Cosmic rays impinging on our atmosphere generate radioactive \({ }^{14} \mathrm{C}\) from \({ }^{14} \mathrm{~N}\) nuclei. \(^{78}\) These \({ }^{14} \mathrm{C}\) atoms soon team up with oxygen to form \(\mathrm{CO}_{2}\), so that plants absorbing \(\mathrm{CO}_{2}\) from the air will have about one in a trillion of their carbon atoms in this form. Animals eating these plants \(^{79}\) will also have this fraction of carbon in their bodies, until they die and stop cycling carbon into their bodies. At this point, the fraction of carbon atoms in the form of \({ }^{14} \mathrm{C}\) in the body declines, with a half life of 5,715 years. If you dig up a human skull, and discover that only one-eighth of the usual one-trillionth of carbon atoms are \({ }^{14} \mathrm{C}\), how old do you deem the skull to be?

Short Answer

Expert verified
Answer: The age of the skull is approximately 17,145 years old.

Step by step solution

01

Identify initial and final fractions

The initial fraction of carbon atoms as \({ }^{14} \mathrm{C}\) when the organism was alive is given to be one in a trillion (1/1,000,000,000,000). In the skull, the final fraction of carbon atoms as \({ }^{14} \mathrm{C}\) is one-eighth of the initial fraction i.e., (1/8)*(1/1,000,000,000,000).
02

Calculate the ratio between initial and final fractions

We need to find the ratio between the final and initial fractions of \({ }^{14} \mathrm{C}\). This ratio is equal to the fraction of carbon atoms left, after a certain number of half-lives. Ratio = (Final fraction) / (Initial fraction) = [(1/8)*(1/1,000,000,000,000)] / (1/1,000,000,000,000)
03

Ratio simplification

Now, let's simplify the ratio: Ratio = (1/8)*(1/1,000,000,000,000) / (1/1,000,000,000,000) = (1/8)
04

Calculate the number of half-lives

We will use the following formula for radioactive decay, which relates the remaining fraction of \({ }^{14} \mathrm{C}\) and the number of half-lives (n): Remaining fraction = (1/2)^n Here, the remaining fraction is the ratio calculated in step 3. Let's plug in the values and solve for n: (1/8) = (1/2)^n Take the logarithm of both sides: log(1/8) = n * log(1/2) Therefore, n = log(1/8) / log(1/2)
05

Calculate the age of the skull

Now that we have the number of half-lives (n) that have passed, we can calculate the age of the skull using the following formula: Age = n * Half-life Plug in the values: Age = (log(1/8) / log(1/2)) * 5,715 years After solving the equation, we get the age of the skull: Age ≈ 17,145 years So, the age of the skull is approximately 17,145 years old.

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Most popular questions from this chapter

A particular fission of \({ }^{235} \mathrm{U}+\mathrm{n}(\) total \(A=236)\) breaks up. One fragment has \(Z=54\) and \(N=86\), making it \({ }^{140} \mathrm{Xe} .\) If no extra neutrons are produced in this event, what must the other fragment be, so all numbers add up? Refer to a periodic table (e.g., Fig. B.1; p. 375 ) to learn which element has the corresponding \(Z\) value, and express the result in the notation \({ }^{\mathrm{A}} \mathrm{X}\).

If a friend creates a nucleus whose half-life is 4 hours and gives it to you at noon, what is the probability that it will not have decayed by noon the following day?

Based on the calculation that 18 TW would require an annual cube of seawater \(300 \mathrm{~m}\) on a side to provide enough deuterium, what is your personal share as one of 8 billion people on earth, in liters? Could you lift this yourself? One cubic meter is \(1,000 \mathrm{~L}\).

A large boulder whose mass is \(1,000 \mathrm{~kg}\) having a specific heat capacity of \(1,000 \mathrm{~J} / \mathrm{kg} /{ }^{\circ} \mathrm{C}\) is heated from \(0^{\circ} \mathrm{C}\) to a glowing \(1,800^{\circ} \mathrm{C}\). How much more massive is it, assuming no atoms have been added or subtracted?

Which of the following is true about the fragments from \(\mathrm{a}^{235} \mathrm{U}\) fission event? a) any number of fragments ( 2 through 235 ) can be produced b) a small number of fragments will emerge ( 2 to 5 ) c) two nearly identical fragments will emerge d) two fragments of distinctly different size will emerge e) the fission is an alpha decay: a small piece having \(A=4\) is emitted

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