P301 Lecture 9 “X-Rays” Experimental results on the absorption of x- rays from C. G. Barkla Phil. Mag. 17 (1909)

P301 Lecture 9 “X-Rays” K  and K  lines from various elements (Mosely)

P301 Lecture 10 “Compton scattering”  = ’  = C [1-cos(  )] where C = h/mc

A. H. Compton Phys. Rev. 21 p483 (1923) Phys. Rev. 22 p409 (1923) P301 Lecture 10 “Compton scattering”

P301 Lecture 9 “JITT question” a.The quantity h/mc as the dimensions of length, and it is called the “Compton Wavelength”. What is the value of the Compton Wavelength for an electron? b.Use this result, and the data in figure 3.21 to estimate the wavelength of a Mo K-α photon. Virtually everyone who answered knew that C =2.43 pm (good, you read the book) (30 answered, 10 did not). The second part caused much confusion: A number noted that this is the wavelength Compton started off with, and this is given in the text as 0.071nm (5). (correct, but it doesn’t make use of the data) Many tried to use equation 3.42, but came up with a somewhat random collection of values (10) (ranging from 16.8fm to 10.9 nm. Most were just confused by this question (14)

P301 Lecture 9 “Compton Scattering question” Use the result from a, and the data in figure 3.21 to estimate the wavelength of a Mo K  ; photon

P301 Lecture 10 “JITT question” a.Explain briefly why there must be two photons emitted (back-to-back) when a particle and its anti-particle annihilate at rest? No answer: 19 5 said to conserve energy (true, but this could be done with only 1 photon) 16 said to conserve energy and momentum (that is the key) b.What do you suppose would happen if the pair annihilated while their center of mass was moving at high speed (say 0.8c) w.r.t. the observer? Energy would be Doppler shift up (3) (and different for the two photons, 5) Angle between would not be 180 o (3) Confused or confusing answers: 10 In fact, the energies of the two would be different (just how different will depend on the angle between the “boost” and the CoM frame direction of the photons), the angle between them will not be 180 degrees (unless the boost is directly lined up with the decay line), and at least one of the photons (and most likely both) will have energies greater than 511keV

P301 Lecture 9 “Discovery of the position” Fig 1 from C. Anderson “The Positive Electron” Phys. Rev. 43 p491 (1933) Note how the curvature increases as the particle passes through the lead plate in the cloud chamber, this provides conclusive evidence for the direction of travel (upward in the figure). By looking at the ionization, and the reduction in momentum due to traverse of the lead, Anderson was able to put limits on the charge and mass of the particle (q<2e m<20 m e ) Track of interest Lead plate 6 mm

P301 Lecture 11 “JITT question” Explain briefly why the planetary model of the atom was rejected by Thomson prior to the results of the Geiger and Marsden experiment. Negative charges repel rather than attract (like gravity) (13 responses) This is a concern, and represents a complicated problem (the “many-body” problem), but it only applies to multi- electron atoms and is not fundamental. Accelerating charges emit radiation, hence the electrons would quickly crash into any central mass (6 responses) This applies even to hydrogen, and requires completely new physics to avoid. Other (4 responses) None: 19

P301 Lecture 11 “Geiger-Marsden experiment” Ra source (  emitter) From Geiger and Marsden PRSL A (1909)

P301 Lecture 11 “Scattering experiments” b = [Z 1 Z 2 e 2 cot(  /2)]/[8  o K]=(4.6) in T&R, [this form holds only for Coulomb forces] NOTES: The scattering angle is related to the “momentum transfer”  p Smaller impact parameter -> larger angle If the ‘target’ is small, then most projectiles’ will not get scattered. Typically we know how many particles are incident on the sample per unit area and time, multiplying this by a “cross-section” gives the rate of scattering.

P301 Lecture 12 “Bohr’s Hypotheses” Bohr formulated the following ad hoc model: 1.Atoms exist only in certain stable “stationary states” 2.The dynamic equilibrium of these stationary states is determined by the laws of classical physics, but the way the atoms interact with the electromagnetic field of light waves is not 3.The emission and absorption of EM waves by atoms takes place ONLY in conjunction with a transition between two stationary states, with the frequency of the emitted light being determined according to the Planck hypothesis: |E 1 – E 2 |= hf 4.The (orbital) angular momentum of the electron in a stationary state can only take on values given by integral multiples of Planck’s constant divided by 2  L n = nh/2  It is this last hypothesis that is the truly new (revolutionary) idea from Bohr himself, the other three are pretty much inescapable and/or had been provided by someone else already.