius of an Au nucleus (styrofoam ball) in the following way: P = _________________________ = _______________________________________ P = ____________ 1.39 m2 We can solve for r (radius) and d (diameter) of the “Au nucleus” / Styrofoam ball: r = ___________________ d = 2r = ___________________ Now we can experimentally find P to determine d, the diameter of the nucleus: P = ____________ = d= II. (FAILURE OF) THE CLASSICAL DESCRIPTION OF AN ATOM How is an e- bound to the nucleus? What is the force between them? Electromagnetic force. We will simplify this by calling it the Coulomb force. So the force between a positively-charge nucleus and a negatively-charged electron is Coulomb force, which can be quantified by Coulomb’s Force Law. For a hydrogen atom: F(r) = ____________ e = absolute value of an electron’s charge (note: nuclear charge = +e for a H atom) r = distance between two charges ε 0 = permittivity constant of a vacuum (8.854 x 10-12 C2J-1m-1) 5 electron -e Consider an H atom (Z=1) with 1 electron and 1 proton. When r → ∞ F(r) = ________ r As r → 0 F(r) = ________ +e H nucleus The closer the electron is to the nucleus, the larger the attractive force between the two charges. • The Coulomb force law tells us the force (F) as a function of r. • The Coulomb force law does not tell us how r changes with ________________. There is a CLASSICAL EQUATION OF MOTION that tells us how the electron and nucleus move under influence of this Coulomb force: Newton’s 2nd Law F = ma Force = mass times acceleration We can rewrite F as a function of velocity, F = m(_______) or distance, F = m(d2r/dt2) We can plug in the Coulomb force law for F, and solve the equation for a given initial value of r. If rinitial is 10 Å (10-10 m), a typical distance for an H atom, the calculation indicates that r = 0 at t = _______ sec! This predicts that the electron should plummet into the nucleus in ______________! What is wrong here? It turns out that the laws of classical mechanics no longer work at this size scale. Need a new kind of mechanics to describe this and other “unsettling” observations. QUANTUM MECHANICS provides a single and comprehensive theory that explains the behavior of matter on an atomic (nanometer or smaller) scale. 6