1. Dec 30, 2008

darren t

[Hey I have also been asked this question, A mouse dropping was found in a house in Ireland and had a 14C:12C ratio that was 57% of a modern day standard. Use this value to estimate the aprox date when the mouse visited the house? The half life of 14C is 5715 year

I used the following equation.

t = half life / 0.693 * ln ( A / Ao)

therefore = 5715 / 0.693 * ln ( 100 / 570

therefore t = 4626.42 years?

What do you guys think ???? Am I doing it right??

2. Dec 30, 2008

HallsofIvy

I think I would have done this:
(1/2)xx= 0.57 so x ln(1/2)= ln(0.57) and x= ln(0.57)/ln(1/2)= 0.811 so the mouse dropping has gone through .811 "half lives". Since each half life is 5715 years, that is .811(5715)= 3961 years, not 4626 years.

Your "0.6943" is, of course, ln(2). What do A and Ao mean in your formula? 100 and 570 seem peculiar values. Shouldn't it be either 100/57= 1./.57 or, perhaps better, 57/100= .57?

3. Dec 30, 2008

darren t

Thanks for getting back to me

A is the amount of radioactivity it has when it is new and Ao refers to how much radioactivity is has now!!

Normally this value would be in Curies ect,. but no other data was given!!

Your right the value should be 57 not 570, that was a typo on my end, Sorry!

4. Dec 30, 2008

darren t

hey:

So i did my calculation again

t = 5715 / 0.693 * ln (100 / 75)

t = 8246.75 * ln (1.33)
t = 8246.75 * 0.28

t = 2309.09 years

Its still does not seem right!!!
I think I might go with your way HallsofIvy :)