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Physics 102: Lecture 24 Heisenberg Uncertainty Principle & Bohr Model of Atom Physics 102: Lecture 24, Slide 1 Heisenberg Uncertainty Principle Recall: Quantum Mechanics tells us outcomes of individual measurements are uncertain ΔpyΔy ≥ Uncertainty in momentum t ((along l y)) h 2 Uncertainty in position iti ((along l y)) Rough idea: if we know momentum very precisely, we lose knowledge of location, and vice versa. Thiss “uncertainty” u ce a y iss fundamental: u da e a : it arises a ses because quantum particles behave like waves! Physics 102: Lecture 24, Slide 2 Electron diffraction El t Electron beam b traveling t li through th h slit lit will ill diffract diff t Single slit diffraction pattern Number of electrons arriving at screen w θ θ electron beam Δpy = p sinθ y x screen Recall single-slit diffraction 1st minimum: sinθ = λ/w w = λ/sinθ = Δy Δp y Δy = p sin θ Physics 102: Lecture 24, Slide 3 λ sin θ = λp = h Using de Broglie λ = h/p Number of electrons arriving at screen py w electron beam Δp y ⋅ w = h y x screen Electron entered slit with momentum along g x direction and no momentum in the y direction. When it is diffracted it acquires a py which can be as big as h/w. The “Uncertainty in py” is Δpy ≈ h/w. An electron passed through the slit somewhere along the y direction The “Uncertainty direction. Uncertainty in yy” is Δy ≈ w. w Physics 102: Lecture 24, Slide 4 Δp y ⋅ Δy ≈ h Number of electrons arriving at screen py w electron beam Δp y ⋅ Δy ≈ h y x screen If we make the slit narrower (decrease w =Δy) the diffraction peak gets broader (Δpy increases). “If we know location very precisely, we lose knowledge of momentum, and vice versa.” Physics 102: Lecture 24, Slide 5 to be precise... Δp y Δy ≥ h 2 Of course if we try to locate the position of the particle alongg the x axis to Δx we will not know its x component p of momentum better than Δpx, where h Δpx Δx ≥ 2 and the same for z. Checkpoint 1 According to the H.U.P., if we know the x-position of a particle, we cannot know its: (1) y-position y position ((2)) x-momentum x momentum (3) Physics 102: Lecture 24, Slide 6 y-momentum (4) Energy Atoms • • • • Evidence for the nuclear atom Today Bohr model of the atom Spectroscopy of atoms Next lecture Q Quantum t atom t Physics 102: Lecture 24, Slide 7 Early Model for Atom • Plum Pudding – negative charges (electrons) in uniformly distributed cloud l d off positive iti charge h lik like plums l in i pudding ddi + + + - + + + + 10 m across? But how can you look inside an atom 10-10 Light (visible) λ = 10-7 m Electron (1 eV) λ = 10-9 m Helium atom λ = 10-11 m Physics 102: Lecture 24, Slide 8 Rutherford Scattering 1911: Scattering He++ (an “alpha particle”) atoms off of gold. Mostly go through, some scattered back! + Plum pudding theory: – electrons in “cloud” of uniformly if l distributed di t ib t d + charge h Æ electric field felt by alpha never gets too large To scatter at large angles, need positive charge concentrated in small region (the nucleus) + - + + + + + + Atom is mostly empty space with a small (r = 10-15 m) positively charged nucleus surrounded by “cloud” of electrons (r = 10-10 m) Physics 102: Lecture 24, Slide 9 Nuclear Atom (Rutherford) Large angle scattering Nuclear atom Classic nuclear atom is not stable! Electrons orbit nucleus -> constant centripetal acceleration Accelerated charges radiate energy Electrons will radiate and spiral into nucleus! Early “quantum” model: Bohr Physics 102: Lecture 24, Slide 10 Need quantum theory Bohr Model is Science fiction The Bohr model is complete nonsense. Electrons do not circle the nucleus in little planetlike orbits. The assumptions injected into the Bohr model have no basis in physical reality reality. BUT the model does get some of the numbers right for SIMPLE atoms… atoms Physics 102: Lecture 24, Slide 11 Hydrogen-Like Atoms single electron with charge -e nucleus with charge +Ze Ze (Z protons) 19 C e=1 1.6 6 x 10-19 E H (Z Ex: (Z=1), 1) He H (Z=2), (Z 2) Li (Z=3), (Z 3) etc t Physics 102: Lecture 24, Slide 12 The Bohr Model Electrons circle the nucleus in orbits Only certain orbits are allowed 2πr = nλ n=1 -e e +Ze Physics 102: Lecture 24, Slide 13 The Bohr Model Electrons circle the nucleus in orbits Only certain orbits are allowed 2πr = nλ = nh/p L = ppr = nh/2π = nħ n=2 -e e +Ze Angular momentum is quantized Physics 102: Lecture 24, Slide 14 v is also quantized in the Bohr model! An analogy: Particle in Hole • Let’s define energy so that a free particle has E=0 • Consider a particle trapped in a hole • To free the particle, need to provide energy mgh • Relative to the surface, energy = -mgh mgh – a particle that is “just free” has 0 energy E 0 E=0 h E=-mgh Physics 102: Lecture 24, Slide 15 An analogy: Particle in Hole • Quantized Energy: only fixed discrete heights of particle allowed • Lowest energy (deepest hole) state is called the h ““groundd state”” E 0 E=0 h Physics 102: Lecture 24, Slide 16 “ground state” For Hydrogen-like atoms: Energy levels (relative to a “just free” E=0 electron): mk 2 e 4 Z 2 13.6 ⋅ Z 2 En = − ≈− eV ( where h ≡ h / 2π ) 2 2 2 2h n n Radius of orbit: 2 2 n2 ⎛ h ⎞ 1 n rn = ⎜ = ( 0.0529 0 0529 nm ) ⎟ 2 Z ⎝ 2π ⎠ mke Z Physics 102: Lecture 24, Slide 17 Checkpoint 2 h 2 1 n2 n2 rn = ( ) = (0.0529nm) 2 2π mke k Z Z Bohr radius If the electron in the hydrogen atom was 207 times heavier (a muon), the Bohr radius would be 1) 207 Times Larger h 2 1 Bohr Radius = ( ) 2 π 2 mke 2)) Same Size 3) 207 Times Smaller This “m” m is electron mass! Physics 102: Lecture 24, Slide 18 ACT/Checkpoint 3 A single electron is orbiting around a nucleus with charge +3. What is its ground state (n=1) energy? (Recall for charge +1, E= -13.6 eV) 1) 2) 3) E = 9 ((-13 13.6 6 eV) E = 3 (-13.6 eV) E = 1 (-13.6 ( 13 6 eV) V) 32/1 = 9 Z2 E n = −13.6eV 2 n Note: This is LOWER energy since negative! Physics 102: Lecture 24, Slide 19 ACT: What about the radius? Z=3, n=1 1. larger than H atom 2 same as H atom 2. 3. smaller than H atom h 2 1 n2 n2 rn = ( ) = (0.0529nm) 2 2π mke Z Z Physics 102: Lecture 24, Slide 20 Summary • Bohr’s Model gives accurate values for electron energy levels... levels • But Quantum Mechanics is needed to describe electrons in atom. atom • Next time: electrons jump between states by emitting or absorbing photons of the appropriate energy. Physics 102: Lecture 24, Slide 21 Some (more) numerology • 1 eV = kinetic energy of an electron that has been accelerated through a potential difference of 1 V 1 eV = qΔV = 1.6 x 10-19 J • h (Planck’s constant) = 6.63 x 10-34 J·s hc = 1240 eV·nm 31 kg • m = mass of electron = 9.1 x 10-31 mc2 = 511,000 eV • U = kke2/r, / so ke k 2 has h units i eV·nm V (like (lik hc) h ) 2πke2/(hc) = 1/137 (dimensionless) “fi structure “fine t t constant” t t” Physics 102: Lecture 24, Slide 22