Question

. A living specimen in equilibrium with the atmosphere contains one atom of 14C (half-life 5 5 730 yr) for every 7.70 3 1011 stable carbon atoms. An archeological sample of wood (cellulose, C12H22O11) contains 21.0 mg of carbon. When the sample is placed inside a shielded beta counter with 88.0% counting efficiency, 837 counts are accumulated in one week. We wish to find the age of the sample. (a) Find the number of carbon atoms in the sample. (b) Find the number of carbon-14 atoms in the sample. (c) Find the decay constant for carbon-14 in inverse seconds. (d) Find the initial number of decays per week just after the specimen died. (e) Find the corrected number of decays per week from the current sample. (f) From the answers to parts (d) and (e), find the time interval in years since the specimen died.

Answer 1

(a) 1.05 × 10²¹ atoms of C

(b) 1.37 × 10⁹ atoms of C-14

(c) k = 3.13 × 10⁻¹² s⁻¹

(d) 951 decays/wk

(e) 9950 yr

Step-by-step explanation:

(a) Number of C atoms

(i) Moles of C

`\text{Moles of C} = \text{0.0210 g C} * \frac{\text{1 mol C}}{\text{12.01 g C}} = 1.749 * 10^(-3)\text{ mol C}`

(ii) Number of C atoms
`\text{No. of atoms} = 1.749 * 10^(-3) \text{ mol C} * \frac{6.022 * 10^(23)\text{ atoms C}}{\text{1 mol C}} =\mathbf{1.05 * 10^(21)} \textbf{ atoms C}`

(b) Number of C-14 atoms
`\text{No. of C-14 atoms} =1.05 * 10^(21) \text{ atoms C} * \frac{\text{1 C-14 atom}}{7.7 * 10^(11)\text{ C atoms}} = \mathbf{1.37 * 10^(9)}\textbf{ C-14 atoms}`

(c) Decay constant

`t_(1/2) = (\ln2)/(k)\\\\k = (\ln 2 )/( t_(1/2)) = \frac{\ln 2 }{ \text{ 5730 yr}} = 1.210 * 10^(-4) \text{ yr}^(-1)\\\\k = 1.210 * 10^(-4) \text{ yr}^(-1) * \frac{\text{1 yr}}{\text{365.25 da}} * \frac{\text{1 da}}{\text{24 h}} * \frac{\text{1 h}}{\text{3600 s}} =\mathbf{3.83 * 10^(-12)} \textbf{ s}^{\mathbf{-1}}`

(d) Corrected decays

The counter is only 88.0 % efficient.

`\text{Corrected decays} = \text{837 detected/wk} * \frac{\text{100 actual}}{\text{88.0 detected }} = \text{951 actual/week}`

(e) Age of specimen

The equation for the activity A of a sample is

A = kN

When the specimen was alive, the activity was

`A = 1.37 * 10^(9)\text{ atoms}*3.834 * 10^(-12) \text{ s}^(-1) * \frac{\text{3600 s}}{\text{1 h}}* \frac{\text{24 h}}{\text{1 da}} * \frac{\text{7 da}}{\text{1 wk}}= \text{3170 atoms/wk}`

Now, the activity of the specimen is 951 atoms/wk.

`\begin{array}{rcl}\ln \left ( (A_(0))/(A) \right )& =& kt\\\\\ln \left ( (3170)/(951) \right )& =&1.21 * 10^(-4) \text{ yr}^(-1)* t\\\\\ln 3.333& =&1.210 * 10^(-4) \text{ yr}^(-1) * t\\\\1.204 & =& 1.210 * 10^(-4) \text{ yr}^(-1)* t\\\\t &=& \frac{\text{1.204}}{1.210 * 10^(-4) \text{ yr}^(-1)* t}\\\\& = & \textbf{9950 yr}\\\end{array}\\\text{The age of the specimen is $\textbf{9950 yr}$}`