YoloShroom
  • YoloShroom
radium has a hlaf life of 1620 years,. what % of the original amount of the radium sample would remain after 1000 years? Halp;... 1/2P=Pe^kt isnt working for me..
Mathematics
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SOLVED
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katieb
  • katieb
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whpalmer4
  • whpalmer4
Show me your work so far...
whpalmer4
  • whpalmer4
that way I don't start explaining it the other way and confuse you unnecessarily...
YoloShroom
  • YoloShroom
Just start frm the begininng. i dont know what im doing.

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DanJS
  • DanJS
1/2P=Pe^kt half the initial amount = initial amount * decay rate e^kt
DanJS
  • DanJS
the P, initial amount can be cancelled \[\huge \frac{ 1}{ 2 }=e^{k*t _{1/2} }\]
YoloShroom
  • YoloShroom
but why is t/2?
DanJS
  • DanJS
they give you the point for the half life , when t=1620, solve for the constant k
DanJS
  • DanJS
then you have a general formula y = P*e^{k*t} for the amount left y, after t years
whpalmer4
  • whpalmer4
\[P(t) = P_0 (\frac{1}{2})^{t/t_{\text{half}}}\]is how I like to figure it out \(P_0\) is the initial amount. \(t_{\text{half}}\) is the half-life \[P(1000) = P_0(\frac{1}{2})^{(1000/1620)} = P_0(0.651897)\] That means you have approximately 65.2% of the initial amount remaining after 1000 years.
DanJS
  • DanJS
y/P is the fraction percent they want
whpalmer4
  • whpalmer4
The exponential form is very useful, but I think it is a little easier for people new to the material to see how it works with a power of 1/2 instead of a power of e...
YoloShroom
  • YoloShroom
Ahh i see!!!! The E is fine :p THANK YoU!
DanJS
  • DanJS
right on, both are good

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