Carbon dating. The half-life of C-14 is about 5730 yr. a) Archeologists find a piece of cloth painted with organic dyes. Analysis of the dye in the cloth shows that only 77% of the C-14 originally in the dye remains. When was the cloth painted? b) A well-preserved piece of wood found at an archeological site has 6.2% of the C-14 that it had when it was alive. Estimate when the wood was cut.

- t = time of decay (or, age of object of interest; i.e., not everything is organic but the 1st order decay equation is good for non-organic objects (rocks) also. Analysts just use a different decay standard => K-40 → Ar-40 + β).

Solving the decay equation for time (t) ...

t = ln(A/A₀)/-k

Applying to problem cases...

Case I => %C-14 remaining = 77%

t = ln(A/A₀)/-k = ln(77/100)/-1.2x10⁻⁴ years = 2178 yrs ~ 2200 yrs

Case II => %C-14 remaining = 6.2%

t = ln(A/A₀)/-k = ln(6.2/100)/-1.2x10⁻⁴ years = 23,172 yrs ~ 23,000 yrs

znk znk · Answer 2 · 2019-06-22T19:02:22+02:00

Answer:

[tex]\boxed{\text{a) 2160 yr ago; b) 23 000 yr ago}}[/tex]

Explanation:

Two important equations in radioactive decay are

[tex](1) \qquad \ln \dfrac{N_{0} }{N_{t}} = kt\\\\(2) \qquad t_{\frac{1}{2}} = \dfrac{\ln2}{k }[/tex]

We use them for carbon dating.

a) The dye

Data:

[tex]t_{\frac{1}{2}} = \text{5730 yr}\\\\N_{t} = 0.77 N_{0}[/tex]

Calculations:

[tex]\text{From Equation (2)}\\\\t_{\frac{1}{2}} = \dfrac{\ln2}{\text{5730 yr}} = 1.21 \times 10^{-4} \text{ yr}^{-1}\\\\\text{From Equation (1)}\\\\\ln \dfrac{N_{0} }{0.77N_{0}} = 1.21 \times 10^{-4}t \text{ yr}^{-1}\\\\\ln\dfrac{1}{0.77} = 1.21 \times 10^{-4}t \text{ yr}^{-1}\\\\-\ln0.77 = 0.261 = 1.21 \times 10^{-4}t \text{ yr}^{-1}\\\\t = \dfrac{0.261}{ 1.21 \times 10^{-4} \text{ yr}^{-1}} = \textbf{2160 yr}\\\\\text{The cloth was painted } \boxed{\textbf{2160 yr}}\text{ ago}[/tex]

b) The wood

Data:

[tex]N_{t} = 0.062 N_{0}[/tex]

Calculations:

[tex]\text{From Equation (1)}\\\\\ln \dfrac{N_{0} }{0.062N_{0}} = 1.21 \times 10^{-4}t \text{ yr}^{-1}\\\\\ln\dfrac{1}{0.0.062} = 1.21 \times 10^{-4}t \text{ yr}^{-1}\\\\-\ln0.062 = 2.78 = 1.21 \times 10^{-4}t \text{ yr}^{-1}\\\\t = \dfrac{2.78}{ 1.21 \times 10^{-4} \text{ yr}^{-1}} = \textbf{23 000 yr}\\\\\text{The wood was cut} \boxed{\textbf{23 000 yr}}\text{ ago}[/tex]