Half- Life Problem Set Solutions
1) Nitrogen-13 decays to carbon-13 with a half-life of 10.0 minutes. Assume that you are
given a starting mass of 2.00 grams of nitrogen-13.
a) How long are four half-lives?
4 x 10min = 40 min
b) How many grams of nitrogen-13 will remain after 42.7 minutes?
0.693
k
 0.0693 min -1
10 min
kt
A 
log   
 A  2.303
 2g  (0.0693 min -1 )(42.7 min)
log  
2.303
A
  2g 
antilog log   antilog 1.285
  A 
2g
 antilog 1.285  10 1.285
A
2g
 19.27
A
2g
A
19.27
A  0.104 g
2) Maganese-56 has a half-life of 2.60 hours. Assuming you start with a sample of 10.0
grams of manganese-56, how much will remain after 24.3 hours?
0.693
k
 0.2665hr -1
2.6 hr
kt
A 
log   
 A  2.303
 10g  (.2665 hr -1 )(24.3 hr)
 
log
2.303
 A 
10g
 antilog 2.812  10 2.812
A
10g
 649.2
A
10g
A
649.2
A  0.015 g
3) The mass of cobalt-60 in a sample is found to have decreased from 5.40 grams to 1.20
grams in a period of 10.6 years. From this information, calculate the half-life of cobalt-60.
 5.4g  (k)(10.6yr)

log

2.303
 1.2g 
(k)(10.6yr)
0.6532 
2.303
k  0.1419
k
0.693
 4.88 yrs
0.1419
4) A patient is administered 20.0 mg of iodine-131. How much iodine-131 will remain in the
patient’s body after exactly 3 weeks (21 days) if the half-life of iodine-131 is 8.00 days?
k
0.693
 0.2665days -1
8 days
kt
A 
log   
 A  2.303
-1
 20 mg  (.0866 days )(21 days)
 
log
2.303
 A 
20 mg
 antilog0.7899   10 0.7899
A
20 mg
 6.164
A
20 mg
A
6.164
A  3.24 mg
5) Suppose you have a sample containing 800. grams of a radioactive substance. If after
exactly 1 hour only 22.7 grams of the original compound remain, what is the half-life of this
isotope?
 800 g  (k)(60 min)

log

2.303
 22.7g 
1.54706  (26.05297 )(k)
k  0.05938 min 1
t1 
2
0.693
0.05938 min -1
 11.7 min
6) You are an archaeologist and you have discovered the remains of an ancient civilization.
In one of the human bones that you find, you determine that of the original 60.6 grams of
carbon-14 present in the bone, only 12.2 grams remain. Knowing that the half-life of
carbon-14 is about 5730 years, what do you determine is the age of the bone?
k
0.693
 1.2094  10  4 yr 1
5730 yr
 60.6 g  (1.2094  10  4 yr 1 )(t)

log

2.303
 12.2g 
.6961128  (0.000052515 )(t)
t  13,256 yrs  13,300 yrs
7) You are a famous paleontologist and an expert in radioactive-dating techniques. One day,
two visitors come to your laboratory and present you with two different fossils. One fossil
is a dinosaur footprint, the other a human jawbone. Both were found at the bottom of a
deep valley cut by a stream through cliffs of sedimentary rock. Your guests are very exited.
Because they found these fossils next to each other near the steam bed, they feel they
have found conclusive evidence that humans and dinosaurs lived at the same time. You are
asked to date the samples to confirm their claims.
a) You first test the human jawbone. You determine that the carbon-14 activity is only
0.224 times that of a human bone today. Estimate the age of the human jawbone.
k
0.693
 1.2094  10  4 yr 1
5730 yr
4
1
 1  (1.2094  10 yr )(t)
log

2.303
 0.224 
t  12,373 yrs  13,000 yrs
b) You next examine the fossil footprint. You discover that the fresh mud that the
dinosaur stepped in had just been covered with a thin layer of volcanic ash. You
study the amount of potassium-40 and argon-40 in the ash. You find that the
sample contains 2.76 mg of potassium-40 and 0.210 mg of argon-40. The half-life of
potassium-40 is 1.3 billion years. What is the age of the fossil footprint?
k
0.693
1.3  10 9 yr
 5.33  10 10 yr 1
 40 
0.21 mg Ar  
  0.21 mg Ar decayed
 40 
 A  0.21 mg  2.76 mg  2.97 mg
 2.97mg
log
 2.76mg
 (5.33  10 10 yr 1 )(t)


2.303

t  137,600,000 yrs  138,000,000 yrs
c) Were your visitors’ conclusions about these fossils’ ages correct? Explain.
No. The human jawbone is 13,000 years old; the dinosaur footprint is 138 million
years old.
d) If they were not, could you explain the fact that they were found together at the
bottom of the valley?
Answers may vary –-- maybe the bone and the footprint were eroded away from
different layers of rock and then carried downstream where they were deposited.