Solve for k using k = 0.693/t1/2, then substitute k into the following:
ln(No/N) = kt
Use No = 8
N = 1
k from above.
t = unknown.
Calculate the age of a plant sample that shows about one-eighth of the carbon 14 of a living sample. The half-life of carbon-14 is about 5760 a.
Help please...
6 answers
Is that
0.693/
t 1/2
0.693 over t one half ?
0.693/
t 1/2
0.693 over t one half ?
yes. k = 0.693/5760 = ? yr^-1
In the start of the equation .. for
In(No/N) I understand the No/N is the percentage that the sample is of carbon 14 compared to a living sample, but how do you get the In ? where does it come from ?
In(No/N) I understand the No/N is the percentage that the sample is of carbon 14 compared to a living sample, but how do you get the In ? where does it come from ?
This is the integrated equation for a first order rate reaction. All of them are
ln(No/N) = kt
I can probably find a place on the web that derives the equation if you really want it. There isn't much to understand about it at this point except how to use it. The problem tells you that the radioactivity is 1/8 of the original; that's the 8/1. No is what you start with and N is what is there currently.
ln(No/N) = kt
I can probably find a place on the web that derives the equation if you really want it. There isn't much to understand about it at this point except how to use it. The problem tells you that the radioactivity is 1/8 of the original; that's the 8/1. No is what you start with and N is what is there currently.
http://en.wikipedia.org/wiki/Rate_equation#First-order_reactions
1 answer