Question

A fossil was analyzed and determined to have a carbon-14 level that is 70 % that of living organisms. The half-life of C-14 is 5730 years. How old is the fossil? Express your answer with the appropriate units. View Available Hint(s) t t t = nothing nothing

Answer 1

Answer: 2948

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

$t_{\frac{1}{2}}=\frac{0.693}{k}$

$k=\frac{0.69}{t_{\frac{1}{2}}}=\frac{0.693}{5730}=1.21\times 10^{-4}years^{-1}$

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant  = $1.21\times 10^{-4}years^{-1}$

t = age of sample  = ?

a = let initial amount of the reactant  = 100

a - x = amount left after decay process = $\frac{70}{100}\times 100=70$

$t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{100}{70}$

$t=2948years$

Thus the fossil is 2948 years old.