The technique known as potassium-argon dating is used to date volcanic rock and

Question

The technique known as potassium-argon dating is used to date volcanic rock and ash, and thus establish dates for nearby fossils, like this hominid skull.(Figure 1) The potassium isotope 40K has a 1.28-billion-year half-life and is naturally present at very low levels. Its most common decay mode is beta-minus decay into the stable isotope 40Ca, but 10.9% of its decays are beta-plus which result in the creation of the stable isotope 40Ar. The high temperatures in volcanoes drive argon out of solidifying rock and ash, so there is no argon in newly formed material. After the rock and ash have completely solidified, however, argon produced in the decay of 40K is trapped, so 40Ar builds up steadily over time. Fairly accurate dating of an object is therefore possible by measuring the ratio of the number of atoms of 40Ar to 40K in nearby volcanic rocks. Part A If in a sample of volcanic rock found near the fossil shown, the ratio of argon to potassium atoms is NArNK= 5.9400×104, what fraction of the original potassium atoms are left? (We have to be very precise here, so please give your answer to six significant figures!) Express your answer using six significant figures. NKN0,K = .89058 SubmitHintsMy AnswersGive UpReview Part Incorrect; Try Again Part B - How old is the sample (and therfore the fossil)? Give your answer in millions of years (to two significant figures). Express your answer using two significant figures. t = 3.35 million years SubmitHintsMy AnswersGive UpReview Part Incorrect; Try Again

Explanation / Answer

(a)

Given that,

NAr / NK = 0.000594

consider 0.000564 mg of Ar come from decay of 1 mg K.

x * (10.9/100) = 0.000594

x = 0.000594 / 0.109

x = 0.00544

(b)

N(t) = No*e^(-kt)

No / 2 = No*e^(-k*1.28*10^9)

k = -ln(1/2) / 1.28*10^9

k = ln(2) / 1.28*10^9

k decay 10.9 % into argon.

time elepsed t = ln (kf / kf + Arf/0.109) / (ln2 / 1.28*10^9)

t = ln (0.998)*1.28*10^9 / -ln2

t = -0.00149 * 1.28*10^9 / -0.693

t = 2.75 million years

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