The half-life of the radioactive isotope Carbon-14 is about 5730 years. a. What percentage of the original amount of Carbon-14 is left after 20,000 years?
b. If an old wooden tool is found in a cave and the amount of Carbon-14 present in it is estimated to be only 42% of the original amount, approximately how old is the tool?
c. Radiocarbon dating is not as easy as these exercises might lead you to believe. With the help of your classmates, research radiocarbon dating and discuss why our model is somewhat over-simplified.

See explanation for details

Explanation:

a) Now from the data given, recall that;

N/No = (1/2)t/t1/2

Where,

N= amount of radioactive material remaining at time t

No= Amount of radioactive material originally present

t= time taken for N amount of radioactive material to remain

t1/2= half life of radioactive material

So,

N/No = (1/2)^20,000/5730

N/No = (1/2)^3.49

N/No = 0.089

Since the fraction N/No is the fraction remaining after t years, then;

Percentage of Carbon-14 left after 20,000 years = 0.089 * 100 = 8.9%

b) From;

0.693/t1/2 = 2.303/t log No/N

Given that N = 0.42 No

Hence;

0.693/5730 = 2.303/t log (No/0.42No)

1.21 * 10^-4 = 0.87/t

t= 0.87/1.21 * 10^-4

t = 7190 years

c) Radiocarbon dating is a method of obtaining the age of an object derived from plants or animals by comparing the Carbon-14 activity of living things with that of the sample under study. Due to contamination, the age of artifacts obtained by Carbon-14 dating may yield outlying figures.

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.