Untitled - Kelly Walsh High School

Nuclear Chemistry 297
that is not associated with a multiple of a half-life. To solve these types of problems,
one must use the mathematical relationships associated with first-order
kinetics. In general, two equations from Chapter 13 are useful:
(1) ln (N t/N 0) kt and (2) t 1/2 (ln 2)/k or t 1/2 0.693/k
In these equations, ln is the natural logarithm, N t is the amount of isotope
radioactive at some time t, N 0 was the amount of isotope radioactive initially,
and k is the rate constant for the decay. If you know initial and final amounts
and if you are looking for the half-life, you would use equation (1) to solve for
the rate constant and then use equation (2) to solve for t 1/2.
What is the half-life of a radioisotope that takes 15 min to decay to 90% of its
original activity?
Solution:
Using equation (1): ln (90/100) k (15 min)
0.1054 k (15 min)
7.03 10 3 min 1 k (unrounded)
Now equation (2): t 1/2 ln 2/7.03 10 3 min 1
t 1/2 0.693/7.03 10 3 min 1
t 1/2 98.5775 99 min
If one knows the half-life and amount remaining radioactive, you can then use
equation (2) to calculate the rate constant, k, and then use equation (1) to solve
for the time. This is the basis of carbon-14 dating. Scientists use carbon-14
dating to determine the age of objects that were once alive.
Suppose we discover a wooden tool and we determine its carbon-14 activity to
have decreased to 65% of the original. How old is the object?
Solution:
The half-life of C-14 is 5730 years. Substituting this into equation (2):
5730 y (ln 2)/k
5730 y 0.693/k
k 1.21 10 4 y 1