24.139 CP - Chemistry: Molecular Nature Of Matter And Change

Question

An ancient sword has a blade from the early Roman Empire, around 100 AD, but the wooden handle, inlaid wooden decorations, leather ribbon, and leather sheath have different styles. Given the following activities, estimate the age of each part. Which part was made near the time of the blade $\mathbf{(}{\mathit{t}}_{\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{ }\mathit{o}\mathit{f}{\mathbf{ }}^{\mathbf{14}}\mathit{C}\mathbf{=}\mathbf{ }\mathbf{5730}\mathit{y}\mathit{r}\mathbf{;}{\mathbf{A}}_{\mathbf{0}}\mathbf{=}\mathbf{15}\mathbf{.}\mathbf{3}\mathit{d}\mathbf{/}\mathit{m}\mathit{i}\mathit{n}\mathbf{.}\mathit{g}\mathbf{)}$?

Part                   (d/min.g)

Handle               10.1

Inlaid wood       13.8

Ribbon               12.1

Sheath                15.0

Step-by-Step Solution

Answer

From the following explanation, it is concluded that the age of the ribbon is closest to the blade.

Step 1: Definition of radioactive Nuclei

A radionuclide (radioactive nuclide, radioisotope, or radioactive isotope) is an “unstable nuclide” having an excess of nuclear energy.

This extra energy can be expelled as:

• Gamma radiation from the nucleus, 
• Transferred to one of its electrons, 
• Released as a conversion electron, or 
• Used to generate and emit a new particle (alpha particle or beta particle) from the nucleus.

Step 2: Explanation for the calculation of time.

An ancient sword has a blade from the early Roman Empire, around 100 A.D. This blade is about $1.9\times {10}^3$years old.

In this problem, we need to identify the rate constant (k) using the formula for half-life t_1/2.

$t_{\frac{1}{2}}=\frac{\ln 2}{k}\to k=\frac{\ln 2}{t_{\frac{1}{2}}}$

 

The known values are substituted.

$$\begin{gathered} k=\frac{\ln 2}{t_{\frac{1}{2}}} \\ =\frac{\ln 2}{5730yr} \\ =\frac{1.2096809\times {10}^{-4}}{yr} \end{gathered}$$

Step 3: Explanation for the calculation of time

Using the expression for finding the ratio of activities, we can solve for the value of t.

$\ln \frac{A_0}{A_t}=kt$

Handle:

The known values are substituted.

$$\begin{gathered} \ln \frac{15.3\frac{disitn.}{\min .g}}{10.1\frac{disitn.}{\min .g}}=\frac{1.2096809x{10}^{-4}}{yr}\left(t\right) \\ t=3433.2806yr \end{gathered}$$

 

Inlaid:

$$\begin{gathered} \ln \frac{15.3\frac{disitn.}{\min .g}}{13.8\frac{disitn.}{\min .g}}=\frac{1.2096809x{10}^{-4}}{yr}\left(t\right) \\ t=852.9782yr \end{gathered}$$

 

Ribbon:

$$\begin{gathered} \ln \frac{15.3\frac{disitn.}{\min .g}}{12.1\frac{disitn.}{\min .g}}=\frac{1.2096809x{10}^{-4}}{yr}\left(t\right) \\ t=1939.7461yr \end{gathered}$$

 

 

Sheath:

$$\begin{gathered} \ln \frac{15.3\frac{disitn.}{\min .g}}{15\frac{disitn.}{\min .g}}=\frac{1.2096809x{10}^{-4}}{yr}\left(t\right) \\ t=163.7012yr \end{gathered}$$