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Period 10 Activity Solutions: Nuclear Reactions10.1 Rates of Radioactive Decay1) 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 theunstable nuclei to decay. After one half-life, the material will be onlyhalf as radioactive and the number of the original nuclei remaining willbe only half what it was originally. After 2 half-lives, only one fourth ofthe original radioactive material will remain.

2) Carbon-14a) What common radioactive isotope is used for archeological dating?

The carbon-14 isotope ( C146 )

b) Why is this isotope unstable?

Carbon-14 is a small nucleus (only 6 protons) with an unequal numberof protons and neutrons.

c) What change to a C146 nucleus would make it a more stable nucleus?

If one neutron decayed into a proton, the nucleus would have equalnumbers of neutrons and protons (7 of each type of nucleon).

d) Write a nuclear equation to show how a C146 nucleus decays to become a more

stable nucleus. What type of ionizing radiation is emitted during this decay?

νν++++→→ −− eNC 01

147

146

Since changing a neutron into a proton adds one positive charge, anegatively charged electron ( −−ββ particle) must be emitted to conserveelectric charge. An antineutrino is also emitted whenever an electronis given off.

e) Where does carbon-14 come from? Write a nuclear reaction to show theformation of C14

6 .

The energy from cosmic rays can turn stable nitrogen-14 in the air intocarbon-14. One proton in the nitrogen atom turns into a neutron. Anantielectron (a positron) and a neutrino are emitted.

νν++++→→ ++ eCN 01

146

147

3) Carbon Datinga) 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 anarcheological object?

Both stable Carbon-12 and unstable Carbon-14 isotopes are present inthe atmosphere (in carbon dioxide). Living organisms absorb bothisotopes of carbon in the process of respiration. After an organismdies, it no longer absorbs any new Carbon-14, and the Carbon-14within it decays. We can accurately estimate the time of an organism'sdeath, if we know the ratio Carbon-12 to Carbon-14 in the atmosphereat the time the organism died, the present ratio of Carbon-12 toCarbon-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-sixtyfourth (1/64) the amount of Carbon-14 that would have been present in livingbone at the time this organism was alive. What is the age of this bone?

Since 62

1641

== we know that 6 half-lives have passed since this

organism died.

6 half lives x 5,568 years = 33,408 years half life

d) Group Discussion Question: An archeological team works in a dig they believe tobe around one million years old. Would Carbon-14 dating be appropriate todetermine the age of this site? Why or why not?

4) Radioactive decay modeled by capacitor dischargeYour instructor will explain how to use a board that simulates radioactive decay withcircuit elements and a capacitor.

a) After charging the capacitor, flip the switch to the right to allow the capacitor todischarge through the resistor. At the same time, start the timer. Measure thevoltage with a multimeter set to measure DC voltage ( V ) every 15 seconds forfour minutes. Record your measurements in the table below.

Timeelapsed

(Min: Sec)

Voltage(volts)

Timeelapsed

(Min: Sec)

Voltage(volts)

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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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 todischarge through the resistor. Start the timer and measure the voltage every 15seconds for 4 minutes. Record your measurements in the table below.

Timeelapsed

(Min: Sec)

Voltage(volts)

Timeelapsed

(Min: Sec)

Voltage(volts)

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

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 alonger time? Why or why not?

No, the background radiation (in this case, background voltage) isalways present.

e) From your graph, find the half-life of the capacitor’s discharge. To do so, you mustsubtract 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 backin the background voltage. This gives ½ the original voltage, correctedfor the background. Find this voltage on your graph. The time elapsedbetween this point and your original point is the half-life – the time it tookfor 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 substancechanges its mass. We have found this mass change too small to measure for physicalchanges and chemical reactions. Next, we consider a situation where mass changes canbe measured – nuclear reactions.

6) Calculating the energy released during nuclear reactions

The mass of a U23892 nucleus is 395.2138 x 10–27 kg. Follow the steps below to find

the binding energy of this nucleus.

a) How many protons does the U23892 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 U23892 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.

Binding energy = (mass difference) c2 = (3.2008 x 10-27 kg ) x(3 x 108m/s)2

= (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 U23892 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 = 7.57 MeV1 joule

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7) Releasing energy from matter

a) When one proton and one neutron come together to form one deuteriumnucleus, 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 EnergyThe 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 ( H21 ). What is the binding energy per

nucleon of deuterium? _1 MeV_

d) The next point to the right is helium ( He42 ). 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 onehelium 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 He42 , 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 fuelcell to form a water molecule, the chemical binding energy given off is approximately2.5 eV per molecule. Approximately how many times greater is the nuclear bindingenergy given off when a proton and a neutron combine to form a nucleus ofdeuterium? (See question 7.d.) Why is it that we cannot measure a change in massin 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 timesgreater than the binding energy of a water molecule.