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Strontium-90 is a radioactive material that decays according to the function A(t)=A0e−0.0244t, where A0 is the initial amount present and A is the amount present at time t​ (in years). Assume that a scientist has a sample of 400 grams of​ strontium-90. ​

(a) What is the decay rate of​ strontium-90?
​(b) How much​ strontium-90 is left after 30 ​years?
​(c) When will only 100 grams of​ strontium-90 be​ left?
​(d) What is the​ half-life of​ strontium-90?

​(a) The decay rate of​ strontium-90 is nothing​%.
​(Type an integer or a decimal. Include the negative sign for the decay​ rate.)

User Andrey Starodubtsev
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Answer:

Explanation:

The decay rate of strontium-90 is -.0244 as given.

For b., we have to use the formula to find out how much is left after 30 years. This will be important for part d.


A(t)=400e^(-.0244(30)) which simplifies a bit to

A(t) = 400(.4809461353) so

A(t) = 192.4 g

For c., we have to find out how long it takes for the initial amount of 400 g to decay to 100:


100=400e^(-.0244t). Begin by dividing both sides by 400:


.25=e^{-.0244t and then take the natural log of both sides:


ln(.25)=lne^{-.0244t . The natural log and the e cancel each other out since they are inverses of one another, leaving us with:

ln(.25) = -.0244t and divide by -.0244:

61.8 years = t

For d., we figured in b that after 30 years, 192.4 g of the element was left, so we can use that to solve for the half-life in a different formula:


A(t)=A_0(.5)^{(t)/(H) and we are solving for H. Filling in:


192.4=400(.5)^{(30)/(H) and begin by dividing both sides by 400:


.481=(.5)^{(30)/(H) and take the natural log of both sides, which allows us to pull the exponent out front. I'm going to include that step in with this one:

ln(.481) =
(30)/(H) ln(.5) and then divide both sides by ln(.5):


(ln(.481))/(ln(.5))=(30)/(H) and cross multiply and isolate the H to get:


H=(30ln(.5))/(ln(.481)) and

H = 28.4 years

User Valentin
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