Respuesta :
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Answer: 90.2 days
Explanation:-
Radioactive decay follows first order kinetics.
Half-life of Iron-59 = 45.1 days
[tex]\lambda =\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{45.1}= 0.0153 days^{-1}[/tex]
[tex]N=N_o\times e^{-\lambda t}[/tex]
N = amount left after time t= [tex]\frac{25}{100}\times a=0.25a[/tex]
[tex]N_0[/tex] = initial amount = a
[tex]\lambda[/tex] = rate constant= [tex]0.015 days^{-1}[/tex]
t= time = ?
[tex]0.25a=a\times e^{- 0.0153 years^{-1}\times t days}[/tex]
[tex]t=90.2days[/tex]
Thus the iron sample is 90.2 years old.
