Question 822989
Find the half-life of a radioactive substance that decays by 2% in 9 years.
:
Using the radioactive decay formula; A = Ao*2^(-t/h), where:
A = amt remaining after t time
Ao = initital amt (t=0)
t = time of decay
h = half-life of substance
:
Let initial amt = 1, then resulting amt after 9 yrs = .98 (2% less)
1*2^(-9/h) = .98
using nat logs
{{{-9/h}}}ln(2) = ln(.98)
{{{-9/h}}} = {{{ln(.98)/ln(2)}}}
{{{-9/h}}} = -.029
h = {{{(-9)/(-.029)}}}
h = 309 yrs is the half life
:
:
Check this on your calc; enter 1*2^(-9/309); results .98 or 98% remains