Chemistry Reference
In-Depth Information
Practice Problem 21.10
(a)
Calculate the time it takes an isotope with a half-life 47.2 years to
disintegrate from
10
15
10
14
1.77
atoms to
5.59
atoms.
10
15
(b)
Calculate the original number of atoms of this isotope if
1.77
atoms
remain after 112 years.
The ratio the original number of atoms of the isotope divided
by the number of atoms (
N
) at time
t,
is equal to the activity ratio, to the mole
ratio, and also to the mass ratio
of the same isotope,
and so those quantities may
be used instead if they are given in a problem. Thus we can calculate a disin-
tegration time, given a half-life and the numbers of atoms or activities or masses
of an isotope at the start and at the end of the period of time.
N
o
/
N
,
(
N
o
)
Extra Help
Activity ratio:
A
o
A
kN
o
kN
N
o
N
Mole ratio:
n
o
N
o
/
Avogadro's
number
N
/
Avogadro's
number
n
EXAMPLE 21.11
N
o
N
Calculate the half-life of an isotope if the activity of a sample was reduced to
45.0% of its original activity in 1.08 years.
Mass ratio:
m
o
m
n
o
(MM)
n(MM)
n
o
n
N
o
N
Solution
Here, the ratio of activities,
A
o
/
A,
is equal to 100/45.0:
a
100
45.0
a
0.693
t
1/2
ln
b
b
(1.08 years)
The left side of this equation is evaluated first:
a
0.693
t
1/2
0.798
5
b
(1.08 years)
Rearranging gives
(0.693) (1.08 years)
0.798
5
t
1/2
0.937 year
This answer is reasonable, because more than one but less than two half-lives
are required.
Practice Problem 21.11
Calculate the half-life of an isotope if a sam-
ple was reduced to 45.0% its original
mass
of the isotope in 1.08 years.
EXAMPLE 21.12
94
(a)
Krypton-94 has a half-life of 1.4 s. Starting with 1.00 mol of
36
Kr
atoms,
94
calculate the number of atoms of
36
Kr
that will remain after 0.500 min.
(b)
Discuss how many atoms would remain after 2.00 min.
Solution
a
N
o
N
a
0.693
t
1/2
a
0.693
1.4 s
(a)
ln
b
b
t
b
(0.500 min) (60 s/min)
14.
9
