Question

Each isotope has a unique half-life. The half-life of an isotope is the time taken for half of the starting quantity to decay (with a ratio of 1:1). After two half-lives, there will be one-fourth of the original parent sample and three-quarters would have decayed to the daughter product (with a ratio of 1:3). After three half-lives, the ratio becomes 1:7, and so forth.
A graph showing months on the x-axis and the amount of parent/daughter on the y-axis. The graph uses four pie charts to demonstrate how parent elements and daughter elements change with each half life for a sample in increments of 4 months.
The graph, for instance, shows that assuming the half-life of a sample is 4 months, then in 4 months, there will be 0.5 gram of the parent element and 0.5 gram of the daughter element will be produced. In month 8 (which is two-half-lives), there will be only 0.25 gram of parent element left and 0.75 gram of daughter element; that is, one-fourth of the parent sample (in red) is left, and in month 12, there is only one-eighth of the parent element.You attend a geology lab where you are asked to estimate the age of a fossil. The ratio of parent to daughter elements in the fossil sample is 1:7. You know that fossils are the remains of living organisms, which have some amount of C-14 isotope. The C-14 isotope, which has a half-life of 5730 years, begins to decay as the organism dies. What would be your estimation of the fossil's age?a. 2865
b. 17,190
c. 5730
d. 11,460
e. 40,110
f. 22,920

Answer 1

Answer:

b. 17,190

Explanation:

Using the formula for C-14 dating,

$t=\dfrac{ln(\dfrac{N}{N_0})t_\frac{1}{2} }{-0.693}$

where Present Value N =1/8 of Parent Sample

Initial Value, $N_0$=1

Half Life, $t_\frac{1}{2}$=5730 years

$t=\dfrac{ln(\dfrac{1/8}{1})5730}{-0.693}$

=17,190

Therefore the fossil is about 17190 years old.