Nuclear Equations Review - SOLUTIONS
1.
Bombardment of aluminum-27 by alpha particles produces phosphorous-30 and one other particle. Write the
nuclear equation for this reaction and identify the other particle.
27
13
1
Al 24   30
15 P  0 n
The other particle formed is a neutron
2.
Plutonium-239 can be produced by bombarding uranium-238 with alpha particles. How many neutrons will be
produced as a by product of each reaction? Write the nuclear equation for this reaction?
238
92
1
U 24   239
94 Pu 3 0 n
Three neutrons are formed for every one nucleus of U-238 bombarded.
3.
When bombarded with neutrons, cobalt-59 is converted to cobalt-60. What is the nuclear equation for this
reaction?
59
27
4.
Co 01n60
27 Co
One method for producing plutonium-238 is by bombarding uranium-238 with deuterium (hydrogen-2), which
produced neptunium-238 and 2 neutrons. The unstable neptunium then decays to form plutonium-238. Write
the nuclear equations for this two-step reaction. What other particle is produced in the second reaction?
238
92
1
U 21H238
93 Np  20 n
238
93
0
U238
94 Pu  1
A beta particle is also produced in the second reaction.
5.
Neutron bombardment of plutonium-239 yields americium-240 and another particle. Write the nuclear equation
and identify the other particle produced.
239
94
0
Pu 01n240
95 Am 1
The other particle produced is a beta particle.
6.
Alpha-particle bombardment of plutonium-239 produced a neutron and another radionuclide. Write the nuclear
equation for this reaction and identify the radionuclide.
239
94
Pu 24 01n  242
96 Cm
The radionuclide formed is Cm-242.
7.
One possible result of the impact of a neutron on a uranium-235 nucleus is the splitting of the uranium into
tellurium-137, zirconium-97 and two other particles. Write the nuclear equation for this reaction and identify
the two other particles.
235
92
1
U 01n 13752Te  97
40 Zr  2 0 n
The two other particles formed are neutrons.
8.
When bombarded with neutrons, lithium-6 produced an alpha particle and an isotope of hydrogen. Write the
nuclear equation for this reaction. What isotope of hydrogen is produced?
6
3
Li 01n 24   31H
The isotope formed is hydrogen-3.
9.
With what particle would you bombard sulfur-32 to produce hydrogen-1 and phosphorus-32? Write the
appropriate nuclear equation?
32
16
32
S 01n11 H 15
P
10. With what particle would you bombard bismuth-209 to produce astatine-211 and 2 neutrons? Express this
reaction in the form of a nuclear equation.
209
83
1
Bi  24  211
85 At  2 0 n
The particle that bombards bismuth-209 is an alpha particle.
Half- Life Review Problems - 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?
k
0.693
 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?
k
0.693
 0.2665hr -1
2.6 hr
kt
A 
log   
A
2.303


1
0g

 (.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
 0.0866days-1
8 days
kt
A 
log    
 A  2.303
20
mg  (.0866 days-1 )(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
9
1.3  10 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.
d)
If they were not, could you explain the fact that they were found together at the bottom of the valley?
No. The human jawbone is 13,000 years old; the dinosaur footprint is 138 million years old.
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.