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)=(.98)^t
(b) Estimate the half-life of strontium.
(c) If presently there is 6 grams of strontium, estimate how many grams of the substance will remain after 38 years.
i need help with b and c
a)
P = (.98)^t , you had that
b)
1/2 = .98^t
.5 = .98^t
take logs of both sides and use log rules ...
log .5 = t(log .98)
t = log .5/log .98 = appr 34.4 years
c) P = 6(.98)^38
= aprr 2.78 g
makes sense, since the half-life of 34.4 years would have reduced it to 3 g , and we went just a few more years.
write a formula,p(t), that models the percent of strontium-90 after t years.
:
The radioactive decay formula
A = Ao*2^(-t/h)
Where
A = resulting amt after t time
Ao - initial amt
h = half-life of substance
t = time of decay
:
To find percent here, you could write it
p(t) = 100*2^(-t/28)
Sure, I'd be happy to help with parts (b) and (c)!
(b) To estimate the half-life of strontium, we need to find the time it takes for half of the substance to decay. In this case, we know that the decay rate is 2 percent per year. So let's set up an equation to find the half-life:
(0.98)^t = 0.5
To solve for t, we can take the logarithm of both sides of the equation:
t * log(0.98) = log(0.5)
Now, we can solve for t:
t = log(0.5) / log(0.98)
Using a calculator, we find that t ≈ 35.16 years. So, the estimated half-life of strontium is approximately 35.16 years.
(c) To estimate how many grams of strontium will remain after 38 years, we can use the fraction P(t) = (.98)^t that we derived in part (a). We know that initially there were 6 grams of strontium, so we can plug in t = 38 into the equation to find the remaining amount:
P(38) = (.98)^38
Using a calculator, we find that P(38) ≈ 3.1115 grams. Therefore, it is estimated that approximately 3.1115 grams of strontium will remain after 38 years.
Remember, these estimates assume the decay rate remains constant at 2 percent per year.
(b) To estimate the half-life of strontium, we can solve the equation P(t) = 0.5 for t, where P(t) is the fraction remaining at time t.
0.5 = (.98)^t
To solve this equation, we can take the logarithm of both sides:
log(0.5) = log((.98)^t)
Since log((.98)^t) can be rewritten as t * log(.98), we have:
log(0.5) = t * log(.98)
Now we can solve for t using basic algebraic steps:
t = log(0.5) / log(.98)
Using a scientific calculator, we find:
t ≈ 34.77 years
Therefore, the estimated half-life of strontium is approximately 34.77 years.
(c) To estimate how many grams of strontium will remain after 38 years, we can use the fraction P(t) = (.98)^t and substitute t = 38 into the equation:
P(38) = (.98)^38
Using a calculator, we find:
P(38) ≈ 0.472
To find the amount of strontium remaining, we multiply the fraction by the initial amount (6 grams):
Amount remaining = 0.472 * 6
Therefore, the estimate for the amount of strontium remaining after 38 years is approximately 2.832 grams.
To estimate the half-life of strontium, we can use the formula:
P(t) = (0.5)^n
Where n is the number of half-lives. Since the decay rate of strontium-90 is 2 percent per year, we can calculate the half-life by finding the number of years it takes for the remaining fraction to reach 0.5.
We'll solve for n in the equation:
(0.98)^t = 0.5
Taking the natural logarithm of both sides:
ln((0.98)^t) = ln(0.5)
Using the logarithmic property:
t * ln(0.98) = ln(0.5)
Now divide both sides by ln(0.98) to solve for t:
t = ln(0.5) / ln(0.98)
Using a calculator, we can find:
t ≈ -34.985 years
Since time cannot be negative in this context, we take the absolute value:
t ≈ 34.985 years
This value represents the approximate half-life of strontium-90.
Moving on to part (c), we can use the formula P(t) = (0.98)^t to estimate how many grams of strontium will remain after 38 years.
P(38) = (0.98)^38
Using a calculator, we can determine:
P(38) ≈ 0.47
This means that approximately 47% of the strontium will remain after 38 years.
To estimate the number of grams remaining, we multiply the percentage by the initial amount of strontium (6 grams):
0.47 * 6 ≈ 2.82 grams
Therefore, it is estimated that around 2.82 grams of strontium will remain after 38 years.