p43 063

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63.

(a) The mass of a

238

U atom is (238 u)(1.661

× 10

24

g/u) = 3.95

× 10

22

g, so the number of uranium

atoms in the rock is N

U

= (4.20

× 10

3

g)/(3.95

× 10

22

g) = 1.06

× 10

19

. The mass of a

206

Pb

atom is (206 u)(1.661

× 10

24

g) = 3.42

× 10

22

g, so the number of lead atoms in the rock is

N

Pb

= (2.135

× 10

3

g)/(3.42

× 10

22

g) = 6.24

× 10

18

.

(b) If no lead was lost, there was originally one uranium atom for each lead atom formed by decay,

in addition to the uranium atoms that did not yet decay. Thus, the original number of uranium
atoms was N

U 0

= N

U

+ N

Pb

= 1.06

× 10

19

+ 6.24

× 10

18

= 1.68

× 10

19

.

(c) We use

N

U

= N

U 0

e

−λt

where λ is the disintegration constant for the decay. It is related to the half-life T

1/2

by λ =

(ln 2)/T

1/2

. Thus

t =

1

λ

ln



N

U

N

U 0



=

T

1/2

ln 2

ln



N

U

N

U 0



=

4.47

× 10

9

y

ln 2

ln



1.06

× 10

19

1.68

× 10

19



= 2.97

× 10

9

y .


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