Question

Determine the activity of the sample of cerium when the sample was 20 seconds old

Answer 1

Answer:

There are $3.779\times 10^{23}$ atoms when the sample of cerium is 20 seconds old.

Explanation:

The decay of isotopes is modelled after the following exponential expression:

$N(t) = N_{o}\cdot e^{-\left(\frac{\ln 2}{t_{1/2}}\right)\cdot t }$ (1)

Where:

$N_{o}$ - Initial number of atoms, dimensionless.

$N(t)$ - Current number of atoms, dimensionless.

$t$ - Time, measured in seconds.

$t_{1/2}$ - Half-time, measured in seconds.

Now we clear the half-time within (1):

$\ln \frac{N(t)}{N_{o}} = -\frac{t\cdot \ln 2}{t_{1/2}}$

$t_{1/2} = -\frac{t\cdot \ln 2}{\ln \frac{N(t)}{N_{o}} }$

If we know that $N_{o} = 5\times 10^{23}$, $N(t) = 2\times 10^{23}$ and $t = 66\,s$, then the half-life of the isotope is:

$t_{1/2}=-\frac{(66\,s)\cdot \ln 2}{\ln \frac{2}{5} }$

$t_{1/2}\approx 49.927\,s$

And the decay equation for the sample is described by:

$N(t) = (5\times 10^{23})\cdot e^{-0.014\cdot t}$

Lastly, if we get that $t = 20\,s$, then the activity of the sample of cerium is:

$N(20) = (5\times 10^{23})\cdot e^{(-0.014)\cdot (20)}$

$N (20) \approx 3.779\times 10^{23}$

There are $3.779\times 10^{23}$ atoms when the sample of cerium is 20 seconds old.