Question

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.

Answer 1

`\boxed{\text{a) 2160 yr ago; b) 23 000 yr ago}}`

Step-by-step explanation:

Two important equations in radioactive decay are

`(1) \qquad \ln (N_(0) )/(N_(t)) = kt\\\\(2) \qquad t_{(1)/(2)} = (\ln2)/(k )`

We use them for carbon dating.

a) The dye

Data:

`t_{(1)/(2)} = \text{5730 yr}\\\\N_(t) = 0.77 N_(0)`

Calculations:

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

b) The wood

Data:

`N_(t) = 0.062 N_(0)`

Calculations:

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

Answer 2

Case I => %C-14 remaining = 77% => Age of artifact = 2200 yrs

Case II => %C-14 remaining = 6.2% => Age of artifact = 23,000 yrs

Step-by-step explanation:

Given:

Half-Life C-14 = 5730 yrs

=> Rate Constant = k = 0.693/t(1/2) = (0.693/5730)yrs⁻¹ = 1.2 x 10⁻⁴ yrs⁻¹

NOTE => All radioactive decay is 1st order kinetics.

=> A = A₀eˉᵏᵗ (classic 1st order decay equation)

- A = remaining activity

-A₀ = initial activity

- k = rate constant

- 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