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:
Icicid
(a) The fraction P of strontium remaining as a function of time t can be found using the formula:
P(t) = (100 - 2t) / 100
Here, t represents the number of years elapsed.
(b) To estimate the half-life of strontium, we can set up the equation:
P(t) = 0.5
Substituting the equation from part (a), we get:
(100 - 2t) / 100 = 0.5
Simplifying the equation, we have:
100 - 2t = 50
Solving for t, we find:
2t = 50
t = 25
Therefore, the half-life of strontium is estimated to be 25 years.
(c) To estimate how many grams of strontium will remain after 11 years, we can substitute t = 11 into the equation from part (a):
P(11) = (100 - 2*11) / 100
Simplifying the equation, we have:
P(11) = (100 - 22) / 100
P(11) = 78 / 100
P(11) = 0.78
To find the grams of strontium remaining, we multiply this fraction by the initial amount of strontium, which is given as 5.5 grams:
Grams remaining = P(11) * 5.5
Grams remaining = 0.78 * 5.5
Grams remaining ≈ 4.29 grams
Therefore, after 11 years, approximately 4.29 grams of strontium will remain.
(a) To write the fraction P of strontium remaining as a function of time t, we can first determine the rate of decay. We are given that strontium-90 decays at a rate of 2 percent per year. This means that after one year, 2 percent of the strontium will decay, leaving 98 percent remaining.
Since we start with 100 percent of the strontium at t=0, we can write the fraction remaining as:
P(t) = 1 - 0.02t
(b) To estimate the half-life of strontium, we need to find the time it takes for the fraction remaining to be half of its initial value (50%). We can set up the equation:
0.5 = 1 - 0.02t
Solving for t, we have:
0.5 = 0.98^t
Using a graphing calculator with the trace function or natural logarithms (as mentioned in hint 1.9), we can find that the value of t is approximately 34.6 years. This is the estimated half-life of strontium-90.
(c) To estimate how many grams of strontium will remain after 11 years, we can use the fraction remaining function P(t) from part (a).
P(11) = 1 - 0.02 * 11
Calculating this, we get:
P(11) = 1 - 0.22 = 0.78
Since there are currently 5.5 grams of strontium, we can multiply this by the fraction remaining to find the estimated amount remaining:
Estimated remaining = 0.78 * 5.5 = 4.29 grams
Therefore, approximately 4.29 grams of the substance will remain after 11 years.
Yes, you do
ds/dt = -.02 s
ds/s = -.02 dt
ln s = -.02 t + c
e^ln s = s = e^(c-.02t) = e^c e^-.02 t
e^c = C any old constant in the general solution
s = C e^-.02 t
when t = 0, s = Si the initial amount
so
s/Si = e^-.02 t
BUT that is P, the fraction remaining so we have part a
b) .5 = e^-.02 t
ln .5 = -.02 t = -.693
t = 34.7 years half life
in 11 years
s = 5.5 e^-.02*11
s = 4.41