Solve Half-Life Problem: 226Ra, 1620 yrs, 1.5g

[audax]

Hello,

I have a relatively simple question. I'm just wondering, with half-life problems, how could I solve something like the following:

Isotope: 226Ra
Half-life: 1620 years.
Initial Quantity: X (Solve for this)
Amount after 1000 years: 1.5g

Where I'm supposed to find the initial quantity.
Wouldn't a part of the equation be something like: .5 = e^1620r, where r is the rate of decay? I'm using natural log to try to get the rate: ( ln(.5) / 1620 ) = r
When I punch that into the calculator, I get huge number and it leads me to believe that I'm messing up somewhere.

 

[hage567]

For your decay constant, (which is what I think you really want. Decay rate* is different, and it isn't interchangeable with decay constant), you have to remember it is exponential decay, so be careful to get +/- stuff right. The decay constant should be a small number.

To now find the initial quantity, you must set up your exponential equation again, showing the relationship for the amount of material remaining from what you started with, using the decay constant you just found.


*Decay rate is also referred to as the activity of a radioactive source. The decay constant doesn't change, it is a property of the isotope. But the decay rate (activity) decreases by half after one half-life.

 

[HallsofIvy]

You know that the basic equation if A= Ce^kt. You are told that the half life is 1620 years. So you know that A(1620)= Ce^1620k= (1/2)C. The C's cancel (of course) so e^1620k= 1/2. Solve that for k. Now use Ce^1000k= 1.5, with known k, to solve for C.

Actually because this is a "half life" problem, it might be better to write the function as $(\frac{1}{2})^x$ rather than $e^x$. (They are interchangeble:$(\frac{1}{2})^x= e^{xln(1/2)}$.)
If T is the half life then the amount is multiplied by 1/2 every T years. There are t/T "periods of T years" in t years and so, in t years, the amount would be multiplied by 1/2 t/T years: with given half life 1620 years, $X= C(\frac{1}{2})^{t/1620}$. Now use $X(1000)= C(\frac{1}{2})^{1000/1629})= 1.5$ to solve for C.