# Half Life and Decay

• May 17th 2009, 02:41 PM
mvho
Half Life and Decay
Hi guys,

New to this forum so let me know If I am doing anything wrong.

My question: Ive decided to take a calculus course after many years of not doing any math.. If you can give me hints in regards to this question it would be greatly appreciated.

A 96-milligram sample of a radioactive substance decays according to the equation
N=96 ·e-0.034 ·t where N is the number of milligrams present after t years.

(a) Find the half-life of the substance to the nearest tenth of a year.

TIA
• May 17th 2009, 02:56 PM
e^(i*pi)
Quote:

Originally Posted by mvho
Hi guys,

New to this forum so let me know If I am doing anything wrong.

My question: Ive decided to take a calculus course after many years of not doing any math.. If you can give me hints in regards to this question it would be greatly appreciated.

A 96-milligram sample of a radioactive substance decays according to the equation
N=96 ·e-0.034 ·t where N is the number of milligrams present after t years.

(a) Find the half-life of the substance to the nearest tenth of a year.

TIA

The general form of exponential decay is $N(t)=N_0e^{-kt}$ where
• $N(t)$ is the amount at a given time t
• $N_0$ is the amount at t=0
• $k$ is a positive constant
• $t$ is time

From the definition of half life $\frac{N_0}{2} = N_0e^{-kt_{1/2}}$ and so $N_0$ will cancel

$\frac{1}{2} = e^{-kt_{1/2}}$

$ln(\frac{1}{2}) = -kt_{1/2}$

Bear in mind that $ln(\frac{1}{2}) = ln(2^{-1}) = -ln(2)$

$-ln(2) = -kt_{1/2}$

$t_{1/2} = \frac{ln(2)}{k}$

From your equation k = 0.034 so the half life is $\frac{ln(2)}{0.034}$
• May 17th 2009, 04:12 PM
mvho
Thank you very much.

Oh man, I need to review my logs

To summarize, I am given K or some variable. Plug known variable in and solve for Time in this case. I understand looking at your steps, but I have a stupid question to ask:

Why do we ln both sides again? Was it to bring down the exponent?

Its been a really, really, long time..
• May 18th 2009, 09:11 AM
stapel
Yes; the only way to get the power down where you could work with it was to take logs. (Wink)
• May 18th 2009, 09:25 AM
e^(i*pi)
Quote:

Originally Posted by mvho
Thank you very much.

Oh man, I need to review my logs

To summarize, I am given K or some variable. Plug known variable in and solve for Time in this case. I understand looking at your steps, but I have a stupid question to ask:

Why do we ln both sides again? Was it to bring down the exponent?

Its been a really, really, long time..

Yeah, $ln(x) = log_e(x)$ and logs and exponentials are inverse operations like adding and subtracting.

In my post everything except the last two lines is deriving the general half life formula so if you don't need to know it's derivation you can stick with

$
t_{1/2} = \frac{ln(2)}{k}
$

k will always be the number next to t and will be a power along with t as well as k > 0. The units will be 1 over the units of k for any exponent must be dimensionless. If k is in per minute, t will be in minutes
• May 18th 2009, 02:22 PM
mvho
Thanks everyone. Never even knew this site existed, will have to share with everyone.

Math is funny; once you get going, it seems all your tools come back pretty quickly.