Suppose strontium-90 decays at a rate of 2 percent per year.
(a) Write the fraction P of strontium remaining, as function of t, measured in years. (Assume that at time t=0 there is 100 % remaining.)
Answer: P(t) =
(b) Estimate the half-life of strontium.
Answer:
Hint: Use your graphing calculator and the trace function. (Or use natural logarithms as in 1.9).
(c) If presently there is 5.5 grams of strontium, estimate how many grams of the substance will remain after 11 years.
Answer:
28.8
(a) To find the fraction P of strontium remaining as a function of time t, we start with 100% of the substance at t=0 and it decays at a rate of 2 percent per year.
Since a rate of 2 percent per year means that 2% of the substance is lost every year, we can express it as 0.02. Therefore, the fraction remaining after one year would be (1 - 0.02) = 0.98, or 98% remaining.
To find the fraction remaining after t years, we can raise 0.98 to the power t. This is because each year, the process of losing 2% is repeated t times. So, the fraction P(t) is:
P(t) = (0.98)^t
(b) To estimate the half-life of strontium, we need to find the value of t for which P(t) is equal to 0.5 (50% remaining).
Using a graphing calculator or a table of values, we can evaluate P(t) for different values of t until we find the value of t that gives P(t) = 0.5.
(c) To estimate the number of grams of strontium remaining after 11 years, we need to multiply the initial amount (5.5 grams) by the fraction remaining after 11 years.
Let's say the initial amount of strontium is P0 grams. The amount remaining after 11 years is given by:
Amount remaining = P0 * P(11)
Substituting the given value of P0 as 5.5 grams and using the fraction remaining from the previous part, we will have the estimated amount of strontium remaining after 11 years.