+78 m(t)=85e^(-t/409) describes the amount of a radioactive isotope as a function of time t [ years]. After how long the amount has dropped to 1/6 of the original ? The response can be written as t=alnb where a and b are integers and a=1
+129850 m(t)=85e^(-t/409) describes the amount of a radioactive isotope as a function of time t [ years]. After how long the amount has dropped to 1/6 of the original ? The response can be written as t=alnb where a and b are integers and a=1
We are solving this :
85(1/6) = 85e^(-t/409) divide through by 85
(1/6) = e^(-t/409) take the natural log of each side
ln (1/60 = ln e^(-t/409) and we can wriite
ln (1/6) = (-t/409)*ln e and ln e = 1, so we can eliminate this
ln (1/6) = -t/ 409 multiply both sides by -409
ln(1/6) * ( -409) = t
(-409)[ ln 1 - ln 6] = t
(-409)(-ln 6) = t
(409)(ln 6) = t = about 732.83 years
CPhill: When you know the half-life of a radioactive material, first you have to figure out its "k"=constant of decay=ln(1/2) / 409 =-0.0016947363......Then the equation becomes:
85/6 =85* e^(-0.0016947363t), solve for t:
t=1,057.25 years.
If you know the half-life, as is the case here, then this procedure is used, which gives the same answer:85/6 =85 * 2^(-t/409), solve for t:
t=1,057.25 years.
It makes sense to notice that since the half-life is 409 years, it will take about 2.58 x 409=1,057.25 to go down to 1/6 of the original amount. At 732.83 years, it will go down to 2^(-732.82/409)=29% of the original amout or about 24.55 units.
