Question 925599
{{{N(t)=s*e^(-kt)}}}

N(t), amount in grams at time t
s, initial amount (at time 0, when the plutonium was released?); think "s"tart;
e, the natural logarithm base
k, a constant
t, time passage in years


The half-life fact allows you to find the value for k.
In general, based on the decay equation
{{{1/2=1*e^(-k*24110)}}}, although this is really for Plutonium-239.
{{{ln(1/2)=ln(1*e^(-k*24100))}}}
{{{ln(1/2)=ln(1)+ln(e^(-k*24100))}}}
{{{ln(1/2)=0+(-kt)*ln(e)}}}
{{{ln(1/2)=-kt}}}
{{{-ln(2)=-kt}}}
{{{kt=ln(2)}}}, combining two steps; and recall that for here, half-life is...
k...
{{{highlight_green(k=ln(2)/24110)}}}
OR
{{{highlight_green(k=2.8749*10^(-5))}}}


You should have no trouble with question part (b), because now you have the function  {{{highlight(N(t)=s*e^((-2.8749*10^(-5)*t)))}}}; and you are given s=20, t=5000.


The way to handle question part (c) is to solve the decay model equation function for time, {{{t}}}.