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The figure shows two images of Jesus. Left image is very faint and hardly visible but the right image shows a much clearer picture.
Part of the Shroud of Turin, which shows a remarkable negative imprint likeness of Jesus complete with evidence of crucifixion wounds. The shroud first surfaced in the 14th century and was only recently carbon-14 dated. It has not been determined how the image was placed on the material. (credit: Butko, Wikimedia Commons)

How old is the shroud of turin?

Calculate the age of the Shroud of Turin given that the amount of 14 C size 12{"" lSup { size 8{"14"} } C} {} found in it is 92% of that in living tissue.

Strategy

Knowing that 92% of the 14 C remains means that N / N 0 = 0 . 92 size 12{N/N rSub { size 8{0} } =0 "." "92"} {} . Therefore, the equation N = N 0 e λt size 12{N=N rSub { size 8{0} } e rSup { size 8{ - λt} } } {} can be used to find λt size 12{λt} {} . We also know that the half-life of 14 C is 5730 y, and so once λt size 12{λt} {} is known, we can use the equation λ = 0 . 693 t 1 / 2 size 12{λ= { {0 "." "693"} over {t rSub { size 8{1/2} } } } } {} to find λ size 12{λ} {} and then find t size 12{t} {} as requested. Here, we postulate that the decrease in 14 C is solely due to nuclear decay.

Solution

Solving the equation N = N 0 e λt size 12{N=N rSub { size 8{0} } e rSup { size 8{ - λt} } } {} for N / N 0 size 12{N/N rSub { size 8{0} } } {} gives

N N 0 = e λt . size 12{ { {N} over {N rSub { size 8{0} } } } =e rSup { size 8{-λt} } } {}

Thus,

0 . 92 = e λt . size 12{0 "." "92"=e rSup { size 8{ - λt} } } {}

Taking the natural logarithm of both sides of the equation yields

ln 0 . 92 = –λt size 12{"ln "0 "." "92""=-"λt} {}

so that

0 . 0834 = λt . size 12{ - 0 "." "0834"= - λt} {}

Rearranging to isolate t size 12{t} {} gives

t = 0 . 0834 λ . size 12{t= { {0 "." "0834"} over {λ} } } {}

Now, the equation λ = 0 . 693 t 1 / 2 size 12{λ= { {0 "." "693"} over {t rSub { size 8{1/2} } } } } {} can be used to find λ size 12{λ} {} for 14 C size 12{"" lSup { size 8{"14"} } C} {} . Solving for λ size 12{λ} {} and substituting the known half-life gives

λ = 0 . 693 t 1 / 2 = 0 . 693 5730 y . size 12{λ= { {0 "." "693"} over {t rSub { size 8{1/2} } } } = { {0 "." "693"} over {"5730"" y"} } } {}

We enter this value into the previous equation to find t size 12{t} {} :

t = 0 . 0834 0 . 693 5730 y = 690 y. size 12{t= { {0 "." "0834"} over { { {0 "." "693"} over {"5730"" y"} } } } ="690"" y"} {}

Discussion

This dates the material in the shroud to 1988–690 = a.d. 1300. Our calculation is only accurate to two digits, so that the year is rounded to 1300. The values obtained at the three independent laboratories gave a weighted average date of a.d. 1320 ± 60 size 12{"1320" +- "60"} {} . The uncertainty is typical of carbon-14 dating and is due to the small amount of 14 C size 12{"" lSup { size 8{"14"} } C} {} in living tissues, the amount of material available, and experimental uncertainties (reduced by having three independent measurements). It is meaningful that the date of the shroud is consistent with the first record of its existence and inconsistent with the period in which Jesus lived.

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There are other forms of radioactive dating. Rocks, for example, can sometimes be dated based on the decay of 238 U . The decay series for 238 U ends with 206 Pb , so that the ratio of these nuclides in a rock is an indication of how long it has been since the rock solidified. The original composition of the rock, such as the absence of lead, must be known with some confidence. However, as with carbon-14 dating, the technique can be verified by a consistent body of knowledge. Since 238 U has a half-life of 4 . 5 × 10 9 y, it is useful for dating only very old materials, showing, for example, that the oldest rocks on Earth solidified about 3 . 5 × 10 9 size 12{3 "." 5 times "10" rSup { size 8{9} } } {} years ago.

Activity, the rate of decay

What do we mean when we say a source is highly radioactive? Generally, this means the number of decays per unit time is very high. We define activity     R size 12{R} {} to be the rate of decay    expressed in decays per unit time. In equation form, this is

R = Δ N Δ t size 12{R= { {ΔN} over {Δt} } } {}

where Δ N size 12{ΔN} {} is the number of decays that occur in time Δ t size 12{Δt} {} . The SI unit for activity is one decay per second and is given the name becquerel    (Bq) in honor of the discoverer of radioactivity. That is,

1 Bq = 1 decay/s. size 12{1" Bq"="1 decay/s"} {}

Activity R size 12{R} {} is often expressed in other units, such as decays per minute or decays per year. One of the most common units for activity is the curie    (Ci), defined to be the activity of 1 g of 226 Ra , in honor of Marie Curie’s work with radium. The definition of curie is

Questions & Answers

what is angular velocity
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an 8.0 capacitor is connected by to the terminals of 60Hz whoes rms voltage is 150v. a.find the capacity reactance and rms to the circuit
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Nwafor Reply
is the study of matter in relation to energy
Kintu
a submersible pump is dropped a borehole and hits the level of water at the bottom of the borehole 5 seconds later.determine the level of water in the borehole
Obrian Reply
what is power?
aron Reply
power P = Work done per second W/ t. It means the more power, the stronger machine
Sphere
e.g. heart Uses 2 W per beat.
Rohit
A spherica, concave shaving mirror has a radius of curvature of 32 cm .what is the magnification of a persons face. when it is 12cm to the left of the vertex of the mirror
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1.75cm
Ridwan
my name is Abu m.konnek I am a student of a electrical engineer and I want you to help me
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the magnification k = f/(f-d) with focus f = R/2 =16 cm; d =12 cm k = 16/4 =4
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A weather vane is some sort of directional arrow parallel to the ground that may rotate freely in a horizontal plane. A typical weather vane has a large cross-sectional area perpendicular to the direction the arrow is pointing, like a “One Way” street sign. The purpose of the weather vane is to indicate the direction of the wind. As wind blows pa
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the same behavior thru the prism out or in water bud abbot
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If this will experimented with a hollow(vaccum) prism in water then what will be result ?
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What is the far point of a person whose eyes have a relaxed power of 50.5 D?
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What is the far point of a person whose eyes have a relaxed power of 50.5 D?
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rerry Reply
try to read several books on phy don't just rely one. some authors explain better than other.
Ju
And don't forget to check out YouTube videos on the subject. Videos offer a different visual way to learn easier.
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Practice Key Terms 8

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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