After 1 half-life 50% should be gone.
only 17% is gone, so this is just a portion of 1 half-life.
17/50 = .34 half-life
5730 times .34 should be your answer
You find a piece of cloth painted with organic dye. By analyzing the dye, you find that only 83 % of the carbon-14 originally in the dye remains. When was the cloth painted?
3 answers
Nope.
.83=e^(-.693*t/5730)
take ln of each side.
ln(.83)=-.693t/5730
t= -5730*ln(.83)/.693=1541years
which is about 400 years different than the first answer.
.83=e^(-.693*t/5730)
take ln of each side.
ln(.83)=-.693t/5730
t= -5730*ln(.83)/.693=1541years
which is about 400 years different than the first answer.
It might be instructive to note that ALL radio decay is 1st order decay. Such is the value of the 1st order decay equation C(remaining) = C(initial)e^(-kt) where k = rate constant = 0.693/(Half-Life). In this case Half-Life of C-14 = 5730 Years => k = (0.693/5730)yrs^-1. Using the classic 1st order time of decay equation = [(ln(remaining/initial)]/-k = ln(83%/100%)/-(1.21x10^-4) = 1540 yrs.
1 answer