{"id":13493,"date":"2017-01-02T10:14:14","date_gmt":"2017-01-02T10:14:14","guid":{"rendered":"http:\/\/sitepourvtc.com\/?page_id=13493"},"modified":"2022-10-19T16:16:36","modified_gmt":"2022-10-19T16:16:36","slug":"macroscopic-cross-section","status":"publish","type":"page","link":"https:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-engineering-fundamentals\/neutron-nuclear-reactions\/macroscopic-cross-section\/","title":{"rendered":"Macroscopic Cross-section"},"content":{"rendered":"
The difference between the microscopic cross-section<\/strong> and macroscopic cross-section<\/strong>\u00a0is very important and is restated for clarity. The microscopic cross-section<\/strong> represents the effective target area of a single target nucleus<\/strong> for an incident particle. The units are given in barns or cm2<\/sup><\/strong>.<\/p>\n While the macroscopic cross-section<\/strong> represents the effective target area of all of the nuclei<\/strong> contained in the volume of the material, the units are given in cm-1<\/sup><\/strong>.<\/p>\n A macroscopic cross-section is derived from microscopic cross-section<\/strong> and the atomic number density<\/strong><\/a>:<\/p>\n \u03a3=\u03c3.N<\/strong><\/p>\n Here \u03c3<\/strong>, which has units of m2<\/sup>, is the microscopic cross-section. Since the units of N (nuclei density) are nuclei\/m3<\/sup>, the macroscopic cross-section \u03a3 has units of m-1<\/sup>. Thus, it is an incorrect name because it is not a correct unit of cross-sections. In terms of \u03a3t<\/sub> (the total cross-section), the equation for the intensity of a neutron beam can be written as<\/p>\n -dI = N.\u03c3.\u03a3t<\/sub>.dx<\/strong><\/p>\n Dividing this expression by I(x) gives<\/p>\n -d\u0399(x)\/I(x) = \u03a3t<\/sub>.dx<\/strong><\/p>\n Since dI(x) is the number of neutrons that collide in dx, the quantity –d\u0399(x)\/I(x)<\/strong> represents the probability that a neutron that has survived without colliding until x will collide in the next layer dx. It follows that the probability P(x) that a neutron will travel a distance x without any interaction in the material, which \u03a3t characterizes, is:<\/p>\n P(x) = e-\u03a3t<\/sub>.x<\/sup><\/strong><\/p>\n We can derive the probability that a neutron will make its first collision in dx from this equation<\/strong>. It will be the quantity P(x)dx<\/strong>. Suppose the probability of the first collision in dx is independent of its history. In that case, the required result will be equal to the probability that a neutron survives up to layer x without any interaction (~\u03a3t<\/sub>dx) times the probability that the neutron will interact in the additional layer dx (i.e., ~e-\u03a3t<\/sub>.x<\/sup>).<\/p>\n P(x)dx = \u03a3t<\/sub>dx . e-\u03a3t<\/sub>.x<\/sup> = \u03a3t<\/sub> e-\u03a3t<\/sub>.x<\/sup> dx<\/strong><\/p>\n From the equation for the probability of the first collision in dx,<\/strong> we can calculate the mean free path<\/strong> traveled by a neutron between two collisions. The symbol \u03bb usually designates this quantity. It<\/b>\u00a0is equal to the average value of x, the distance traveled by a neutron without any interaction, over the interaction probability distribution.<\/p>\n <\/a><\/p>\n whereby one can distinguish \u03bbs<\/sub>, \u03bba<\/sub>, \u03bbf<\/sub><\/strong>, etc. This quantity is also known as the relaxation length<\/strong>\u00a0because it is the distance in which the intensity of the neutrons that have not caused a reaction has decreased with a factor e.<\/p>\n For materials with a high absorption cross-section<\/a>, the mean free path is very short,<\/strong> and neutron absorption occurs mostly on the material\u2019s surface<\/strong>. This surface absorption is called self-shielding<\/strong> because the outer layers of atoms shield the inner layers.<\/p>\n Most materials are composed of several chemical elements and compounds. Most chemical elements contain several isotopes<\/strong> of these elements (e.g.,, gadolinium<\/a> with its six stable isotopes). For this reason, most materials involve many cross-sections. Therefore, to include all the isotopes within a given material, it is necessary to determine the macroscopic cross-section for each isotope and then sum all the individual macroscopic cross-sections.<\/p>\n In this section, both factors (different atomic densities<\/strong> and cross-sections<\/strong>) will be considered in calculating the macroscopic cross-section of mixtures<\/strong>.<\/p>\n First, consider Avogadro\u2019s number N0<\/sub> = 6.022 x 1023<\/sup><\/strong>, which is the number of particles (molecules, atoms) contained in the amount of substance given by one mole. Thus if M is the molecular weight<\/strong>, the ratio N0<\/sub>\/M<\/strong> equals the number of molecules in 1g of the\u00a0mixture. The number of molecules per cm3<\/sup> in the material of density \u03c1 and the macroscopic cross-section for mixtures is given by the following equations:<\/p>\n Ni<\/sub> = \u03c1i<\/sub>.N0<\/sub> \/ Mi<\/sub><\/strong><\/p>\nMean Free Path<\/h2>\n
Macroscopic Cross-section of Mixtures and Molecules<\/h2>\n