Question 938336
{{{p=I*e^(-kt)}}}
t, time in days
p, amount after time t
k, decay constant
I, initial amount


{{{ln(p)=ln(I)+ln(e^(-kt))}}}
{{{-kt*ln(e)+ln(I)=ln(p)}}}
{{{-kt=ln(p)-ln(I)}}}
{{{highlight_green(kt=ln(I)-ln(p))}}}-----You can use this form of the equation both to find k using the half-life information, and then to find t using the part (b) material quantities data.


{{{k=(1/138)ln(2/1)}}}, using time units of DAYS.
{{{highlight(k=0.00502)}}}


Be careful in part (a).   Convert 5 years into how many days, and use the time in number of days, because that is what k was based on.
{{{5*365.25=t}}} in days.


(a)
{{{p=200*e^(-(0.00502)(5*365.25))}}}