physics 2d lecture slides lecture 14 feb 3 rd 2004
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Physics 2D Lecture Slides Lecture 14: Feb 3 rd 2004 Vivek Sharma - PDF document

Brian Wecht, the TA is back! Pl. give all regrade requests to him Quiz 4 is This Friday Physics 2D Lecture Slides Lecture 14: Feb 3 rd 2004 Vivek Sharma UCSD Physics Where are the electrons inside the atom? Early Thought: Plum pudding


  1. Brian Wecht, the TA is back! Pl. give all regrade requests to him Quiz 4 is This Friday Physics 2D Lecture Slides Lecture 14: Feb 3 rd 2004 Vivek Sharma UCSD Physics

  2. Where are the electrons inside the atom? Early Thought: “Plum pudding” model � Atom has a homogenous distribution of Positive charge with electrons embedded in them (atom is neutral) Positively charged e - matter e - e - e - + Core e - e - e - + e - or e - e - e - e - e - e - e - e - e - ? e - e - • How to test these hypotheses? � Shoot “bullets” at the atom and watch their trajectory. What Kind of bullets ? •Indestructible charged bullets � Ionized He ++ atom = α ++ particles •Q = +2e , Mass M α =4amu >> m e , V α = 2 x 10 7 m/s (non-relavistic) [charged to probe charge & mass distribution inside atom] Plum Pudding Model of Atom • Non-relativistic mechanics (V α /c = 0.1) • In Plum-pudding model, α -rays hardly scatter because – Positive charge distributed over size of atom (10 -10 m) – M α >> M e (like moving truck hits a bicycle) – � predict α -rays will pass thru array of atoms with little scatter (~1 o ) Need to test this hypothesis � Ernest Rutherford

  3. Probing Within an Atom with α Particles Most α particles pass thru gold foil with nary a deflection • SOME ( ≅ 10 -4 ) scatter at LARGE angles Φ • Even fewer scatter almost backwards � Why • “ Rutherford Scattering” discovered by his PhD Student (Marsden)

  4. Rutherford Discovers Nucleus (Nobel Prize) Force on α -particle due to heavy Nucleus α particle trajectory is hyperbolic Scattering angle is related to impact par. •Outside radius r =R, F ∝ Q/r 2 ⎛ ⎞⎛ θ ⎞ kq Q = ⎜ α ⎟⎜ ⎟ •Inside radius r < R, F ∝ q/r 2 = Qr/R 2 Impact Parameter b cot 2 2 ⎝ ⎠ ⎝ ⎠ m v α α •Maximum force at radius r = R

  5. Rutherford Scattering: Prediction and Experimental Result 2 2 4 k Z e NnA ∆ = n 2 ⎛ ⎞ 1 ϕ 2 2 4 ⎜ ⎟ 4 R m v Sin ( / 2) α α ⎝ ⎠ 2 •# scattered Vs φ depends on : •n = # of incident alpha particles •N = # of nuclei/area of foil •Ze = Nuclear charge • K α of incident alpha beam •A= detector area Rutherford Scattering & Size of Nucleus ∝ distance of closest appoach r size of nucleus 1 α 2 Kinetic energy of = K = 2 m v α α β α particle will penetrate thru a radius r nucleus until all its kinetic energy is used up to do work AGAINST the Coulomb potent ial of the Nucleus: ( )( ) Ze 2 e 1 = = 2 K = m v 8 MeV k α α β 2 r 2 2 kZe ⇒ = r K α = For K =7.7.MeV, Z 13 α Al 2 2 kZ e ⇒ = = × − 15 r 4.9 10 m K α nucleus - 15 Size of Nucleus = 10 m -10 Siz e of Ato m = 1 0 m

  6. Dimension Matters ! -15 Size of Nucleus = 10 m -10 Size of Atom = 10 m •how are the electrons located inside an atom •How are they held in a stable fashion •necessary condition for us to exist ! •All these discoveries will require new experiments and observations Rutherford Atom & Classical Physics

  7. Continuous & Discrete spectra of Elements Visible Spectrum of Sun Through a Prism

  8. Emission & Absorption Line Spectra of Elements Kirchhoff’ Experiment : “D” Lines in Na D lines darken noticeably when Sodium vapor introduced Between slit and prism

  9. Emission & Absorption Line Spectrum of Elements •Emission line appear dark because of photographic exposure Absorption spectrum of Na While light passed thru Na vapor is absorbed at specific λ The Rapidly Vanishing Atom: A Classical Disaster ! Not too hard to draw analogy with dynamics under another Central Force Think of the Gravitational Force between two objects and their circular orbits. Perhaps the electron rotates around the Nucleus and is bound by their electrical charge M M Q Q ⇒ 1 2 1 2 F= G k 2 2 r r Laws of E&M destroy this equivalent picture : Why ?

  10. Bohr’s Bold Model of Atom: Semi Quantum/Classical -e 1. Electron in circular orbit m e around proton with vel=v 2. Only stationary orbits allowed . Electron does not F V radiate when in these stable +e (stationary) orbits 3. Orbits quantized: r M e v r = n h/2 π (n=1,2,3…) – 4. Radiation emitted when electron “jumps” from a stable orbit of higher energy 2 e � stable orbit of lower = − U r ( ) k r energy E f -E i = hf =hc/ λ 5. Energy change quantized 1 = 2 KE m v • f = frequency of radiation e 2 Reduced Mass of 2-body system -e m e General Two body Motion under a central force F V +e reduces to r m e Both Nucleus & e - revolve around their common center of mass (CM) • Such a system is equivalent to single particle of “reduced mass” µ that • revolves around position of Nucleus at a distance of (e - -N) separation ฀ µ= ( m e M)/(m e +M), when M>>m, µ= m (Hydrogen atom) ฀ Ν ot so when calculating Muon (m µ = 207 m e ) or equal mass charges rotating around each other (similar to what you saw in gravitation)

  11. Allowed Energy Levels & Orbit Radii in Bohr Model 2 Radius of Electron Orbit : 1 e − 2 E=KE+U = m v k r = e � 2 mvr n Force Equality for Stable Orbit � n ⇒ = v , ⇒ Coulomb attraction = CP Force mr 2 2 2 1 ke e m v = = 2 substitute in KE= 2 e m v k r e 2 2 r r 2 � 2 n 2 2 m v e ⇒ = = ∞ ⇒ = = r , n 1 ,2 ,.... e K E k n 2 mk e 2 2 r = ⇒ n 1 B ohr Radius a 2 e 0 E = KE+U= - 2 k r Total En erg y � 2 2 1 = = × − 10 a 0.529 10 m 0 2 ⇒ mk e Negative E Bound sy stem = = ∞ 2 In ge neral r n a n ; 1 ,2,... . Thi s much energy must be added to n 0 Quantized orbits of rotat io n the system to break up the bound atom Energy Level Diagram and Atomic Transitions − 2 ke = + = E K U n 2 r = 2 since r a n , n =quantum number n 0 − 2 ke 13.6 = = − = ∞ E eV , n 1, 2, 3.. n 2 2 2 a n n 0 → Interstate transition: n n i f ∆ = = − E h f E E i f ⎛ ⎞ − 2 ke 1 1 = − ⎜ ⎟ ⎜ ⎟ 2 2 2 a n n ⎝ ⎠ 0 i f ⎛ ⎞ 2 ke 1 1 = − ⎜ ⎟ f ⎜ ⎟ 2 2 2 ha n n ⎝ ⎠ 0 f i ⎛ ⎞ 2 1 f ke 1 1 = = − ⎜ ⎟ ⎜ ⎟ λ 2 2 c 2 hca n n ⎝ ⎠ 0 f i ⎛ ⎞ 1 1 = R − ⎜ ⎟ ⎜ ⎟ 2 2 ⎝ n n ⎠ f i

  12. Hydrogen Spectrum: as explained by Bohr ⎛ ⎞ 2 2 ke Z = −⎜ ⎟ E n 2 ⎝ ⎠ 2 a n 0 Bohr’s “R” same as the Rydberg Constant R derived emperically from photographs of the spectral series Another Look at the Energy levels ⎛ ⎞ 2 2 ke Z = −⎜ ⎟ E n 2 ⎝ ⎠ 2 a n 0

  13. Bohr’s Atom: Emission & Absorption Spectra photon photon Some Notes About Bohr Like Atoms • Ground state of Hydrogen atom (n=1) E 0 = -13.6 eV • Method for calculating energy levels etc applies to all Hydrogen- like atoms � -1e around +Ze – Examples : He + , Li ++ • Energy levels would be different if replace electron with Muons • Bohr’s method can be applied in general to all systems under a central force (e.g. gravitational instead of Coulombic) Q Q M M = → 1 2 1 2 If change U r ( ) k G r r Changes every thing: E, r , f etc "Importance of constants in your life"

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