Answer:
Rate constant: [tex]\lambda= 3.65 \cdot 10^{-04} d^{-1} [/tex]
Time to deacy to 1/8 of its original cocentration: [tex]t = 5697.1 d = 15.6 years[/tex]
Explanation:
To find the rate constant for the decay process, we use the next equation:
[tex]t_{\frac{1}{2}}=\frac{Ln(2)}{\lambda}[/tex]
where λ: constant for the decay process, and [tex]t_{\frac{1}{2}}[/tex]: half-life for the decay
[tex]\lambda=\frac{Ln(2)}{t_{\frac{1}{2}}} [/tex]
[tex]\lambda=\frac{Ln(2)}{1.9\cdot 10^{3}} [/tex]
[tex]\lambda= 3.65 \cdot 10^{-04} d^{-1} [/tex]
Now, to calculate the time for a sample of 60Co to decay to 1/8th of its original concentration, we use the exponential decay equation:
[tex]N_{(t)}=N_{0} \cdot e^{-\lambda \cdot t}[/tex]
where [tex]N_{0}[/tex] : initial quantity of the 60Co and [tex]N_{(t)}[/tex] : is the quantity has not yet decayed after a time t.
[tex]\frac{N_{0}}{8}=N_{0} \cdot e^{-\lambda \cdot t}[/tex]
[tex]Ln(\frac{1}{8})=-\lambda \cdot t[/tex]
[tex]t = -\frac{Ln(\frac{1}{8})}{\lambda} [/tex]
[tex]t = -\frac{Ln(\frac{1}{8})}{3.65 \cdot 10^{-04}} [/tex]
[tex]t = 5697.1 d = 15.6 years [/tex]
Have a nice day!
