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Radiocarbon dating suggests that the eruption eruption occurred around 1360 BC,

ID: 1021359 • Letter: R

Question

Radiocarbon dating suggests that the eruption eruption occurred around 1360 BC, through other records place the eruption of Thera in the year 1628 BC. What is the percent difference in the^14C (t_v2 - 5730 yr) decay rate in biological sample from these two dates? Assuming that the year is 2007, What is the^14C/^12 C ratio for the grain harvested in 2007? Radiocarbon dating of blackened grains from the site of ancient Jericho provides date of 1315 BC plusminus 13 years for the fall of the city, what is the^14 C/^12 C ratio in the blackened grains compared with that of the grain harvested in 2007? Calculate the percent decrease in^14 C/^12 C ratio comparing grain from 1315 BC to 2007 AD. (^14 C/^12_2007 AD = (^14 C/^12_1315 BC = % decrease =

Explanation / Answer

(a): Given half-life, t1/2 = 0.693 / k = 5730 yr

=> k = 0.693 /  t1/2 = 0.693 / 5730 yr = 1.21x10-4 yr-1  

Since radioactive decay is first order reaction,

kt = ln(Ao / At)

When t1 = 1360 BC = 1360 + 2007 = 3367 yr

=> kt1 = ln(Ao / At1)

=> 1.21x10-4 yr-1 x 3367 yr = ln(Ao / At1)

=> Ao / At1 = exp (1.21x10-4 yr-1 x 3367 yr) = 1.503 ---- (1)

When t2 = 1628 BC = 1628 + 2007 = 3635 yr

=> kt2 = ln(Ao / At2)

=> 1.21x10-4 yr-1 x 3635 yr = ln(Ao / At2)

=> Ao / At2 = exp (1.21x10-4 yr-1 x 3635 yr) = 1.552 ---- (2)

Dividing equation(2) by equation(1)

=> At1 / At2 = 1.552 / 1.503 = 1.0326

=> At1 =  1.0326xAt2

Hence percentage difference = 3.26 % (answer)

(b): Since the grain is living during harvest, the ratio of C14 / C12 at 2007 will be equal to the ratio in normal air which is 1.0x10-12 (1 parts per trillion)

=> (14C / 12C)2007 AD = 1.0x10-12 (answer)

As the time passes the above ratio will start decreasing

Now when t = 1315 BC = 1315 + 2007 = 3322 yr

Applying integrated rate law for 1st order

kt = ln(N0 / Nt)

=> 1.21x10-4 yr-1 x 3322 yr = ln(No / Nt)

=> No / Nt = exp (1.21x10-4 yr-1 x 3322 yr) = 1.49447

=> Nt = 0.6691xNo

Hence (14C / 12C)1315 BC = 0.6691x10-12 = 6.69x10-13 (answer)

PErcentage decrease = [(1 - 0.6691) / 1]x100 = 33.1 % (answer)