The manganese content of a certain stainless steel is to be verified by an activation measurement. The activity induced in a sample of volume V by neutron capture during a time t is given by (4.57) A foil of area 1 cm2 and thickness 2 mm is irradiated in a thermal neutron flux of 3 × 1012/(cm2 s) for 2 h. Counts taken immediately afterward yield an activity of 150 mCi for the induced Mn-56, half-life 2.58 h. Assuming that the atom number density of the alloy is 0.087 × 1024/cm3 and that the cross-section for capture in Mn-55 is 13.3 barns, find the percent of Mn in the sample. 4.18 For fast neutrons in uranium-

Complete Question

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percent of Mn in the sample is 1.12%

Explanation:

     The formula to obtain the number of manganese N is

 Here

    φ is the flux

    V is the volume

    t is the time

    A is the initial activity of Manganese

    N is the number of manganese

   The yield an activity  (rate  constant)

 From the question we are given that half live is 2.58 h = 9288 sec

 Hence

     The volume is V = Area × Thickness

  The formula for percent of manganese in the sample

            Let it be denoted by W

  Substituting   for N and for atom number density , in equation 3

              =  1.12%

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a ball spinning on a finger   >   kinetic energy > moving energy

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a   ball flying through the air   > kinetic energy   >   moving energy

hope that                                         : )