A beam of monochromatic x-rays with a wavelength of 0.490 nm are incident on the

Question

A beam of monochromatic x-rays with a wavelength of 0.490 nm are incident on the surf arc of a crystal The atoms of the crystal arc arranged in a simple cubic structure wish a spacing of 0.375 nm between atoms. Determine the angle of incidence relative to the surface for which Bragg scattering will occur for the set of crystal planes oriented at 45 degree to the surface as shown in the figure below. Alpha particles with an energy of 5.6 MeV arc incident on a copper foil 2.5 * 104 cm thick. The density of copper is 8.96 g/cm1 What is the impact parameter for scattering at 60 degree? What fraction of the protons scatter at an angle grater than 60 degree?

Explanation / Answer

Given: Monochromatic x- ray wavelength = 0.49nm = 0.49 x10-9 m Inter atomic distance of the atoms d = 0.375 nm = 0.375 x 10-9 m In order to occur the bragg's scatering 2d sin = n for forst order n = 1 2(0.375x10-9) sin = (1)(0.49 x10-9)                           = 40.79o since , set of crystal planes oriented through 45o with respect to the surface of the crystal thus, The angle of incidence should be i = 45- 40.79 = 4.21o plz .., remaing post seperetely.., Monochromatic x- ray wavelength = 0.49nm = 0.49 x10-9 m Inter atomic distance of the atoms d = 0.375 nm = 0.375 x 10-9 m In order to occur the bragg's scatering 2d sin = n for forst order n = 1 2(0.375x10-9) sin = (1)(0.49 x10-9)                           = 40.79o since , set of crystal planes oriented through 45o with respect to the surface of the crystal thus, The angle of incidence should be i = 45- 40.79 = 4.21o plz .., remaing post seperetely..,

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