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.

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

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

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

Explanation:

Two important equations in radioactive decay are

We use them for carbon dating.

a) The dye

Data:

Calculations:

b) The wood

Data:

Calculations:

let that person have a turn to speak : )

the answer is b. dilution