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%

spectral lines lyman is one set

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic =   - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day