The radioactive isotope carbone-14 is used as a tracer in medical and biological research. Compute the half-life of carbone-14 given that the decay constant k is $- 1.2097 \times 10 ^ { - 4 }$ (The units for k here are such that your half-life answer will be in years.) Please round the answer to one decimal place.
A) 5,728.7years
B) 5,729.2years
C) 5,758.5years
D) 5,730.3years
E) 5,729.9years

Question

Quizplus · Accepted Answer

The half-life of a radioactive isotope is given by the formula:$$t_{1/2} = \frac{ln(2)}{k}$$where:* $t_{1/2}$ is the half-life* $ln$ is the natural logarithm* $k$ is the decay constantSubstituting the given values into the formula, we get:$$t_{1/2} = \frac{ln(2)}{- 1.2097 \times 10 ^ { - 4 }} = 5,729.9 \text{ years}$$Therefore, the half-life of carbon-14 is 5,729.9 years, rounded to one decimal place.

The Radioactive Isotope Carbone-14 Is Used as a Tracer in Medical