What is the next step after finding the k-value in half-life calculations?

AI Thread Summary
After determining the k-value using the half-life of an element, the next step in age calculations involves using the equation y(t) = e^{kt}. To find the age of a rock, the initial abundance of the element at time zero is crucial, but this information is often not provided. In carbon dating, the current atmospheric level of C14 is used as the initial abundance since its creation and decay rates are balanced. For other isotopes, relative concentrations of isotopes may be utilized, though the specifics can vary. Understanding the initial conditions is essential for accurate age determination.
mateomy
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Maybe this is a dumb question...

Let's say I want to figure out the age of a rock and I have the half life of an element.
If I have this equation
<br /> y(t)=e^{kt}
where first I figure out the k-value using the half life. That part I get.

Now that I have the k-value I 're-use' the formula but instead of put 0.5 for y(t) as before I put what exactly? Do I use the natural abundance of the element? This is assuming I wasn't given any information except 'figure out how old this rock is using element X'.

Thanks.
 
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In your equation, y is a ratio of abundances at different times. So to determine t you need to know the abundance at time 0.
 
How would you know that if you weren't told?
 
mateomy said:
How would you know that if you weren't told?
Depends. Take carbon dating. The theory is that C14 is continually created high in the atmosphere at a steady rate as a result of radiation, so the level in the atmosphere is constant. Creation rate matches decay rate. Once captured at ground level, the decay continues but creation stops. So the initial level is taken to be the current atmospheric level. I don't know whether known variations in atmospheric concentrations of carbon are taken into account.
For other isotopes, I believe there's some way to use the relative concentrations of a mix of isotopes, but I don't know how this works.
 
Hmmmm...

Okay, thanks.
 
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