1. Leaf fossils in shale contain 21 ppm of 14C and 651 ppm of 14N. What elements

Question

1. Leaf fossils in shale contain 21 ppm of 14C and 651 ppm of 14N.
What elements do C and N stand for?
Determine the age of the fossils.
The half-life of 14C is 5730 yrs.

2. If the above leaf fossils contain 24 ppm 14C and 640 ppm of 14N, determine the age of the fossils.
Here the number of half-lives needed is not an integer number, but is a fractional number. In this situation you must use a mathematical expression, or a graph, to find the solution. The mathematical expression is as follows:
Age in half-lives = 3.322 X log10[(p+d)/p]. Apply this mathematical expression to determine the age.

3. A sandstone contains grains with 1.9 ppm of 238U and 0.24 ppm 206Pb. How old is the rock? The half-life of 238U (the parent) as it decays to 206Pb (the daughter) is 4.46 Gyr (4.46 billion years; also written as 4.46 x 109yrs). Sandstone is composed of cemented grains that pre-existed the time of sedimentation and subsequent cementation. You have to take this into account when composing your answer.

Explanation / Answer

1

14C-carbon

14N-Nitrogen

Radio acitve decay is given by

N=No(1/2)t/t1/2

21 =651 (1/2)t/5730

ln(21/651) = (t/5730)*ln(1/2)

t=28387.5 years

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