Carbon dating

Question

Half life radioactive decay dating. 

 

Prehistoric cave paintings were discovered in France. The paint contained 7% of the original carbon-14. The half-life of carbon-14 is 5730 years. How old are the paintings?

Answer 1

Find the decay rate at half life , A = A_o e^kt, the final A is half of A_o

1/2 A_o = A_o e^5730k, 1/2 = e^5730k, take ln of both sides

 

ln(1/2) = ln e^5730k, ln(1)-ln(2) = 5730k, -ln2 = 5730k, k=-1.209x10^-4

 

knowing that only 7% of the original ¹⁴C is left, then

 

0.07A_o = A₀ e^{-1.209x10-4 t} , 0.07 = e^{-1.209x10-4 t}, taking ln of both sides

 

ln 0.07= ln e^{-1.209x10-4 t}, 1n 0.07 = -1.209x10^-4 t ,

 

t = ln 0.07/ -1.209x10^-4 = 21996 yrs 

 

 

Answer 2

fraction remaining = 0.5n where n = # of half lives that have elapsed

fraction remaining = 7% = 0.07

0.07 = 0.5n

log 0.07 = n log 0.5

-1.1549 = -0.3010 n

n = 3.837 = number of half lives elapsed

age of paintings = 3.837 half lives x 5730 years/half live = 21,985 years old

Answer 3

0.07a₀ = a₀(0.5)^t/5730

0.07 = (0.5)^t/5730

ln 0.07 = (t/5730)(ln 2)

 

Can you calculate and answer?