Question 1057609
Suppose a sample of a radioactive substance weighs 38 mg.
 One year later, the sample weighs 21.5 mg.
 What is the half-life of this substance? (Round your answer to two decimal places.)
:
Use the radioactive decay formula: A = Ao*2^(-t/h), where:
A = Amt of remaining after t time
Ao = Initial amt 
t = time of decay
h = half-life of the substance
in this problem
A = 21.5
Ao = 38
t = 1 year
h = half life of substance
:
{{{38*2^(-1/h) = 21.5}}}
{{{2^(-1/h) = 21.5/38}}}
{{{2^(-1/h) = .5658}}}
using Natural logs
{{{(-1/h)ln(2) = ln(.5658)}}}
{{{(-1/h)*.6931 = -.5695)}}}
{{{-1/h = (-.5695)/.6931)}}}
{{{-1/h = -.8216}}}
h = {{{(-1)/(-.8216)}}}
h = 1.22 yrs is the half life
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Check this with a calc enter: 38*2^(-1/1.22) ~ 21.5