A radioactive isotope has a half-life of 3000 years. If a sample of this isotope origanally has a mass 30g, what equation would model the mass of this sample over time? What would its mass be after 5 hours?
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Let A(t) represent the amount of
isotope at time t. The formula is:
A(t) = Pert
At the start, when time = t = 0,
then A(0) = 30
So we substitute 0 for t:
A(0) = Per*0
and we substitute 30 for A(0)
30 = Pe0
30 = P(1)
30 = P
So now the formula
A(t) = Pert
becomes
A(t) = 30ert
Nw we read >>...half-life of 3000 years...<<
which means that in 3000 years the isotope will
have reduced to half of its original mass of
30 grams. That means that A(3000) = 15 grams.
So we substitute 3000 for t
A(t) = 30ert
A(3000) = 30er(3000)
and then substitute 15 for A(3000)
A(3000) = 30er(3000)
15 = 30er(3000)
Divide both sides by 30
= e3000r
0.5 = e3000r
Taking the natural log of both sides,
we have
ln(0.5) = 3000r
Use a calculato to get the left side:
-.6931471806 = 3000r
Divide both sides by 3000
-.00023104906 = r
So the formula
A(t) = 30ert
becomes
A(t) = 30e-0.00023104906t
That is the equation that models the mass,
which was the first part of your problem.
Now for the second part, we only need to plug
in 5 hours. But we must change that to
years. So
Change 5 hours to days by dividing by 24, getting
5 hours = 0.2083333333 days,
Now divide that by 365.25 to change it to years:
5 hours = 0.0005703855807 years
So plug that into
A(t) = 30e-.00023104906t
A(0.0005703855807) = 30e-.00023104906(0.0005703855807)
A(0.0005703855807) = 29.99999605 grams.
As we might guess, if it takes 3000 years for
30 grams to decrease to 15 grams, we wouldn't
expect it to have noticeably decreased from the
original 30 grams after only 5 hours!
Edwin