5 Steps to a 5 AP Chemistry

264  Step 4. Review the Knowledge You Need to Score High

decay is not a linear process, you cannot use the chart to predict how much would still

be radioactive at the end of 12 days or at some time (or amount) 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 that were presented in the Kinetics chapter.

In general, two equations are used:

In these equations, the ln is the natural logarithm; Atis the amount of isotope radioac-

tive at some time t; Aois the amount initially radioactive; and kis the rate constant for the

decay. If you know initial and final amounts and 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 t1/2.

For example: What is the half-life of a radioisotope that takes 15 min to decay to 90%

of its original activity?

Using equation (1):

If one knows the half-life and amount remaining radioactive, equation (2) can be used

to calculate the rate constant kand equation (1) can then be used to solve for the time. This

is the basis of C-14 dating, which is used to determine the age of objects that were once alive.

For example, suppose a wooden tool is discovered and its C-14 activity is determined

to have decreased to 65% of the original. How old is the object?

The half-life of C-14 is 5730 yr. Substituting this into equation (2):

Substituting this rate constant into equation (1):

Mass–Energy Relationships

Whenever a nuclear decay or reaction takes place, energy is released. This energy may be in

the form of heat and light, gamma radiation, or kinetic energy of the expelled particle and

recoil of the remaining particle. This energy results from the conversion of a very small

amount of matter into energy. (Remember that in nuclear reactions there is no conservation

ln yr )

0.4308

65 100 1 21 10

121

/(.^41

(.

=− ×

=− ×

−−t

− 110

3600

−− 41

=

yr

yr

)t

t

5730 2

0 6931

121

yr ln

5730 yr

=

=

=×

/

./

.

k

k

k 110 −−^41 yr

Now equation (2):t1/2 ln min

t

=×2 7 02 10−−^31

1

/.

//

/

./. min

.min

2

31

12

0 693 7 02 10

98 7

=×

=

−−

t

Using equation (1) ln90 100 15 min

0 1054

/

.

=−()

−=−

k

kk

k

15

702 10^3

min

.min

()

×=−−

()^22 tk1/2=ln /

()^1 ln [ ]AAto−ln [ ]