SOLUTION: Please help me answer this word problem:
The rate at which a radioactive isotope disintegrates is proportional to the amount present. If a 30g sample will contain only 20g afte
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Question 1015192: Please help me answer this word problem:
The rate at which a radioactive isotope disintegrates is proportional to the amount present. If a 30g sample will contain only 20g after ten minutes, what is the half-life of this isotope, correct to the nearest tenth of a minute? Answer this question by first finding the amount y of this isotope as a function of time t(in minutes) in the form y=y0e^kt.
Thank-you, your help is always appreciated!
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The rate at which a radioactive isotope disintegrates is proportional to the amount present. If a 30g sample will contain only 20g after ten minutes, what is the half-life of this isotope, correct to the nearest tenth of a minute? Answer this question by first finding the amount y of this isotope as a function of time t(in minutes) in the form
y = yo*e^(kt)
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20 = 30*e^(k*(1/6))
----
e^(k/6) = 2/3
-----
k/6 = ln(2/3) = -0.4055
-----
k = -2.4328
------
Substitute for "k" in the formula:
y = yo*e^(-2.4328t)
---
Note: Half-life means y = yo/2
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Then e^(-2.4328t) = 1/2
----
-2.4328t = ln(1/2) = -0.6932
----
time = -0.6932/-2.4328 = 0.2849 hrs = 17min 5.7sec
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Cheers,
Stan H.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Substitute the values you know:
So
Take the log of both sides
\ =\ \ln\left(\frac{2}{3}\right))
\ -\ \ln(3)}{10})
Now that you know the value of 
, solve for 
in
Take the log
\ -\ \ln(2)\ =\ -\ln(2))
Substitute value of and solve for
\right)}{\ln(2)\ -\ \ln(3)})
The rest is just calculator work.
John
My calculator said it, I believe it, that settles it
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