# Exponential function

• Apr 17th 2008, 04:46 PM
Snowboarder
Exponential function

The half-life of the radioactive element Radium-226 is 1590 years. This means that after 1590 years only half of the original radioactive material will have disintegrated. Because the
rate of decay is proportional to the amount of material present, the function that describes this decay will be exponential.

If the initial amount of Radium-226 present was 100g, write an exponential function, call it R(t), that describes the decay over t years, at a decay rate of k.
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• Apr 17th 2008, 05:12 PM
Jhevon
Quote:

Originally Posted by Snowboarder
The half-life of the radioactive element Radium-226 is 1590 years. This means that after 1590 years only half of the original radioactive material will have disintegrated. Because the
rate of decay is proportional to the amount of material present, the function that describes this decay will be exponential.

If the initial amount of Radium-226 present was 100g, write an exponential function, call it R(t), that describes the decay over t years, at a decay rate of k.
.

$\displaystyle k = \frac {\ln 2}{T}$ where $\displaystyle T$ is the half life.

From there it is simple. the equation is of the form $\displaystyle R(t) = R_0e^{-kt}$

you know what $\displaystyle R_0$ is right?

I hope you can derive the formulas I gave you above. If not, it is a good exercise to try
• Apr 17th 2008, 05:24 PM
Snowboarder
thx a lot
So is it something like:

50 = 100e^{-k*1590} ???
• Apr 17th 2008, 05:38 PM
Jhevon
Quote:

Originally Posted by Snowboarder
thx a lot
So is it something like:

50 = 100e^{-k*1590} ???

no.

$\displaystyle R_0$ is the initial amount, so it is 100

$\displaystyle R(t)$ stays as it is, it is just the amount after time t

as i said, $\displaystyle k = \frac {\ln 2}{1590}$, so there should be no "k" appearing in your formula

$\displaystyle t$ is just time, it stays as is.

Please look up what each variable represents.
• Apr 17th 2008, 06:02 PM
Snowboarder
i need also show that k = -0.000436 so

Ro/2 = Roe^-kt
ln(0.5) = -1590k
k = ln(0.5)/(-1590)
k = - 0.000436

is it correct??
Thank you a lot