Relative Dating and FossilsRelative Dating and FossilsRelative datingRelative datingLaw of SuperpositionLaw of Superposition Index FossilsIndex FossilsWhat are fossils?What are fossils?

Preserved remains of ancient organismsPreserved remains of ancient organismsHow are fossils formed?How are fossils formed?

Dead organisms are buried by sediments, if Dead organisms are buried by sediments, if conditions are perfect, they are perseveredconditions are perfect, they are persevered

Why is the fossil record incomplete?Why is the fossil record incomplete? It takes perfect conditions to make fossils, so It takes perfect conditions to make fossils, so

many organisms were never perseveredmany organisms were never persevered

Radioactive DatingRadioactive DatingUse radioactive decay to find absolute ages Use radioactive decay to find absolute ages

of rocks and fossilsof rocks and fossilsHalf-life: the amount of time required for Half-life: the amount of time required for

half the atoms in a sample to decayhalf the atoms in a sample to decayAfter 1 half-life: ½ atoms remainAfter 1 half-life: ½ atoms remainAfter 2 half-lives: ½ x ½ = ¼ atoms remainAfter 2 half-lives: ½ x ½ = ¼ atoms remainAfter 3 half-lives: ½x ½ x ½ = ⅛ atoms remainAfter 3 half-lives: ½x ½ x ½ = ⅛ atoms remainAfter n half-lives: ½After n half-lives: ½nn atoms remain atoms remain

Different elements have different half-livesDifferent elements have different half-livesCarbon-14 (Carbon-14 (1414C) is very common and has a C) is very common and has a

half life of about 5730 yearshalf life of about 5730 years

Radioactive DecayRadioactive Decay ““Cold” elements won’t harm youCold” elements won’t harm you ““Hot” elements give off radiation (causing Hot” elements give off radiation (causing

decay) which may harm youdecay) which may harm you Organic compounds have a mix of Organic compounds have a mix of 1414C (hot) and C (hot) and

1212C (cold)C (cold) 1414C decays into C decays into 1414N which is cold and can escape N which is cold and can escape

into the atmosphereinto the atmosphere Atomic number= protonsAtomic number= protons

(C= 6, N= 7)(C= 6, N= 7) Atomic mass= protons + neutronsAtomic mass= protons + neutrons 14= 6 protons + 8 neutrons (& 6e14= 6 protons + 8 neutrons (& 6e--)) One neutron decays into a proton and an electronOne neutron decays into a proton and an electron 14= 7 protons + 7 neutrons (& 7e14= 7 protons + 7 neutrons (& 7e--))

As time goes by, the amount of As time goes by, the amount of 1212C increases C increases compared to the amount of compared to the amount of 1414CC

Sample ProblemSample Problem 1414C has a half-life of 5730 yrs, we can use C has a half-life of 5730 yrs, we can use

it to date things back about 60,000 yrsit to date things back about 60,000 yrsRemember: the amount of an element Remember: the amount of an element

remaining= ½remaining= ½nn, n= number of half-lives , n= number of half-lives passedpassed

If you are made of 1% If you are made of 1% 1414C when you die, C when you die, how much how much 1414C will be left after 5,730yrs? C will be left after 5,730yrs? 5,730 yrs x 1 half-life/5,730yrs = 1 half-life5,730 yrs x 1 half-life/5,730yrs = 1 half-life½½11 x 1% = 0.5% x 1% = 0.5%

11,460 yrs?11,460 yrs?11,460 yrs x 1 half-life/5,730yrs = 2 half-lives11,460 yrs x 1 half-life/5,730yrs = 2 half-lives½½22 x 1% = 0.25% x 1% = 0.25%

PracticePractice A fossil contains 12.5% of the A fossil contains 12.5% of the 1414C that would C that would

have been present at the time of death. How have been present at the time of death. How old is the fossil?old is the fossil? 0.5 x 0.5 x 0.5 = 0.1250.5 x 0.5 x 0.5 = 0.125 3 half-lives have passed3 half-lives have passed 3 half-lives x 5,730yrs/ 1 half-life = 17,190yrs3 half-lives x 5,730yrs/ 1 half-life = 17,190yrs CHEATER METHOD! (USE AT YOUR OWN RISK!)CHEATER METHOD! (USE AT YOUR OWN RISK!)

½½nn = 0.125 = 0.125 n log ½ = log 0.125n log ½ = log 0.125 n = log 0.125 / log ½ = 3n = log 0.125 / log ½ = 3

A fossil is tested and found to be 28,650 years A fossil is tested and found to be 28,650 years old. How much of the original old. How much of the original 1414C is still present?C is still present? 28,650 yrs x 1 half-life/5,730yrs = 5 half-lives28,650 yrs x 1 half-life/5,730yrs = 5 half-lives ½½55 = 0.03125 or 3.125% = 0.03125 or 3.125%