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Chapter 21: Problem 42

A wooden artifact from an Indian temple nas a ' of 42 counts per minute as compared with an activity of 58.2 counts per minute for a standard zero age. From the half-life of \({ }^{14} \mathrm{C}\) decay, 5715 years, calculate the age of the artifact.

Short Answer

Expert verified
The age of the wooden artifact from the Indian temple is approximately 1960.95 years.

Step by step solution

01

Identify the given information

We are given the following information: - Activity of the artifact: 42 counts per minute - Activity of a standard zero age: 58.2 counts per minute - Half-life of \({ }^{14} \mathrm{C}\) decay: 5715 years
02

Understand the radioactive decay formula

The formula for radioactive decay is given by: \[ N_t = N_0 e^{-\lambda t} \] where - \( N_t \) is the remaining number of radioactive atoms at time \(t\), - \( N_0 \) is the initial number of radioactive atoms, - \(\lambda\) is the decay constant, and - \(t\) is the time.
03

Calculate the decay constant

We can find the decay constant (\(\lambda\)) using the half-life formula: \[ T_{1/2} = \frac{\ln(2)}{\lambda} \] Where \(T_{1/2}\) is the half-life. Now, let's calculate the decay constant: \begin{align*} \lambda &= \frac{\ln(2)}{T_{1/2}} \\ &= \frac{\ln(2)}{5715} \\ &\approx 1.21 \times 10^{-4} \,\mathrm{year}^{-1} \end{align*}
04

Find the age of the artifact

Now, let's plug in the given information into the radioactive decay formula: \begin{align*} \frac{N_t}{N_0} &= e^{-\lambda t} \\ \end{align*} Where \( N_t \) is the remaining activity, which is 42 counts per minute, and \( N_0 \) is the initial activity, which is 58.2 counts per minute. \begin{align*} \frac{42}{58.2} &= e^{- (1.21 \times 10^{-4})t} \\ 0.721 &= e^{- (1.21 \times 10^{-4})t} \end{align*} Taking the natural logarithm of both sides, we get: \[ \ln(0.721) = - (1.21 \times 10^{-4})t \] Now, let's solve for \(t\): \begin{align*} t &= \frac{\ln(0.721)}{-(1.21 \times 10^{-4})} \\ &\approx 1960.95 \, \mathrm{years} \end{align*} The age of the artifact is approximately 1960.95 years.

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Most popular questions from this chapter

Tests on human subjects in Boston in 1965 and \(1966,\) following the era of atomic bomb testing, revealed average quantities of about \(2 \mathrm{pCi}\) of plutonium radioactivity in the average person. How many disintegrations per second does this level of activity imply? If each alpha particle deposits $8 \times 10^{-13} \mathrm{~J}\( of energy and if the average person weighs \)75 \mathrm{~kg},$ calculate the number of grays and sieverts of radiation in 1 yr from such a level of plutonium.

A portion of the Sun's energy comes from the reaction $$ 4{ }_{1}^{1} \mathrm{H} \longrightarrow{ }_{2}^{4} \mathrm{He}+2{ }_{1}^{0} \mathrm{e} $$ which requires a temperature of \(10^{6}\) to \(10^{7} \mathrm{~K}\). Use the mass of the helium-4 nucleus given in Table 21.7 to determine how much energy is released per mol of hydrogen atoms.

One nuclide in each of these pairs is radioactive. Predict which is radioactive and which is stable: \((\mathbf{a}){ }_{20}^{40} \mathrm{Ca}\) and \({ }_{20}^{45} \mathrm{Ca},\) (b) \({ }^{12} \mathrm{C}\) and \({ }^{14} \mathrm{C}\) (c) lead-206 and thorium-230. Explain your choice in each case.

Chlorine has two stable nuclides, \({ }^{35} \mathrm{Cl}\) and ${ }^{37} \mathrm{Cl}\(. In contrast, \){ }^{36} \mathrm{Cl}$ is a radioactive nuclide that decays by beta emission. (a) What is the product of decay of \({ }^{36} \mathrm{Cl}\) ? (b) Based on the empirical rules about nuclear stability, explain why the nucleus of \({ }^{36} \mathrm{Cl}\) is less stable than either ${ }^{35} \mathrm{Cl}\( or \){ }^{37} \mathrm{Cl}$.

A laboratory rat is exposed to an alpha-radiation source whose activity is \(14.3 \mathrm{mCi}\). (a) What is the activity of the radiation in disintegrations per second? In becquerels? (b) The rat has a mass of $385 \mathrm{~g}\( and is exposed to the radiation for \)14.0 \mathrm{~s}$, absorbing \(35 \%\) of the emitted alpha particles, each having an energy of $9.12 \times 10^{-13} \mathrm{~J}$. Calculate the absorbed dose in millirads and grays. (c) If the RBE of the radiation is \(9.5,\) calculate the effective absorbed dose in mrem and Sv.
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