Activity 48
Nuclear Chemistry: Rates of Radioactive Decay
What Do You Think?
Radioactive iodine-131 is used to treat thyroid cancer. A patient is given 20.0 mg of iodine-131 in
the form of NaI, and after 8 days only 10 mg remain. How long, from the time of the initial dose,
do you think it will it take until only 5 mg remain?
(A) 12 days
(B) 16 days
(C) 20 days
(D) cannot be determined from the information given
Learning Objectives
•
Understand how the level of radioactivity decreases with time
•
Understand how radioactivity can be used to determine the age of materials
Success Criteria
•
Relate the amount of radioactive material remaining after some period of time to the half-life of
the radioactive isotope and the rate constant for the decay
•
Estimate the age of materials from their radioactivity
Prerequisites
•
Activity 28: Rates of Chemical Reactions
•
Activity 46: Nuclear Chemistry: Radioactivity
Activity 48 —Nuclear Chemistry: Rates of Radioactive Decay
325
Tasks
1. Examine the model and complete Table 48.1 by entering N, the number of unstable nuclides
remaining at each point in time, and ln(N), in the last two columns, respectively.
Model
In the chart below, ten unstable nuclides are represented by white circles. They decay spontaneously
by some mechanism to produce stable nuclides, which are indicated by dark circles.
Table 48.1 Time Evolution of Nuclear Decay
Time
(mins)
N
ln(N)
0
10
2.30
2.6
7
1.95
5.0
5
1.60
8.7
3
1.10
11.6
2
0.69
16.6
1
0.00
Original and Product Nuclides
2. Make two graphs of the data (see the next page): Graph 1 with N plotted on the y-axis and time
plotted on the x-axis, and Graph 2 with ln(N) plotted on y-axis and time plotted on the x-axis.
Quality graphs will have titles, labels on the axes, data point shown with small circles around
them, and a smooth line drawn through the data points.
3. On your graphs mark the points on the x-axis where the fraction remaining is equal to
1 , 1 , and 1 .
2 4
8
326
Foundations of Chemistry
Graph 1:
10.00
9.00
8.00
7.00
6.00
5.00
4.00
3.00
2.00
1.00
0.00
0.00
5.00
10.00
15.00
20.00
10.00
15.00
20.00
Graph 2:
2.50
2.00
1.50
1.00
0.50
0.00
0.00
5.00
Activity 48 —Nuclear Chemistry: Rates of Radioactive Decay
327
Key Questions
1. From Graph 1, how long does it take for half of the radioactive nuclides in the model to decay?
This time is called the half-life.
2. Does the time it takes for half of the radioactive nuclides to decay depend on the point in time
taken as the starting point? Explain how the information in Graph 1 supports your answer.
3. How can the data in the model be used to identify radioactive decay as one of the following
types of reactions: zero order, first order, or second order?
4. What is the mathematical equation that represents the straight line that can be drawn through the
data points in Graph 2?
5. How can the rate constant for this radioactive decay be obtained from Graph 2?
Exercises
Use the following information for Exercises 1–5: Two radioactive isotopes, A and B, have decay
rate constants kA and kB, respectively, where kA is larger than kB. NAo and NBo are the number of
nuclides present at t = 0 for each of these isotopes. The decay rate R is the number of decay events
per second, where
R = ∆N/∆t = –k N
328
Foundations of Chemistry
1. Sketch one graph with a curve for NA(t) and a curve for NB(t) to compare the number of
radioactive nuclides present as a function of time for both of these isotopes. Start at t = 0 with
the same number of A isotopes and B isotopes.
B
A
2. Sketch one graph with a curve for ln [NA(t)] and a curve for ln [NB(t)] as a function of time
to compare these functions for the two isotopes. Again, start at t = 0 with same number of A
isotopes and B isotopes.
B
A
3. Identify the isotope A or B with the longer half-life.
4. Identify the isotope A or B with the faster decay rate when the amounts of A and B present are
equal.
Activity 48 —Nuclear Chemistry: Rates of Radioactive Decay
329
5. Using your answer to Key Question 4, derive the following relationship between the half-life
and the rate constant: t½ = ln(2)/k.
ln(N) = ln(No ) -kt
when t = t1/2 then N =
No
2
N 
ln  o  = ln(No ) - kt1/2
 2 
N 
ln  Noo 
 2  = ln(2)
t1/2 =
k
k
6. Derive your answer to Key Question 4 from the fact that the data in Graph 1 is described by an
exponential function: N = Noe–kt
N = No e − kt
ln(N) = ln(No ) - kt
7. Iodine-123 has a half-life of 13.3 hours. Using your answer to Exercise 5, calculate a value for
the decay rate constant, k, in units of s–1.
k =
ln(2)
13.3 hr
k =
0.0521 1hr 1min
= 1.45 × 10 -5 s -1
hr 60 min 60 s
8. Determine the half-life of a radioactive nuclide if, after 2.5 hrs, only 1/32 of the initial amount
remains unchanged.
9. Estimate how many half-lives must pass until less than 1% of a radioisotope remains.
Note: After the students make an estimation without doing the calculation, you could have them
calculate exactly when 1% of the material remains. You might point out that knowing how long
it takes for the level of radioactivity to decrease to 1% or 0.1% of the initial value in a patient
is important in medical applications.
Calculation:
x
1
1 
  =
100
2
2 x = 100
x log(2) = log(100) = 2
2
2
x =
=
= 6.64
log(2)
0.301
330
So 6.64 half-lives are needed until
1% of the material remains.
Foundations of Chemistry
What do you think now?
A patient is given 20.0 mg of iodine-131 in the form of NaI. After 8 days only 10 mg remain. How
long do you think it will it take until only 5 mg remain?
(A) 12 days
(B) 16 days
(C) 20 days
(D) cannot be determined from the information given
How long will it take until only 0.1% of the iodine-131 remains?
Problems
1. A 0.133 pg sample of iron-59 produces 242 β-particles per second. What is the half-life of this
isotope?
Rate = kN
242 s -1 = kN
N =
0.133 × 10 -12 g
(6.03 × 10 23 /mol) = 1.36 × 10 9
59.0 g/mol
242 s -1
k =
= 1.78 × 10 -7 s -1
9
1.36 × 10
ln(2)
t1/2 =
= 3.89 × 10 6 s
-7
-1
1.78 × 10 s
1hr 1day
3.89 × 10 6 s
= 45.0 days
3600 s 24 hr
2. A wooden tool is found at an archaeological site. Estimate the age of the tool using the following
information. A 100 gram sample of the wood emits 1120 β-particles per minute from the decay
of carbon-14. The decay rate of carbon-14 in living trees is 15.3 per minute per gram. Carbon-14
has a half-life of 5730 years.
lnN = lnNo - kt
t =
1  No 
ln
k  N 
Since Rate = k N, the ratio of the rates equals
t
1
= 1/2 = 8267 yrs
k
ln(2)
No
N
No
15.3 min -1 g -1
=
= 1.37
N
11.2 min -1 g -1
t = 8267 yrs ln(1.37) = 2580 yrs
Activity 48 —Nuclear Chemistry: Rates of Radioactive Decay
331
3. Potassium-40 decays into argon-40 by electron capture with a half-life of 1.28 × 109 years. This
decay forms the basis of the potassium-argon dating method that is used with igneous rocks.
This method assumes that no argon was trapped in the rock when it was formed from the molten
material.
A meteorite was found to have equal amounts of 40K and 40Ar.
a) Of the potassium-40 that was originally present, what fraction has decayed?
b) How long ago was this meteorite formed?
Half the potassium decayed so the time must be the half-life of 1.28 × 109 yrs, but here is
the calculation anyway.
ln N = lnNo -kt
t =
1  No 
ln
k  N 
Since half has decayed the ratio
t
1
= 1/2 = 1.85 × 10 9 yrs
k
ln(2)
No
= 2
N
t = (1.85 × 10 9 yrs) ln(2) = 1.28 × 10 9 yrs
c) Another meteorite had a 40K to 40Ar ratio of 0.33. How long ago was this meteorite
formed?
The current ratio is 1/3. If there were 4 units of potassium present initially and 3 units
decayed leaving 1 unit then the K/Ar ratio would be 1/3.
So No / N = 4/1.
Another way to look at this is to say
No = NK + N Ar and N Ar = 3NK so No = 4NK and No /NK = 4
lnN = lnNo -kt
1  No 
ln
k  N 
t
1
= 1/2 = 1.85 × 10 9 yrs
k
ln(2)
t =
t = (1.85 × 10 9 yrs) ln(4) = 2.56 × 10 9 yrs
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Foundations of Chemistry