Answer:
There are [tex]3.779\times 10^{23}[/tex] atoms when the sample of cerium is 20 seconds old.
Explanation:
The decay of isotopes is modelled after the following exponential expression:
[tex]N(t) = N_{o}\cdot e^{-\left(\frac{\ln 2}{t_{1/2}}\right)\cdot t }[/tex] (1)
Where:
[tex]N_{o}[/tex] - Initial number of atoms, dimensionless.
[tex]N(t)[/tex] - Current number of atoms, dimensionless.
[tex]t[/tex] - Time, measured in seconds.
[tex]t_{1/2}[/tex] - Half-time, measured in seconds.
Now we clear the half-time within (1):
[tex]\ln \frac{N(t)}{N_{o}} = -\frac{t\cdot \ln 2}{t_{1/2}}[/tex]
[tex]t_{1/2} = -\frac{t\cdot \ln 2}{\ln \frac{N(t)}{N_{o}} }[/tex]
If we know that [tex]N_{o} = 5\times 10^{23}[/tex], [tex]N(t) = 2\times 10^{23}[/tex] and [tex]t = 66\,s[/tex], then the half-life of the isotope is:
[tex]t_{1/2}=-\frac{(66\,s)\cdot \ln 2}{\ln \frac{2}{5} }[/tex]
[tex]t_{1/2}\approx 49.927\,s[/tex]
And the decay equation for the sample is described by:
[tex]N(t) = (5\times 10^{23})\cdot e^{-0.014\cdot t}[/tex]
Lastly, if we get that [tex]t = 20\,s[/tex], then the activity of the sample of cerium is:
[tex]N(20) = (5\times 10^{23})\cdot e^{(-0.014)\cdot (20)}[/tex]
[tex]N (20) \approx 3.779\times 10^{23}[/tex]
There are [tex]3.779\times 10^{23}[/tex] atoms when the sample of cerium is 20 seconds old.
