Answer:
116 years
Step-by-step explanation:
For this type of problems we have to use the formula that model the radioactive decay, that it's:
[tex]N = N_{0} e^{-\lambda t}[/tex],
where [tex]N_{0}[/tex] correspond to the initial amount of mass of the element, [tex]\lambda[/tex] correspond to the decay constant of the element, [tex]t[/tex] correspond the time measured in years and [tex]N[/tex] correspond to the amount of mass left after [tex]t[/tex] years.
In this case the original amount of mass is unknown but we don't need it to resolve the question, so let's make use of the half-life data that we've been given:
If the original amount of mass is [tex]N_{0}[/tex] then after [tex]t[/tex] = 29 years the amount of mass left is the half of [tex]N_{0}[/tex].
[tex]N = N_{0}\frac{1}{2} = N_{0}e^{-29\lambda }[/tex], we cancel equal terms on both sides and obtain that [tex]e^{-29\lambda} = \frac{1}{2}[/tex], now we take natural logarithm on both sides (due to exponential function and logarithmic function are inverses) and then we obtain [tex]\lambda[/tex] by dividing for -29, so [tex]\lambda = \frac{-log(\frac{1}{2})}{29}[/tex]
Now we know the formula that model the radioactive decay for Strontium-90:
[tex]N = N_{0}e^{\frac{log(\frac{1}{2})}{29}t}[/tex]
So we've been asked for the time in years were the mass left will be the 1/16 of it's original mass,
By that time [tex]t[/tex] in years we have that [tex]N = N_{0}\frac{1}{16} = N_{0}e^{\frac{log(\frac{1}{2})}{29}t}[/tex], we cancel equal terms and take natural logorithm on both sides [tex]log(\frac{1}{16}) = \frac{log(\frac{1}{2})}{29}t}[/tex], finally we obtain after dividing that [tex]t = \frac{29 log(\frac{1}{16})}{log(\frac{1}{2})} = 116[/tex] years
