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:

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

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