Question 181467
the level of thorium in a sample decreases by a factor of one-half every 2 million years. A meteorite is discovered to have only 8.6% of its original thorium remaining. How old is the meteorite?
:
The half-life formula: A = {{{Ao*(2^(-t/h))}}}
Where
A = final amt
Ao = initial amt
t = time years
h = half-life
:
In this problem, we assume the initial amt is 1 and final amt .086 (8.6%),
time and half-life in millions of years.
{{{Ao*(2^(-t/h))}}} = A
{{{1*(2^(-t/2))}}} = .086
we can ignore the 1
{{{(2^(-t/2))}}} = .086
using logs here
{{{log(2^(-t/2))}}} = log(.086)
log equiv of exponents
{{{(-t/2)}}}*log(2) = log(.086)
Find the logs
.301*{{{-t/2}}} = -1.0655
Multiply both sides by 2
-.301t = -2.131
t = {{{(-2.131)/(-.301)}}}
t = 7.08 million years to decay to 8.6%
:
:
Check on a calc: enter: 2^(-7.08/2) = .086 ~ 8.6%

:
did this help you? any questions