Question

Question 3.1 The differential cross-section of the coherent scattering intensity for a free electron is expressed by the following equation with respect to per unit plane angle (radian), if the classical electron radius is set to r. os20) 2n sin (m²/rad) (1) Estimate me which is called the Thomson classical scattering coefficient by the integration about angle. (2) Express the Thomson classical scattering coefficient in barn unit. (1barn = 1 x 10-28 m²). (3) When X-rays penetrate an aluminum foil of thickness 1 mm, calculate the probability of the coherent scattering induced from free electrons. The density of aluminum is 2.70 x 106 g/m?.

Answer 1

Here, I have used simple calculations based on integration and used the concepts of coherent scattering and scattering cross section. Thus, I have calculated the required results.
Question 3.1:- The 323) enental scattering cross section is given as doe – The -eos-0). 27 ring (m²/rad) do Integrating we get Pe [ 2a vinø dp + S 20 cost o min do] = 772 Srino do + "coso sin odo] = tre [-consel het, eas & 57 rinodo=-dz = The [-(Cost-coso) [lor and port] + +(33)'] = 7[-+-+- = Tre [-(-1-1] = 277 -- 2* * (282410) =-49196610-30 Thomson lamiel sentering en E = 2*1 - (Amis 01) Now, classical elechon nadies r = 2-82 x 10 ton The sadene of the co-eflictent will be = 27x (2.82 x 1016 - 49.966 x 103 m = 0.49366 x 1028 m² 0.49966 baron [I baren= 1x102 m

X-rays penetrate the Aluminiuon Joit af thickness, t = 1 mm = 103 m density of aluminium, P = 2.70 x 100 g/m = 2.70x10 Kg/m Atomie mass of aluminium. IMAL = 26.98 u = 26.98* 1.668 10 kg. Probability of coherent scattering induced et e - 2.70X10° x 10 x 0. 1996681028 26.98×1-668 10-27 0.0030 1223 (Anis) (3)