Engineering Chemistry S Datta

NUCLEAR CHEMISTRY 59
Pb
or, 1 = 1 +
U
or,
F
HG
∴ e λt = 1.5
206
238
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e –λt
1 = (1 + 0.5) e –λt = 1.5 e –λt
or, λt = 2.303 × log 10
1.5 = 2.3 × 0.1761
∴
∴
0693 .
45 . × 10 9 t = 0.23 × 0.1761
t = 2.63 × 10 9 years.
Example 9. A sample of U 238 is found to contain 11.9 g of U 238 and 10.3 g of Pb 206 .
Calculate the age of the ore. (t 1/2
= 4.5 × 10 9 years)
Sol. Mole of U 238 = 119 .
238
Mole of Pb 206 = 10 . 3
206
We have, t = 2 . 303 t
0.
693
= 0.05
= 0.05
1/2
log
F
HG
2. 303 × 4.
5 × 10
=
0.
693
2. 303 × 4.
5 × 10
=
0.
693
Pb
1 +
U
9
9
log
206
238
F
HG
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1 005 .
+
005 .
log 2 = 4.5 × 10 9 years.
Example 10. The amount of 6
C 14 isotope in a piece of fresh wood is found to be 1/6th of
the total carbon. Calculate the age of wood. (t 1/2
of C = 5577 yrs).
Sol. We have,
t = 2 . 303 t
0.
693
1/2
1
log
16 /
2.
303 × 5577
= log 6
0.
693
2. 303 × 5577 × 0.
7782
= = 20,170 yrs.
0.
693
Example 11. Half-life period of C 14 is 5760 years. An old piece of wood in a museum has
a disintegration rate which is 25% of the disintegration rate of an equal weight of a new piece of
wood. What is the age of the old piece of wood?
Sol. We have,
2.
303 × 5760 1
t = log
0.
693 14 /
2. 303 × 5760 × 0.
6021
= = 11,525 yrs.
0.
693
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