The element strontium-90 is radioactive. The percent of strontium -90, A(t), left in a sample can be modelled by the half-life function A(t) = A0 (1/2)^t/29, where t represents the time, in years, after the initial time, and A0 represents the initial amount, 100% of the strontium.
B) After 20 years ___% of a sample of strontium-90 will remain. Round your answer to one decimal place.
First, if you mean
A(t)=A0(1/2)^(t/29)
the parentheses are mandatory because of the PEMDAS rule.
After 20 years, t=20
so
A(20)=(1/2)^(20/29)
pull out your calculator and find the correct answer. Do not forget the parentheses this time.
Post your answer for checking if you have difficulties.
By the way, if you are not familiar with the PEMDAS rule, google the keyword PEMDAS and read any of the articles.
P=parentheses
E=exponentiation
MD=multiplication/division
AS=addition/subtraction
I get the answer of 3%. I hope this is right...
I don't get this. What is that half life function supposed to calculate. You have two unknowns; i.e., t and Ao AND is Ao the percent? Can't be. Is B a separate problem? Just look up the half life and plug into ln(No/N) = kt.
DrBob222, yes, you're right.
The problem should read as is,
A(t)=(1/29)(1/2)^t
with t in years.
I got carried away!
On second thought, I guess it really IS math. I see mathmate plugged in 20 for t but that isn't clear to me in the problem.
haha, on second thought, the 29 should be the denominator of t, as in my first post. It modifies the half-life of the radioactive material.
Sorry!
3% does not sound right.
In
A(20)=(1/2)^(20/29)
you need to calculate
(1/2)=0.5
(20/29)=0.6897
first because they are within parentheses.
After that, you can calculate the exponentiation.
wow...missing parentheses sure caused a dustup
To find the percent of strontium-90 remaining after 20 years, we can use the half-life function A(t) = A0 (1/2)^(t/29).
Substituting t = 20 into the equation, we have:
A(20) = A0 (1/2)^(20/29)
Since A0 is the initial amount, which is 100%, we can simplify the equation to:
A(20) = (1/2)^(20/29)
Now, we can calculate the value of A(20):
A(20) = (0.5)^(20/29)
A(20) ≈ 0.772
Therefore, after 20 years, approximately 77.2% of a sample of strontium-90 will remain.