Question

Q4D.1 (Requires completion of chapter R9.) If we shine X-rays (which are very high-frequency electromagnetic waves) at a target, both the wave and photon models of electromagnetic radiation predict that the X-rays will be scattered in all directions by the target. The wave model predicts that the scattered X-rays will have the same wave- length as that of the incoming X-rays (since the incoming waves will wiggle the electron at the same frequency, so the scattered waves emitted by the electron in all direc- tions must have the same frequency as the bobbing elec- n that emits them). The photon model, on the other hand, predicts that the wavelength of the scattered X-rays will depend on the angle in a specific way. Our purpose here is to derive this angle dependence Let's define our coordinate system so that the initial X-ray photons move in the +x direction. Let the initial energy of the X-ray photons be Eo and the mass of an elec- tron be m. Imagine that a photon strikes the electron and scatters at an angle 0, while the electron scatters at an angle P, as shown below. BEFORE AFTER: photon,y energy Eo o photon energy E electron at rest scattered electron

Answer 1

As per the equation given in the equation of matrices, m+E, - E+ m2 + p.1) Photon incident is given as E = ln , and photon scattered is given as 'n'. According to the relativistic conservation of momentum as given, hv + m,c= hv' cp + mic) Rearrange the above expression, we get, hv + moc = "+(p: +mc") E and I' E E. + moc? = E +(p:+mc) According to the De Broglie Law, for photon, PE h_hc E + m = E+ Hp+m? Second row of matrices, E - Ecos + P.cos. Third row of matrices as per the question given, In y direction there is no momentum of incident photon so the equation becomes, 0- sin -, sino OF 0 - Esin +p, sino Momentum conservation states that as photon moves in x direction so projection of scattered photon and electron momentum will be equal to the incident photon momentum. pap+p. www.cose + P.lcos E, - Ecose + P. coso се

b). According to the condition given in part b of the question, E = E cose +|p.cos 0 Esine+p.sino p=p'+ p. P. = p-p squaring both sides P.1 = (p-p')(P-p') = p +p-2 pp'cos e p=E, p=E Ip.p = E + E2-2 EE cose Hence proved. According to the matrices given in the question, the m+E = E+ /m2 +\p?!) (m +Eo – E) * = (v/m2 +1930) = m+&+E+2mE, - 2mE - 2EE, -m? =\p./ = 2mE,+E+E-2E(m+ E) = p.1 Hence proved. From equation (1) as given in the question,

bv + m,c* = ''+ vcp2 +m c* P+E = p'c+VE +p?c? On Squaring both sides,we get E;+c(p-p")*+2CE (P-p') = E + pc? p = p + po-2 pp - 25.(P-') p? = p?+p" - 2 pp.cos e E (P-p) - pp' (1-cose) Multiply each term with hc / pp' E, and taking 2 = h p = h 1-2 = 42 se mac (1 cose)