# 31.5 Half-life and activity  (Page 3/16)

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## How old is the shroud of turin?

Calculate the age of the Shroud of Turin given that the amount of ${}^{\text{14}}\text{C}$ found in it is 92% of that in living tissue.

Strategy

Knowing that 92% of the ${}^{\text{14}}\text{C}$ remains means that $N/{N}_{0}=0\text{.}\text{92}$ . Therefore, the equation $N={N}_{0}{e}^{-\mathrm{\lambda t}}$ can be used to find $\mathrm{\lambda t}$ . We also know that the half-life of ${}^{\text{14}}\text{C}$ is 5730 y, and so once $\mathrm{\lambda t}$ is known, we can use the equation $\lambda =\frac{0\text{.}\text{693}}{{t}_{1/2}}$ to find $\lambda$ and then find $t$ as requested. Here, we postulate that the decrease in ${}^{\text{14}}\text{C}$ is solely due to nuclear decay.

Solution

Solving the equation $N={N}_{0}{e}^{-\mathrm{\lambda t}}$ for $N/{N}_{0}$ gives

$\frac{N}{{N}_{0}}={e}^{-\mathrm{\lambda t}}\text{.}$

Thus,

$0\text{.}\text{92}={e}^{-\mathrm{\lambda t}}\text{.}$

Taking the natural logarithm of both sides of the equation yields

$\text{ln}\phantom{\rule{0.25em}{0ex}}0\text{.}\text{92}=\mathrm{–\lambda t}$

so that

$-0\text{.}\text{0834}=-\mathrm{\lambda t}\text{.}$

Rearranging to isolate $t$ gives

$t=\frac{0\text{.}\text{0834}}{\lambda }\text{.}$

Now, the equation $\lambda =\frac{0\text{.}\text{693}}{{t}_{1/2}}$ can be used to find $\lambda$ for ${}^{\text{14}}\text{C}$ . Solving for $\lambda$ and substituting the known half-life gives

$\lambda =\frac{0\text{.}\text{693}}{{t}_{1/2}}=\frac{0\text{.}\text{693}}{\text{5730 y}}\text{.}$

We enter this value into the previous equation to find $t$ :

$t=\frac{0\text{.}\text{0834}}{\frac{0\text{.}\text{693}}{\text{5730 y}}}=\text{690 y.}$

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. $\text{1320}±\text{60}$ . The uncertainty is typical of carbon-14 dating and is due to the small amount of ${}^{\text{14}}\text{C}$ 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.

There are other forms of radioactive dating. Rocks, for example, can sometimes be dated based on the decay of ${}^{\text{238}}\text{U}$ . The decay series for ${}^{\text{238}}\text{U}$ ends with ${}^{\text{206}}\text{Pb}$ , 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 ${}^{\text{238}}\text{U}$ has a half-life of $4\text{.}5×{\text{10}}^{9}$ y, it is useful for dating only very old materials, showing, for example, that the oldest rocks on Earth solidified about $3\text{.}5×{\text{10}}^{9}$ 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     $R$ to be the rate of decay    expressed in decays per unit time. In equation form, this is

$R=\frac{\Delta N}{\Delta t}$

where $\text{Δ}N$ is the number of decays that occur in time $\text{Δ}t$ . 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,

$1\phantom{\rule{0.25em}{0ex}}\text{Bq}=\text{1 decay/s.}$

Activity $R$ 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 ${}^{\text{226}}\text{Ra}$ , in honor of Marie Curie’s work with radium. The definition of curie is

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