SRL Archive - Mathematics

To histroically date objects less than 50,000 years old, carbon 14 dating is used. Carbon 14 has a half life of 5730 years. Before death animals and vegetation have a reading for carbon 14 of 12.5 counts per minute on a Geiger counter. Show that the decay rate of carbon 14 is -1.21 *10^(-4). If a piece of wood from an excavation site has a reading of 7 counts per minute, show that the wood's age is approximately 4800 years.

I did the first part anf ound the rate.
The second part, I don't even know where to start. What does it mean?:huh:

You lost me at Carbon, sorry bro. :)

It all about using A = Ao * e^(k*t)

Amount equals initial amount times the constant e raised to the rate k times t as in time.

From: duckysd
It all about using A = Ao * e^(k*t)

Amount equals initial amount times the constant e raised to the rate k times t as in time.

What you've said there is correct, A = InitialA * e^(-λt), where λ is called the "decay constant".

It is known that λ * Half Life = ln 2.

So from that, we are able to find the decay constant.

Substitute λ into A = InitialA * e^(-λt), where initialA is the count rate prior to the object being dead, and A is the count rate at the particular time you are trying to find, then use logarithms to find t.

Hopefully that helps.

seeing as you said you got the first part, ill skip that. After solving and stuff using pert/etc… youll end up with A = 12.5 * e^(ln2 *t/5730) orrr… to simplify thats A=12.5*2^(t/5730). (ill do this both ways with base e and base 2)
From here its saying that A = 7. soo… 7= 12.5 *e^(ln2 *t /5730) = 12.5 *2^(t/5730)
Divide both sides by 12.5… 7/12.5 = .56 = e^(ln2 *t /5730) = 2^(t/5730)
Now is where each method gets a little different… In case e, take the natural log of both sides… in case 2, take log2 of both sides… (for simplicity sake seeing as you probably keep things with e, we will switch to that way right meow.)
ln(.56) = ln2 *t /5730.
multiply both sides by 5730… 5730 ln(.56) = ln2*t
divide both sides by ln2… 5730 ln(.56)/ln2 = t. soo… using a calculator for those natural logs and we get… t= -4793.15
4793 is pretty close to 4800, so you have just shown it…

As a side note, i ended up with a negative value for my t, if in the original equation, which i assume you did, made t negative, the final value for t would be positive. I like to keep t positive, because if i get a negative t, that just means its in the past. but do it which ever way your teacher wants you to do it.