A sample of a vegetable dye scraped from a cave painting was converted to CO2 for C-14 determination. The radioactivity of the CO2 was measured as 3.76 counts per minute per gram of carbon( c.p.m). The C-14 in a living organism has 15.3 c.p.m. and the half-life of C-14 is 5760 yr. How old is the painting in years? (Answer to 3 significant figures and use e notation for power of 10, e.g. 1.23e2 for 1.23 x 102)
2007-08-09 16:13:40 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
Your homework problem person is real lazy. If you divide 3.76 cpm [the dye activity] by 15.3 cpm
[current activity], you get about 1/4. That's TWO half-lives (hint, hint)
2007-08-09 16:23:42 · answer #1 · answered by cattbarf 7 · 0⤊ 1⤋
Sorry this is not a time scale that C14 testing has 3 significant figures of accuracy. That would be scientific fraud.
Unless this is a homework assignment, uh , then?
It would be academic fraud
2007-08-09 23:18:50 · answer #2 · answered by Anonymous · 0⤊ 2⤋