SOLUTION: The half-life of radioactive potassium is 1.3 billion years. If 10 grams is present now, how much will be present in 100 years? The equation used was {{{A(t)=Aoe^kt}}} My work

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Question 1102181: The half-life of radioactive potassium is 1.3 billion years. If 10 grams is present now, how much will be present in 100 years?
The equation used was A%28t%29=Aoe%5Ekt
My work was
5=10e%5Ek1.3 where 1.3 is 1.3 billion when entered in calculator
0.5=e%5Ek1.3
ln0.5=lne%5Ek1.3
ln0.5=k1.3
k=ln0.5%2F1.3
k=-5.33190
The I subbed in the new value of k to find grams left after 100 years
A%28t%29=Aoe%5Ekt
A%28100%29=10e%5E%28-5.33190%29%28100%29
A%28100%29=0
Therefore, there are 0 grams left after 100 years.
I just wanted to know where I went wrong or if its ok?

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


If you pay attention to the real problem, instead of just plugging numbers into equations, you have to know that your answer is not right. If the half life is 1.3 billion years, only a very tiny amount will decay in 100 years; virtually all of it will remain.

The way you start the problem is okay, except that the value you got for k has the decimal point in the wrong place; it should be -0.533190.

But the problem with your work is that, by using 1.3 in your calculation for the value of k, your units of time are billions of years.

So in finding the number of grams left after 100 years, where you show

A%28100%29=10e%5E%28-5.33190%29%28100%29

both of the "100"s should be "(100/10^9)", or 10^-7.

Do that calculation and you will see that nearly all of the original 10 grams remains after 100 years.