Particle pool

1 Particle pool Adolf Cusmariu [email protected] In a celebrated Effect [1], a never-at-rest photon bouncing at angle q off an - if ever - stationary single electron undergoes a wavelength increase unexpected from classical electrodynamics, the impact also recoiling the electron at angle f ; see graphic below (photon in red, electron in blue). The positive change in wavelength Dl is Dl = l(q) - l 0 = (h/mec)(2sin2( q/2 )) where q = Photon bounce angle h = Planck’s constant c = Speed of Light me = Electron rest-mass Compton with his famous formula. To recover the usual form for the Effect, from trigonometry cos(q) = cos(q/2 + q/2) = cos2(q/2) - sin2(q/2) hence 1 - cos(q) = 1 - cos2(q/2) + sin2(q/2) = 2sin2(q/2) which yields the justly famous expression q f l(q) l0 2 Compton used Molybdenum K a x-rays rays with incoming wavelength l0 = 0.0709 nm ≈ 17KeV. Now, three Effects compete for attention during X-ray photon/electron interaction - in energy as a function of emitter Atomic Number: The Photoelectric, Compton, and Electron/Positron Pair production. Two on-line sources display essentially the same regional dominance (see below). The energy region for the Effect overlaps the Photoelectric Effect region at Molybdenum’s Atomic Number of 42; small wonder it took Compton years of careful experiments to detect (only) his Effect using the chosen energy level. C.T.R. Wilson and his Cloud Chamber subsequently discovered the recoiled electrons (see below), an equally significant and predicted phenomenon, strongly emphasized by Compton. Mo(Z) = 42 Compton (17 KeV) Mo(Z) = 42 Compton (17 KeV) Dl = l(q) - l 0 = (h/mec)(1 - cos( q )) = l c(1 - cos( q )) l c = 0.002426 nm 3 Compton and Wilson shared the 1927 Nobel Prize, nominated by Einstein himself, among others. Einstein and Compton during an event at the University of Chicago in 1940. Two on-line sources for Compton’s experimental results are shown below; not exactly identical profiles! 4 Thus l0 = 0.0709 nm = 17.487 KeV l (0o) = l0 l (45o) = 0.0715 nm = 17.340 KeV l (90o) = 0.0731 nm = 16.961 KeV l (135o) = 0.0749 nm = 16.553 KeV A rewrite of Compton’s formula yields (in principle) the rest-mass of the electron: me = (h/c)(1 - cos( q ))/( l(q) - l0 ) an astonishing result on the face of it: electron rest-mass estimation from photon recoil angles! Set A( q ) = (1 - cos( q ))/( l(q) - l0 ) = 1/ l c = 412.201 5 A( q ) should be a constant independent of the photon recoil angle q and the wavelength l(q ); but how really so, for the (reportedly) measured (angle, wavelength) sets above?; we get A(45o) = 488.155 A(90o) = 454.545 A(135o) = 426.776 Not a particularly impressive agreement; why the difference? Either the (single?) recoiled electron wasn’t really stationary - contrary to the assumptions - or measured photon recoiling wavelengths, or angles, (or both), were a bit off as reported. The incoming x-ray wavelength of 0.0709 nm will be assumed accurate. The images (p.4 above) show (spline?) curve fits through measured values, although re-scaling to check the estimates is difficult as tickmark spacing isn’t obvious; as it happens, the wavelength measurement at 135o does allow scaling (see enlarged image below), also showing the sampling rate wasn’t uniform. From this image (p.4 above) the measured wavelength datapoint of 0.0749 nm appears exact, so it will be taken as accurate; the claim, however, of 135o as recoil angle is difficult to accept in vertical scale given the location of its neighbors. Still, the pairing (130o, 0.0749 nm) ⟹ A(130o) = 410.696 is very close indeed to the expected value of 412.201; an angular error of 5o is perhaps acceptable. Similarly, the pairing (85o, 0.0731 nm) ⟹ A(85o) = 414.929 is again close to the ideal, showing a numerically similar angular error. Finally, the pairing (41o, 0.0715 nm) ⟹ A(41o) = 408.817 also fits. More exact pairings were estimated as: .0700 .0720 .0740 .0760 nm 6 (41.177, 0.0715 nm) ⟹ A(41.177o) = 412.201 (84.6547o, 0.0731 nm) ⟹ A(84.6547o) = 412.201 (130.415o, 0.0749 nm) ⟹ A(130.415o) = 412.201 What if instead the recoil wavelength l(q) measurements were a bit off? The pairings below work now. (45o, 0.07161056 nm) ⟹ A = 412.201 (90o, 0.073326 nm) ⟹ A = 412.210 (135o, 0.07504144 nm) ⟹ A = 412.201 By contrast, only (laboriously) minute changes in the scattering wavelengths - while keeping the angular measurements as reported - brought agreement; perhaps the Bragg spectrometer was a bit off. The bounce angle f of the recoiling electron satisfies [2] cot( f ) = (1 + l c/ l0 )tan( q /2) Thus f = arccot{(1 + l c/ l0 )tan( q /2)} Note the recoiling electron’s explicit angular independence of the recoiling photon’s wavelength l(q) ; after all, the electron is a true particle in this setting. Angularly, though, the two particles do seem to ‘know’ each other! In fact, some algebra will wrap it all up neatly in one package; first, from the above expression cot( f )/tan( q /2) = (1 + l c/ l0 ) while from Compton’s formula l(q) / l0 = 1 + ( l c/ l0 )(1 - cos( q )) = 1 + ( l c/ l1 ) - ( l c/ l1 )cos( q ) hence l(q) / l0 + ( l c/ l0 )cos( q ) = 1 + ( l c/ l0 ) Altogether then To verify experimentally this particular form of the Effect, angular data on the recoiling electron would also have been needed! The ‘corrected’ photon recoil angles of {41.177o, 84.6547o, 130.415o} would correspond to electron recoils of {68.0861o, 45.7141o, 23.3317o}. l(f,q) = lo cot( f )/tan( q /2) - l ccos( q ) 7 For 0 < q < p /2 these two recoil angles are plotted below; (photon in red, electron in blue) They actually meet around 600 (58.86080 more precisely) where the photon wavelength is ≈#0.07207 nm. Compton playing banjo to students at Washington University in St. Louis, 1949. References [1] A. H. Compton, ‘A Quantum Theory of the Scattering of X-Rays by Light Elements’, Phys. Rev, 21, 1923. [2] Wikipedia article on the Compton Effect.