Radioactive Decay

Nov 2009
76
1
A radioactive material is known to decay at a yearly rate proportional to the amount at each moment. There were 2000 grams of the material 10 years ago. There are 1990 grams right now. What is the half-life of the material?

The equation to solve is P = P0e^kt. Where P = population, P0 = initial population, k is a constant and t = time.

To find k:
1990 = p0e^0k
-------------------
2000 = p0e^-10k

Once I divide these out I get 1990/2000 = e^10k.
To find k take the natural log of both sides:-

ln 1990/2000 = 10k. Therefore k = 1/10*ln 1990/2000.

To solve for the half life:

1000 = 2000e^(1/10 ln 1990/2000)t.

When I work through this I get:-
t = 10* (ln 1/2)/(ln (1990/2000)).

However the correct answer is:-
t = 10* (ln 2)/(ln (2000/1990)).

Much appreciated if someone can show where I am going wrong.
 
Mar 2010
116
41
Bratislava
When I work through this I get:-
t = 10* (ln 1/2)/(ln (1990/2000)).

However the correct answer is:-
t = 10* (ln 2)/(ln (2000/1990)).
Both answers equal to the same number
Just use the fact that ln(1/x)=-ln(x) and you can see that your number is equal to
t = 10* (-ln 2)/(-ln (2000/1990)) = 10* (ln 2)/(ln (2000/1990)).
 
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