# Finding half life given decay rate

• Apr 25th 2012, 11:54 PM
jkort13
Finding half life given decay rate
Hey guys, I'm working on a problem right now and I'm having some trouble.
The decay of a certain chemical is 9.3% per year. Using the exponential decay model P(t) = P0-kt d where k is the decay rate, and P0 is the original amount of chemical find the half-life.

Thanks!
• Apr 26th 2012, 01:39 AM
biffboy
Re: Finding half life given decay rate
Just out of interest did you get the answer 7.1 years?
• Apr 26th 2012, 01:48 PM
jkort13
Re: Finding half life given decay rate
Yes that is the correct answer.
• Apr 26th 2012, 05:00 PM
HallsofIvy
Re: Finding half life given decay rate
I'm a bit surprized- that formula doesn't make sense. If \$\displaystyle P(t)= P_0^{kt d}\$, then the "initial value" is \$\displaystyle P(0)= 1\$, not \$\displaystyle P_0\$. Did you mean \$\displaystyle P(t)= P_0e^{kt d}\$? And if \$\displaystyle P_0\$ is the initial value, t is the time, and k is the decay rate, what is "d"?

More common is the formula \$\displaystyle P(t)= P_0e^{kt}\$ for decay. In one year, we will have \$\displaystyle P(1)= P_0e^k\$. If the chemical decays 9.3% per year, P(1) should equal \$\displaystyle (1- .093)P_0= .907P_0\$ so \$\displaystyle P_0e^k= .907P_0\$. That is, k must satisfy \$\displaystyle e^k= .907\$ so k= ln(.907)= -.0976.

Once you know that, the "half life" is the time until \$\displaystyle P_0\$ becomes \$\displaystyle P_0/2\$. Solve \$\displaystyle P_0e^{-.0976t}= P_0/2\$ which is the same as \$\displaystyle e^{-.0976t}= 1/2\$. Solve that for t.