The function below represents the amount y of plutonium-238 after t years.
y= a(0.5)^(t/ x)
Here, a is the initial amount and x is the length of the half-life (in years). We know that the half-life of plutonium-238 is 88 years, and a scientist is studying a 3-gram sample. This makes a= 3 and x= 88.
y= 3(0.5)^(t/ 88)
This function represents the amount y of plutonium-238 after t years.
Now we can find the yearly percent increase r, since the decay factor of this function is equal to (1+r).
0.9922 = (1- r) ⇒ r = 0.0078 or 0.78 %
The yearly percent decrease is about 0.78 %.
c
To draw the graph of y= 3(0.5)^(t/88), we will make a table of values.
x
3(0.5)^(t/88)
y= 3(0.5)^(t/88)
1
3(0.5)^(1/88)
≈ 2.98
3
3(0.5)^(3/88)
≈ 2.93
5
3(0.5)^(5/88)
≈ 2.88
7
3(0.5)^(7/88)
≈ 2.84
9
3(0.5)^(9/88)
≈ 2.79
The points ( 1, 2.98), ( 3, 2.93), ( 5, 2.88), ( 7, 2.84), and ( 9, 2.79) are on the graph of the function y= 25 000(1.055)^t.
Let's now plot and connect them with smooth curves.
From the graph, we can see that the amount remaining after 12 years is about 2.67 grams.
Checking Our Answer
The exact value of the remaining amount can be found by substituting t=12.
