Answer:
t = 15825.11 years
Step-by-step explanation:
From the given information:
We can make use of the activity of the sample present after time "t" to determine the age of the sample.
This can be expressed by the formula:
[tex]A = A_o e^{-\lambda t}[/tex] --- (1)
where;
[tex]A_o[/tex]= activity of the sample at time t = 0
[tex]\lambda[/tex] = disintegration constant
[tex]\lambda = \dfrac{0.693}{T_{1/2}}[/tex]
If we replace the value of [tex]\lambda[/tex] into equation (1), we have:
[tex]A = A_o e^{ \Bigg [ -{ \dfrac{0.693}{T_{1/2}} \Bigg ] } t}[/tex]
[tex]\dfrac{A}{A_o} = e^{ \Bigg [ -{ \dfrac{0.693}{T_{1/2}} \Bigg ] } t}[/tex]
By rearrangement:
[tex]t = \dfrac{-T_{1/2} In (\dfrac{A}{A_o})}{0.693}[/tex]
[tex]t = - \dfrac{\left(5730\ \cdot\ \ln\left(\frac{\left(1770\right)}{0.8\cdot10^{3}\cdot15}\right)\right)}{0.693}[/tex]
t = 15825.11 years
