Question 433433
A sample of a radioactive substance decayed to 91% of its original amount after a year.
 (Round your answers to two decimal places.)
:
The decay formula: A = Ao*2^(-t/h)
Where
A = resulting amt after t time
Ao = initial amt
h = half-life the of substance
t = time (yrs)
:
(a) What is the half-life of the substance?
Let initial amt = 1
1*2(-1/h) = .91
Using nat logs
ln(2^(-1/h)) = ln(.91)
{{{-1/h}}}*ln(2) = ln(.91)
{{{-1/h}}}= {{{ln(.91)/ln(2)}}} 
{{{-1/h}}} = -.13606
h = -1/.13606 
h = 7.35 yrs is the half-life
:
(b) How long would it take the sample to decay to 30% of its original amount?
1*2^(-t/7.35) = .3
Doing it the same way with nat logs
{{{-t/7.35}}} = {{{ln(.3)/ln(2)}}}
{{{-t/7.35}}} = -1.737
t = -7.35 * -1.737
t = +12.77 yrs for 30% to remain