Download Period 10 Activity Solutions: Nuclear Reactions

2/09/04
Period 10 Activity Solutions: Nuclear Reactions
10.1 Rates of Radioactive Decay
1) Half-Life
What is the half-life of a radioactive source?
The half-life of a radioactive source is the time required for half of the
unstable nuclei to decay. After one half-life, the material will be only
half as radioactive and the number of the original nuclei remaining will
be only half what it was originally. After 2 half-lives, only one fourth of
the original radioactive material will remain.
2) Carbon-14
a) What common radioactive isotope is used for archeological dating?
The carbon-14 isotope ( 146 C )
b) Why is this isotope unstable?
Carbon-14 is a small nucleus (only 6 protons) with an unequal number
of protons and neutrons.
c) What change to a
14
6C
nucleus would make it a more stable nucleus?
If one neutron decayed into a proton, the nucleus would have equal
numbers of neutrons and protons (7 of each type of nucleon).
d) Write a nuclear equation to show how a 146 C nucleus decays to become a more
stable nucleus. What type of ionizing radiation is emitted during this decay?
14
6C
→
14
7N
+
0
−1 e
+ ν
Since changing a neutron into a proton adds one positive charge, a
negatively charged electron ( β − particle) must be emitted to conserve
electric charge. An antineutrino is also emitted whenever an electron
is given off.
e) Where does carbon-14 come from? Write a nuclear reaction to show the
formation of 146 C .
The energy from cosmic rays can turn stable nitrogen-14 in the air into
carbon-14. One proton in the nitrogen atom turns into a neutron. An
antielectron (a positron) and a neutrino are emitted.
14
7N
→
14
6C
+
0
+1 e
+ ν
3) Carbon Dating
a) How long does it take for one-half of the carbon-14 nuclei in a sample to decay?
5,568 years
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b) How can the half-life of carbon-14 allow us to determine the age of an
archeological object?
Both stable Carbon-12 and unstable Carbon-14 isotopes are present in
the atmosphere (in carbon dioxide). Living organisms absorb both
isotopes of carbon in the process of respiration. After an organism
dies, it no longer absorbs any new Carbon-14, and the Carbon-14
within it decays. We can accurately estimate the time of an organism's
death, if we know the ratio Carbon-12 to Carbon-14 in the atmosphere
at the time the organism died, the present ratio of Carbon-12 to
Carbon-14 in the fossil, and the half-life of Carbon-14.
c) A sample of bone from an archeological site was found to contain only one-sixty
fourth (1/64) the amount of Carbon-14 that would have been present in living
bone at the time this organism was alive. What is the age of this bone?
Since
1
1
= 6 we know that 6 half-lives have passed since this
64
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organism died.
6 half lives x 5,568 years
half life
= 33,408 years
d) Group Discussion Question: An archeological team works in a dig they believe to
be around one million years old. Would Carbon-14 dating be appropriate to
determine the age of this site? Why or why not?
4) Radioactive decay modeled by capacitor discharge
Your instructor will explain how to use a board that simulates radioactive decay with
circuit elements and a capacitor.
a) After charging the capacitor, flip the switch to the right to allow the capacitor to
discharge through the resistor. At the same time, start the timer. Measure the
voltage with a multimeter set to measure DC voltage ( V ) every 15 seconds for
four minutes. Record your measurements in the table below.
Time
elapsed
(Min: Sec)
Voltage
(volts)
Time
elapsed
(Min: Sec)
0:00
2:15
0:15
2:30
0:30
2:45
0:45
3:00
1:00
3:15
1:15
3:30
1:30
3:45
1:45
4:00
2:00
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Voltage
(volts)
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b) Make a graph on the grid below of the voltages you measured versus time elapsed.
c) Use your graph to find the half-life of the capacitor discharge. _________________
5) Radioactive decay modeled by capacitor discharge with background radiation
Next, we use the equipment to produce a more realistic model of radioactive decay.
Your instructor will explain how to change the settings on the board.
a) After charging the capacitor, flip the switch to the right to allow the capacitor to
discharge through the resistor. Start the timer and measure the voltage every 15
seconds for 4 minutes. Record your measurements in the table below.
Time
elapsed
(Min: Sec)
0:00
Voltage
(volts)
Time
elapsed
(Min: Sec)
2:15
0:15
2:30
0:30
2:45
0:45
3:00
1:00
3:15
1:15
3:30
1:30
3:45
1:45
4:00
Voltage
(volts)
2:00
b) What is the smallest voltage reading you obtained? __ approximately 3 volts__
c) What does this voltage represent? Why didn’t your graph decrease further?
This voltage represents the background radiation.
d) Would your graph decrease below this voltage if you had measured the voltage for a
longer time? Why or why not?
No, the background radiation (in this case, background voltage) is
always present.
e) From your graph, find the half-life of the capacitor’s discharge. To do so, you must
subtract the background radiation, find the time for the voltage to decrease by ½,
and then add in the background radiation.
Pick a data point on your graph as your starting point for the voltage.
Subtract the background voltage. Divide the result in half. Then add back
in the background voltage. This gives ½ the original voltage, corrected
for the background. Find this voltage on your graph. The time elapsed
between this point and your original point is the half-life – the time it took
for ½ of the capacitor’s charge to be released.
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10.2 Mass as a Form of Energy, E = Mc2
From Einstein’s equation, E = Mc2, we know that changing the energy of a substance
changes its mass. We have found this mass change too small to measure for physical
changes and chemical reactions. Next, we consider a situation where mass changes can
be measured – nuclear reactions.
6) Calculating the energy released during nuclear reactions
–27
The mass of a 238
kg. Follow the steps below to find
92 U nucleus is 395.2138 x 10
the binding energy of this nucleus.
a) How many protons does the 238
92 U nucleus have? _92 protons_
Find the mass of these protons. (The mass of one proton = 1.6726 x 10-27 kg)
mass of 92 protons = 92(1.6726 x 10-27 kg) = 153.8792 x 10-27 kg
b) How many neutrons does the 238
92 U nucleus have? _146 neutrons_
Find the mass of these neutrons. (The mass of one neutron = 1.6749 x 10-27 kg.)
mass of 146 neutrons = 146(1.6749 x 10-27 kg) = 244.5354 x 10-27 kg
c) Find the total mass of these nucleons.
153.8792 x 10-27 kg
+ 244.5354 x 10-27 kg
398.4146 x 10-27 kg
d) What is the difference in mass between the total mass you found in part c)
and the mass of the U-238 nucleus, which is 395.2138 x 10 –27 kg?
–
398.4146 x 10-27 kg
395.2138 x 10-27 kg
3.2008 x 10-27 kg
e) Calculate the binding energy of the U-238 nucleus in joules.
2
Binding energy = (mass difference) c
= (3.2008 x 10-27 kg ) x(3 x 108m/s)
= (3.2008 x 10-27 kg ) x (9 x 1016m2/s2 = 2.8807 x 10–10 J
f) Find the binding energy per nucleon of a
238
92 U
nucleus.
2.8807 x 10–10 J = 0.0121 x 10–10 J = 1.21x 10–12 J
238 nucleons
nucleon
nucleon
g) Convert the binding per nucleon from joules into units of megaelectron volts (MeV).
(1 joule = 6.25 x 1012 MeV)
1.21 x 10–12 joules x 6.25 x 1012 MeV
1 joule
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= 7.57 MeV
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7) Releasing energy from matter
a) When one proton and one neutron come together to form one deuterium
nucleus, 3.2 x 10–13 joules of energy are given off. What is this energy called?
__the binding energy of the nucleus__
b) What is the process of nucleons binding called? _fusion_
c) Calculate the number of electrons volts (eV) in 3.2 x 10–13 joules. (Hint: 1 eV =
1.6 x 10–19 J). _2 x 106 eV_
d) How many megaelectron volts (MeV) is this? __2 MeV__
e) How many MeV is this per nucleon? _1_
10.3 Nuclear Binding Energy
The figure below graphs the binding energy of nuclei versus the mass number
(A) of the nucleus.
8) Binding energy
a) According to the graph, which nucleus is the most stable? _iron (Fe)_
b) Why is it the most stable?
Iron is the most stable because it has the greatest binding energy.
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c) The leftmost point of the graph is deuterium ( 21 H ). What is the binding energy per
nucleon of deuterium? _1 MeV_
d) The next point to the right is helium ( 42 He ). What is the binding energy per
nucleon of helium? _7 MeV_
e) How much energy is given off per nucleon if two deuterium nuclei fuse to form one
helium nucleus? 7 MeV – 1 MeV = 6 MeV per nucleon
f) How much energy is given off per helium nucleus formed?
Since there are 4 nucleons in 42 He , the total energy released is 4 nucleons
times 6 MeV/nucleon = 24 MeV.
g) Group Discussion Question: When hydrogen and oxygen atoms combine in a fuel
cell to form a water molecule, the chemical binding energy given off is approximately
2.5 eV per molecule. Approximately how many times greater is the nuclear binding
energy given off when a proton and a neutron combine to form a nucleus of
deuterium? (See question 7.d.) Why is it that we cannot measure a change in mass
in chemical reactions but we can measure a change in mass in nuclear reactions?
The binding energy of a deuterium nucleus is approximately 4 x 105 times
greater than the binding energy of a water molecule.
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