Question

The cloth shroud from around a mummy is found to have \(a^{14} \mathrm{C}\) activity of \(9.7\) disintegrations per minute per gram of carbon as compared with living organisms that undergo \(16.3\) disintegrations per minute per gram of carbon. From the halflife for \({ }^{14} \mathrm{C}\) decay, \(5715 \mathrm{yr}\), calculate the age of the shroud.

Short answer

The age of the shroud is approximately \(4978\) years.

Step-by-step solution

Identify the given data

We are given the following information: - The activity of the shroud, \(A_s = 9.7 \ \text{disintegrations/min/gram}\). - The activity of living organisms, \(A_o = 16.3 \ \text{disintegrations/min/gram}\). - The half-life of \({ }^{14}\mathrm{C}\), \(t_{1/2} = 5715 \ \text{years}\).

Calculate the decay constant

We will use the half-life formula to find the decay constant, denoted by \(\lambda\). The half-life formula is: \[t_{1/2} = \frac{\ln{2}}{\lambda}\] Solving for \(\lambda\) gives: \[\lambda = \frac{\ln{2}}{t_{1/2}}\] Now, plug in the given half-life and find the decay constant: \[\lambda = \frac{\ln{2}}{5715} \approx 1.21 \times 10^{-4} \ \text{yr}^{-1}\]

Set up the decay formula

Now, we will use the radioactive decay formula to find the time that has passed, denoted as \(t\): \[\frac{A_s}{A_o} = e^{-\lambda t}\] Plug in the given activities and the decay constant calculated in Step 2: \[\frac{9.7}{16.3} = e^{- (1.21 \times 10^{-4})t}\]

Solve for time, \(t\)

To solve for \(t\), we will first take the natural logarithm of both sides: \[\ln{\left(\frac{9.7}{16.3}\right)} = - (1.21 \times 10^{-4})t\] Next, divide both sides by the decay constant to isolate \(t\): \[t = \frac{\ln{\left(\frac{9.7}{16.3}\right)}}{- (1.21 \times 10^{-4})}\] Now, plug in the numbers and find the value of \(t\): \[t \approx \frac{\ln{(0.5951)}}{- (1.21 \times 10^{-4})} \approx 4977.95 \ \text{years}\]

Finding the age of the shroud

Now that we have found the time elapsed, we can conclude that the age of the shroud is approximately \(4978\) years.

Key concepts

Half-life Calculation

Half-life is a crucial concept in understanding the process of radioactive decay, including radiocarbon dating, and helps determine how long it takes for half of a given amount of a radioactive isotope to decay. This predictable rate of decay allows scientists to date ancient objects.
To calculate the half-life ( t_{1/2} ), the equation t_{1/2} = \frac{\ln{2}}{\lambda} is used, where \( \lambda \) is the decay constant.

• This formula is derived from the natural logarithmic constant which represents the continuous rate of decay.
• The value \( \ln{2} \) is approximately 0.693 and is a fixed natural logarithmic value indicating the decay over each half-life period.

Let's apply this formula in the context of our exercise: given a half-life of 5715 years for \(^{14}C\), helps us find the decay constant which can be further used to analyze ancient carbon samples like the shroud in the exercise.

Decay Constant

The decay constant ( \lambda ) is a parameter that quantifies the rate at which a radioactive isotope decays per unit time. It is essential for understanding how fast or slow a particular radioactive substance will reduce its quantity over time.
By using the half-life relation, the decay constant can be determined as \lambda = \frac{\ln{2}}{t_{1/2}} .

• This constant provides a way to directly relate the activity of radioactive material to the amount remaining.
• A larger decay constant means faster decay, while a smaller one indicates slower decay rates.

In the original problem, substituting the half-life of \(^{14}C\) gave us \lambda = 1.21 \times 10^{-4} \, \text{yr}^{-1} , which is key to calculating the age of the shroud using the radioactive decay analysis.

Radioactive Decay Formula

The radioactive decay formula is pivotal in calculating the age of an artifact by comparing the activity levels of remaining and original material. It is expressed as \( \frac{A_s}{A_o} = e^{-\lambda t} \). This formula connects the decay constant, the initial and remaining activity of the sample, and the time that has passed.

• \( A_s \) represents the current activity, while \( A_o \) is the original activity.
• \( e \) denotes Euler's number, a constant approximately equal to 2.718.

To solve for the age \( t \), one can take logs of both sides resulting in \( \ln{\left(\frac{A_s}{A_o}\right)} = -\lambda t \) and subsequently calculate \( t \) given the decay constant and activities.
By inserting the observed activities from the original question and solving, we estimate that the shroud is roughly 4978 years old. This formula is remarkable as it transforms the data of disintegrations per minute per gram into estimations of historical time elapsed since the death of an organism.