Asked by Beth
When any radioactive dating method is used, experimental error in the measurement of the sample's activity leads to error in the estimated age. In an application of the radiocarbon dating technique to certain fossils, an activity of 0.15 Bq per gram of carbon is measured to within an accuracy of 15 percent. Find (a) the age of the fossils and (b) the maximum error (both in years) in the value obtained. Assume that there is no error in the 5730-year half-life of 6C14 nor in the value of 0.23 Bq per gram of carbon in a living organism.
I understand that I have to use the equation A=Ao EXP(-wavelength x T) and once i plug in the answer it always turns out to be 7300 or so which is the incorrect answer. Therfore allwing me not to be able to solve B. If anyone can help me in where my calculations went wrong that would be greatly appriciated.
I understand that I have to use the equation A=Ao EXP(-wavelength x T) and once i plug in the answer it always turns out to be 7300 or so which is the incorrect answer. Therfore allwing me not to be able to solve B. If anyone can help me in where my calculations went wrong that would be greatly appriciated.
Answers
Answered by
Elena
A=Aₒ•e(-λ•t).
λ is the decay constant (not the wavelength!!!)
λ=ln2/T, where T is half-life
A/Aₒ= e(- ln2•t/T).
ln(A/Aₒ)= - ln2•t/T
ln(0.15/0.23)=ln(0.652)=
= - 4.27 = -ln2 •t/5730.
t=4.27•5730/ln2=35299 years
15% •35299/100% =5295 years
Ans. 35299±5295 years
λ is the decay constant (not the wavelength!!!)
λ=ln2/T, where T is half-life
A/Aₒ= e(- ln2•t/T).
ln(A/Aₒ)= - ln2•t/T
ln(0.15/0.23)=ln(0.652)=
= - 4.27 = -ln2 •t/5730.
t=4.27•5730/ln2=35299 years
15% •35299/100% =5295 years
Ans. 35299±5295 years
Answered by
Anonymous
this is wrong
Answered by
Aaron
Elena was so close, it looks like a calculator error when calculating part A. The answer is off by a factor of ten. So 3529.9 years, not 35299.
Answered by
Derp
Good Guy Aaron
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