Complete Question
The complete question is shown on the first uploaded image
Answer:
percent of Mn in the sample is 1.12%
Explanation:
The formula to obtain the number of manganese N is
[tex]N =\frac{A}{\phi V \sigma_a[1-e^{(-\lambda t)}]} ----(1)[/tex]
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
[tex]\lambda[/tex]The yield an activity (rate constant)
[tex]The\ yield\ an\ activity\ (rate \ constant)\ = \frac{0.693}{t_{(1/2)}}----(2)[/tex]
From the question we are given that half live [tex]t_{(1/2)}[/tex] is 2.58 h = 9288 sec
Hence
[tex]\lambda = \frac{0.693}{9288} =7.46*10^{-5}[/tex]
The volume is V = Area × Thickness
[tex]Thickness = 2/10 =0.2cm[/tex]
[tex]V = 0.2cm^3[/tex]
[tex]Substituting \ 3*10^{12} \ for \ flux \ , 9288 sec \ for \ time \ 13.3*10^{-24} \ for \\\sigma _a \ 0.2cm^2 \ for \ V \ 150*10^{-3} \ for \ initial \ activity \ and \ 7.46*10^{-5} \ for \\\ \lambda \ into \ equation \ one[/tex]
[tex]N = \frac{150*10^{-3} * 3.7*10^{10}}{(3*10^{12}*0.2 *13.3 *10^{-24}(1-e^{(7.46^{-5})(7200)}))}[/tex]
[tex]N = 9.789*10^{20}C/m^3[/tex]
The formula for percent of manganese in the sample
Let it be denoted by W
[tex]W = \frac{N}{atom \ number \ density} ----(3)[/tex]
Substituting [tex]N = 9.789*10^{20}C/m^3[/tex] for N and [tex]0.087 *10^{24}[/tex] for atom number density , in equation 3
[tex]W =\frac{9.789*10^{20}}{0.087*10^{24}} *\frac{100}{1}[/tex]
= 1.12%
