The Foundations of Chemistry

Plan

As we did in Example 26-3, we determine the specific rate constant kfrom the known half-

life. The time required to reach the present fraction of the original activity is then calculated

from the first-order decay equation.

Solution

First we find the first-order specific rate constant for^14 C.

t1/2 or k1.21
 10 ^4 y^1

The present^14 C activity, N(disintegrations per unit time), is 0.636 times the original activity,

N 0.

N0.636 N 0

We substitute into the first-order decay equation

ln 

N

N

0

kt

ln 

0.63

N

6

0

N 0

(1.21
10 ^4 y^1 )t

We cancel N 0 and solve for t.

ln (1.21
10 ^4 y^1 )t

0.452(1.21
 10 ^4 y^1 )t or t 3.74
 103 y (or 3740 y)

You should now work Exercises 58 and 62.

EXAMPLE 26-5 Uranium–Lead Dating

A sample of uranium ore is found to contain 4.64 mg of^238 U and 1.22 mg of^206 Pb. Estimate

the age of the ore. The half-life of^238 U is 4.51
 109 years.

Plan

The original mass of^238 U is equal to the mass of^238 U remaining plus the mass of^238 U that

decayed to produce the present mass of^206 Pb. We obtain the specific rate constant, k,from

the known half-life. Then we use the ratio of original^238 U to remaining^238 U to calculate the

time elapsed, with the aid of the first-order integrated rate equation.

Solution

First we calculate the amount of^238 U that must have decayed to produce 1.22 mg of^206 Pb,

using the isotopic masses.

_?_ mg^238 U1.22 mg^206 Pb


2

2

0

3

6

8

m

m

g

g

2

2

0

3

6

8

P

U

b

1.41 mg^238 U

Thus, the sample originally contained 4.64 mg1.41 mg6.05 mg of^238 U.

We next evaluate the specific rate (disintegration) constant, k.

t1/2 so k1.54
 10 ^10 y^1

Now we calculate the age of the sample, t.

0.693



4.51
 109 y

0.693



t1/2

0.693



k

1



0.636

0.693



5730 y

0.693



t1/2

0.693



k

1018 CHAPTER 26: Nuclear Chemistry