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)( (1)/(2) )^{ (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)( (1)/(2) )^{ (t)/(h)`

`5=15( (1)/(2) )^{ (t)/(285)`

Since
`t` is the exponent, we are going to use logarithms to find its value:

`5=15( (1)/(2) )^{ (t)/(285)`

`( (1)/(2) )^{ (t)/(285)}= (5)/(15)`

`( (1)/(2) )^{ (t)/(285)}= (1)/(3)`

`ln( (1)/(2) )^{ (t)/(285)}= ln((1)/(3))`

`(t)/(285) ln( (1)/(2))=ln( (1)/(3) )`

`(t)/(285)= (ln( (1)/(3)) )/(ln( (1)/(2) ))`

`t= (285ln( (1)/(3)) )/(ln( (1)/(2)) )`

`t=451.71`

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