Question

Cerium-144 is a radioactive isotope with a half-life of 285 days. How long would it take for an initial sample of 15 grams of Cerium-144 to decay until only 5 gram remains?
a.
about 452 days

b.
about 855 days

c.
about 570 days

d.
about 407 days

Answer 1

To solve this we are going to use the formula: $A=A_{0}( \frac{1}{2} )^{ \frac{t}{h}$
where 
$A$ is the final amount aster a time $t$
$A_{0}$ is the initial amount 
$t$ is the time 
$h$ is the half-life

We know for our problem that $A=5$, $A_{0}=15$, and  $n=285$, so lets replace those values in our formula:
$A=A_{0}( \frac{1}{2} )^{ \frac{t}{h}$
$5=15( \frac{1}{2} )^{ \frac{t}{285}$

Since $t$ is the exponent, we are going to use logarithms to find its value:
$5=15( \frac{1}{2} )^{ \frac{t}{285}$
$( \frac{1}{2} )^{ \frac{t}{285}}= \frac{5}{15}$
$( \frac{1}{2} )^{ \frac{t}{285}}= \frac{1}{3}$
$ln( \frac{1}{2} )^{ \frac{t}{285}}= ln(\frac{1}{3})$
$\frac{t}{285} ln( \frac{1}{2})=ln( \frac{1}{3} )$
$\frac{t}{285}= \frac{ln( \frac{1}{3}) }{ln( \frac{1}{2} )}$
$t= \frac{285ln( \frac{1}{3}) }{ln( \frac{1}{2}) }$
$t=451.71$

We can conclude that the correct answer is: a. about 452 days.