Radiochemical Dating Chp 24

Iceman Austria/Italy Border dated: 3000-3500 BCE Horse Cave Painting France 13000 BCE

Carbon Dating-to age once living artifacts. Constantly forming C-14 mixes with stable C-12 and C-13 in Earth's biosphere. Plants take in carbon atoms in photosynthesis and animals eat the plants. The ratio of C-14, C-12, and C-13 remains constant as long as organism is alive. After death, no more carbon is ingested and the C-14 starts to decay. C-14 has a half life of 5730 years. The ratio of unstable C-14 to stable C-12 and C-13 decreases. Scientists measure this ratio and compare it to the constant atmospheric ratio to determine the object's age. C-14 dating is limited to dating objects up to about 45,000 years. https://www.youtube.com/watch?v=Kcuz1JiMk9k

Example 2 pg 872-Calculating the Amount of Remaining Isotope Krypton-85 is used in indicator lights of appliances.The half-life of Krypton-85 is 11y. How much of a 2.00mg sample remains after 33 years?

Example 2 pg 872-Calculating the Amount of Remaining Isotope Krypton-85 is used in indicator lights of appliances.The half-life of Krypton-85 is 11y. How much of a 2.00mg sample remains after 33 years? 33 yrs = 3 half lives 11 yrs 3 halfs lives = ½ x ½ x ½ = 1/8 1/8 x 2.00 mg = 0.250 mg

The half life of cobalt-5 is 270 days. How much of a 5 The half life of cobalt-5 is 270 days. How much of a 5.000 mg sample will remain after 810 days?

The half life of cobalt-5 is 270 days. How much of a 5 The half life of cobalt-5 is 270 days. How much of a 5.000 mg sample will remain after 810 days? 810 days = 3 half lives 270 days 3 half lives = 1/8 1/8 x 5.000 mg = 0.625 mg

9. Bandages can be sterilized by exposure to gamma radiation from cobalt-60 which has a half-life of 5.27 y. How much of a 10.0 mg sample of cobalt-60 is left after 1 half life? Two half lives? Three? ½ x 10.0 mg = 5.00 mg ¼ x 10.0 mg = 2.50 mg 1/8 x 10.0 mg = 1.25 mg

10. If the passing of 5 half lives leaves 25 mg of a strontium-90 sample, how much was present in the beginning?

10. If the passing of 5 half lives leaves 25 mg of a strontium-90 sample, how much was present in the beginning? ½ x ½ x ½ x ½ x ½ = 1/32 32 x 25 mg = 800 mg