Respuesta :
Answer:
decay constant is 1.290 × [tex]10^{-3}[/tex] s-1
Number is 0.870 × [tex]10^{8}[/tex] nuclei
activity is 1.1223 × [tex]10^{5}[/tex] Ci
Explanation:
Given data
isotope N1 = 6.030×10^8 nuclei
half life T/2 = 537 s
to find out
decay constant and How many nuclei remain and activity of the sample at the beginning
solution
we know that half life and decay constant relationship that is
N = N1 × [tex]e^{-kt}[/tex] ...............1
here k is decay constant and we know dN / dt is equal to -k × N
and T/2 = 0.693/k
so k = 0.693 / (T/12)
put here T/2
k = 0.693 / 537 = 1.290 × [tex]10^{-3}[/tex] s-1
so decay constant is 1.290 × [tex]10^{-3}[/tex] s-1
and
by equation 1 we get
N = N1 × [tex]e^{-kt}[/tex]
N = 6.030×10^8 × [tex]e^{-1.290 × [tex]10^{-3}[/tex] 1.50×10^3}[/tex]
N = 0.870 × [tex]10^{8}[/tex] nuclei
and
we know Activity of sample =k×N
so
activity = 0.870 × [tex]10^{8}[/tex] × 1.290 × [tex]10^{-3}[/tex]
activity = 1.1223 × [tex]10^{5}[/tex] Ci
