1

Nuclear Chemistry

I. Radioactivity. A. Review Isotopic Notation. 1. Proton, neutron and electron model. a. Very approximate and incorrect model. However, it can be used to keep tract of, but not explain, radioactive decay and nuclear stability. Mass Charge Particle Symbol kg amu (u) Coulombs Electron Units

Proton 11 H 1.67252x10-27 1.007276 1.6022x10-19 +1

Neutron 10 n 1.67496x10-27 1.008665 0 0

Electron 0-1 e 9.1095x10-31 0.000549 -1.6022x10-19 -1

2. Protons and neutrons comprise the nucleus = nucleons. 3. Atomic number = Z = Charge on nucleus in electron charge units = number of protons in the nucleus. Mass number = A = Total number of nucleons in nucleus = mass in amu rounded to nearest whole number = number of protons + number of neutrons. 4. Isotopic notations.

AZ(symbol)

Examples: 14 6 C {6 protons

8 neutrons } in nucleus + 6 electrons outside of nucleus

238 92 U { 92 protons

146 neutrons } in nucleus + 92 electrons outside of nucleus.

B. Radioactive Decay. 1. Neutrons are important in stabilizing the nucleus. In general, the greater the number of protons in the nucleus, the greater the number of neutrons required to stabilize it. a. For the lighter elements, up to Z = 20 (Ca) the stable, nonradioactive, nuclei have neutron to proton ratios, n/p = 1.

b. As Z increases the n/p ratio increases to a value of approximately 2.5 for 209 83 Bi.

c. All isotopes with Z > 83 are radioactive. 2. Isotopes outside this "belt of stability" will undergo radioactive decay and change there n/p ratio until a stable isotope is formed.

2

Proton/Neutron Ratio and Stability

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3. Types of radioactive decay. a. Alpha particle decay.

1) Alpha particle is a helium-4 nucleus. Symbol is 42 He or

42 α.

2) In alpha particle decay, Z goes down by 2 and A goes down by 4. Alpha decay is the only way a radioactive isotope can decrease its mass number.

3) Example: 238 92 U ---->

42 He +

234 90 Th

Note in a balanced nuclear equation the sum of the Z's and the sum of the A's on both sides of the equation must be the same. 4) Very important in the decay of the heavier elements b. Beta decay.

1) Beta particle is an electron that is emitted from the nucleus. Symbol is 0-1 e or β-.

2) In beta decay, Z goes up by one and A remains unchanged. A neutron is converted into a

proton (10 n --->

11 H + 0-1 e ).

3) Example; 14 6 C ----->

0-1 e +

14 7 N

c. Gamma decay.

1) Gamma ray is a very high energy photon. Symbol 00 γ or γ .

2) In gamma decay, neither Z nor A changes. Gamma radiation accompanies alpha and beta decay. It is the way a nucleus can lose energy. 3) Example: 60Co* ---> 60Co + γ where the asterisk (*) denotes an excited state of the nucleus. d. Positron emission. 1) Positron is a particle that is identical to an electron except that it has a positive

charge. It is the antimatter equivalent of the electron. Symbol 01 e or β+.

2) In positron emission, Z goes down by one and A remains unchanged. A proton is converted

into a neutron ( 11 H --->

10 n +

01 e).

3) Example: 2211 Na ---->

2210 Ne +

01 e

4) Positrons and electron will annihilate one another. The annihilation reaction is:

0-1 e +

01 e ---> 2 γ

e. Electron capture or K capture. 1) In K capture an electron from the lowest energy shell of the atom, the K shell, is captured by the nucleus. 2) In K capture, Z goes down by one and A remains unchanged. A proton is converted into a

neutron ( 11 H +

0-1 e --->

10 n).

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3) Example: 4422 Ti +

0-1 e ---->

4421 Sc

4. The most common forms of radioactive decay found in naturally occurring radioactive isotopes are alpha, beta, and gamma decay. K capture is also found but is not common. Positron emission is not found in naturally occurring isotopes. C. Naturally Occurring Radioactive Isotopes. 1. Heavier atoms with Z>83 are all radioactive. Most of the heavier isotopes are members of one of three series. a. The Uranium-radium Series.

205

210

215

220

225

230

235

240

A

80 82 84 86 88 90 92 94

Z

The uranium-radium series

4n + 2 series

1) The only way that the mass number changes is through alpha decay, therefore the mass numbers of all the isotopes will differ by multiples of four.

The heaviest isotope in the series is 238 92 U. Its mass number (238) is divisible by four with

two left over, that is 238 = 4 (59)+2. Therefore, all daughter isotopes will have mass numbers that are divisible by four with two left over. This series is also referred as the 4n+2 series.

2) Initial steps and half-lives.

238 92 U ------>

234 90 Th +

42 He t1/2 = 4.5x109 yr.

234 90 Th ------->

234 91 Pa +

0-1 e t1/2 = 24.4 da.

234 91 Pa ------->

234 92 U +

0-1 e t1/2 = 1.14 min.

234 92 U ------->

230 90 Th +

42 He t1/2 = 2.7x105 yr.

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3) Net decay: 238 92 U ----->

206 82 Pb + 8

42 He + 6

0-1 e

The slowest step is the first step with t1/2 = 4.5x109 yr, this is the effective half-life for the entire series. b. The uranium-actinium series.

205

210

215

220

225

230

235

240

A

80 82 84 86 88 90 92 94

Z

Uranium-actinium series

4n+3 series

1) The heaviest member is

235 92 U. Note that 235 = 4 (58)+3 . Therefore, all daughters in this

this series will have mass numbers that are divisible by four with three left over. The series is also called the 4n+3 series. 2) Initial steps and half-lives.

235 92 U ----->

231 90 Th +

42 He t1/2 = 7.1x108 yr.

231 90 Th ----->

235 91 Pa +

0-1 e t1/2 = 24.6 hr.

235 91 Pa ----->

227 89 Ac +

42 He t1/2 = 3.2x104 yr.

227 89 Ac ----->

227 90 Th +

0-1 e t1/2 = 13.5 yr.

3) Τhe net decay is 235 92 U ----->

207 82 Pb + 7

42 He + 4

0-1 e

The slowest step is the first step with a t1/2 = 7.1x108 yr. This is the effective half-life of the series.

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c. The Thorium series.

205

210

215

220

225

230

235

A

80 82 84 86 88 90 92

Z

Thorium series

4n series

1) The heaviest member is

232 90 Th. Note that 232 = 4 (58). Therefore, all the daughter

isotopes will have mass numbers that are divisible by four with zero left over. The series is also called the 4n series.

2) Initial steps and half-lives.

232 90 Th ------>

228 88 Ra +

42 He t1/2 = 1.39x1010 yr.

228 88 Ra ------>

228 89 Ac +

0-1 e t1/2 = 6.7 yr.

228 89 Ac ------>

228 90 Th +

0-1 e t1/2 = 6.1 hr.

228 90 Th ------->

224 88 Ac +

42 He t1/2 = 1.9 yr.

3) Net decay: 232 90 Th ------>

208 82 Pb + 6

42 He + 4

0-1 e

The slowest step is the first with a t1/2 = 1.39x1010 yr. The is the effective half-life of the series. d. For completeness there should be another series whose mass number would be divisible by

four with one left over. The 4n+1 series does not exist in nature but it has been synthesized in nuclear reactors. This series is called the neptunium series.

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205

210

215

220

225

230

235

240

245A

80 82 84 86 88 90 92 94 96Z

Neptunium series4n+1 series

1) Initial steps and half-lives.

237 93 Np ------->

233 91 Pa +

42 He t1/2 = 2.25x106 yr.

233 91 Pa ------->

233 92 U +

0-1 e t1/2 = 27.4 da.

233 92 U ------->

229 90 Th +

42 He t1/2 = 1.62x105 yr.

229 90 Th ------->

225 88 Ra +

42 He t1/2 = 7.0x103 yr.

2) The net decay: 237 93 Np ------>

209 83 Bi + 7

42 He + 4

0-1 e

The slowest step is the first step with a t1/2 = 2.25x106 yr. This is much faster than the steps in the other series. It has long since disappeared from the earth's crust. 2. Other naturally occurring radioactive isotopes. a. There are some radioactive elements found in the earth's crust that are not members of one of

the heavy atom series. Some have half-lives that make them useful in radiochemical dating.

b. 14 6 C is a naturally occurring radioactive isotope that decays by beta emission with a half-life of

5730 yrs. It is being continually produced in the upper atmosphere by the interaction of

cosmic rays with 14 7 N.

The production and decay reactions are;

Production 14 7 N +

10 n ------>

14 6 C +

11 H

decay 14 6 C ----->

14 7 N +

0-1 e

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c. 31 H (tritium)

31 H ---->

32 He +

0-1 e t1/2 = 12.3 yr.

d. 4019 K.

4019 K undergoes two modes of decay,

4019 K ----->

4020 Ca +

0-1 e (89%)

4019 K +

0-1 e --->

4018 Ar (11%)

The half-life for 4019 K decay is 1.28x109 yr. Although K capture is not the main mode of decay,

the 4018 Ar isotope is fairly unique and the

40Ar40K ratio is used in dating minerals.

D. Artificial or Induced Radioactivity. 1. Bombardment reactions. a. In 1934 Frédéric and Iréne Joliot-Curie reported that B, Mg, and Al could be made radioactive when bombarded with alpha particles. This was the first instance of the synthesis of new isotopes. The nuclear reactions were as follows:

10 5 B +

42 He ---->

13 7 N +

10 n

2713 Al +

42 He ---->

3015 P +

10 n

2412 Mg +

42 He ---->

2714 Si +

10 n

All of these "artificial" radioactive isotopes decay by positron emission with half lives of 10 min. for 13N, 2.5 min. for 30P, and 4.2 s. for 27Si. b. These nuclear reactions can be abbreviated as:

10 5 B(α,n)

13 7 N [this is called a alpha-in-neutron-out reaction];

2713 Al(α,n)

3015 P; and

2412 Mg(α,n)

2714 Si.

c. A large number of artificial radioactive isotopes have been made by bombarding stable nuclei with light charged particles (alpha particles, deuterium, protium, boron, etc.) 1) Such reactions are called nuclear transformation reactions and the process is called nucleosynthesis. 2) Since the transformation reactions require the merging of two nuclei, they have extremely high activation energies, and are usually carried out using particle accelerators, such as, cyclotrons.

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2. Transuranium elements. a. Uranium is the heaviest naturally occurring element. All elements with Z's greater than 92 have been produced by nucleosynthesis. Some examples are shown in the following Table. Selected Transuranium Elements Z Symbol Synthetic Method Half-life

93 Np 238 92 U +

10 n ---->

239 93 Np +

0-1 e 2.35 da.

94 Pu 239 93 Np ----->

239 94 Pu +

0-1 e 86.4 yr.

95 Am 239 94 Pu +

10 n --->

240 95 Am +

0-1 e 458 yr.

96 Cm 239 94 Pu +

42 He --->

242 96 Cm +

10 n 4.5 hr.

99 Es 238 92 U + 15

10 n --->

253 99 Es + 7

0-1 e 20 da.

101 Md 253 99 Es +

42 He --->

256101 Md +

10 n 1.5 hr.

103 Lr 252 98 Cf +

10 5 B --->

257103 Lr + 5

10 n 8 s.

104 Rf 249 98 Cf +

12 6 C --->

257104 Unq + 4

10 n 4.5 s.

105 Db 249 98 Cf +

15 7 N --->

260105 Unp + 4

10 n ----

106 Sg 249 98 Cf +

18 8 O --->

263106 Unh + 4

10 n <10-2s.

109 Mt 209 83 Bi +

5826 Fe --->

266109 Une +

10 n -----

b. Symbols and nomenclature of the transuranium elements.

1) Until a name can be agreed on, IUPAC has systemized the "naming" of the newly discovered elements. They are just called by their atomic number. Therefore, element with the atomic number of 112 is just called element-112.

2) A more formal way of expressing the atomic number and naming is to use a three letter symbol that gives its atomic number. The roots are as follows:

digit name digit Name digit Name 0 nil 4 quad 8 oct 1 un 5 pent 9 enn 2 bi 6 hex 3 tri 7 sept 3) Examples: element 104 = Unq = unnilquadium (the elemental ium ending is appended last). element 112 = Uub = ununbium

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II. Half-Life and the Rates of Decay. A. Review. 1. Radioactive decay is a first order process. Therefore, the half-life is independent of the amount. 2. Consider a radioactive isotope whose half-life = t1/2. After a time, t = n t1/2, has elapsed, the

fraction remaining f = (12 )n

.

3. The useful equations are:

number of half-lives elapsed = n = t

t1/2

fraction remaining = f = (12 )n

amount remaining = (f) (original amount)

B. 14 6 C. Radiocarbon Dating.

1 As discussed above, 14C is being produced in the upper atmosphere and is decaying with a half-

life of 5740 yrs. Since 14 6 C is being produced and consumed at fairly constant rates, a small

steady state concentration of 14 6 C exists is nature, so that the

14 6 C12 6 C

ratio in nature is about 1

1012

2. The 14 6 C can react with O2 in the atmosphere to form 14CO2 . The CO2 can be taken up by

plants and thus enters the food chain. Therefore, all living organisms are somewhat radioactive

due to the 14 6 C. The amount of

14 6 C in living organisms is enough to give about 15.3 counts per

min. per gram of carbon on a Geiger Counter. All living organisms are continually

exchanging CO2, and thus maintain a constant level of 14 6 C. They should give 15.3 counts per

min. per g of C. When the organism dies, the 14 6 C is trapped, and will decay with a half-life of

equal to 5730 yrs; the counts would slowly decrease from the 15.3 counts per min. per g. of C found for living organisms. This can be used to date old carbonaceous samples. 3. Example. The Lescaux Cave near Montignac, France contains some remarkable drawings made by prehistoric man. A sample of charcoal from campsite in the cave gave 2.34 counts per min. per g. of C. Given that freshly made charcoal gives 15.3 counts per min. per g of C, calculate the age of the campsite. 1) Since radioactive decay is kinetically a first order process, at a time

t = (n) (t1/2), the fraction remaining, f = (12 )n

. 2) Fraction of

14 6 C remaining =

2.34 counts15.3 counts = 0.153 = (1

2 )n

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∴ (0.5)n = 0.153 or n log(0.5) = log(0.153)

∴ n = log(0.153)log(0.5) =

-0.8153-0.3010 = 2.71

∴ t = (2.71) (5730 yrs.) = 15,520 yrs. C. Other dating methods. 1. The radium-uranium series begins with 238U and ends with 206Pb. The slowest step in the series is:

238 92 U ----->

234 90 Th +

42 He with a t1/2 = 4.5x1010 yr. This is basically the half-life for the series.

That is if you start with 1 mole of 238 92 U in 4.5x109 yrs, you will have 0.5 mole of

238 92 U and 0.5

mole of 206 82 Pb. Therefore, the ratio of

238U206Pb in a uranium containing mineral can be use to date the

mineral. 2. Example: Suppose that a rock sample was analyzed and found to contain 1.25% 238U and 0.62% 206Pb. Calculate the age of the rock. A 100 g sample of the mineral would contain 1.25 g 238U and 0.62 g 206Pb. Assuming that all the 206Pb can from the decay of 238U, then

mole of 238U = 1.25 g

238 g/mol = 5.25x10-3 mol

mole of 206Pb = 0.62 g

206 g/mol = 3.01x10-3 mol Total moles of 238U originally present = 8.26x10-3 mol

∴ the fraction remaining, f = 5.25x10-3

7.26x10-3 =0.636 = (12 )

n

n = log0.636log0.5 =

-0.1968-0.3010 = 0.645

age = n x t1/2 = (0.645)(4.5x109yr) = 2.9x109 yrs.

3. In the same way, the 235U207Pb ratio, the

232Th208Pb ratio, the

40K40Ar ratio, and the

87Rb87Sr ratio

(t1/2 = 5x1011 yr) have also been used in dating rocks. D. Rates of disintegration and the biological effects of radiation. 1. The fundamental unit of radiation is the Curie (Ci). a. 1 Ci = 3.70x1010 disintegrations per second. This is equivalent to the radiation produced by 1 g of radium. b. 1 Becquerel (Bq) = 1 disintegration per second (1 Ci = 3.70x1010 Bq.). 2. Absorbed doses of radiation. a. rad (radiation absorbed dose) = amount of radiation that is equivalent to 1x10-5 J/g of irradiated material.

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b. The biological effectiveness of the radiation depends on the type of radiation and the type of tissue being irradiated. The biological effect of the radiation can be given by a parameter called RBE (relative biological effectiveness). 1) α particles have extremely high energies and are about 20 times as effective (damaging) than are β and γ particles, which are about equal in there biological effectiveness. Therefore, the RBE of β and γ radiation is set equal to 1, the RBE of α particles is 20. 2) Some tissues are more susceptible to radiation damage than are others. Whole body radiation is given a RBE = 1, some tissues, such as bone marrow have much greater RBE's. 3) The product of 1 rad x 1 RBE = 1 rem (roentgen equivalent man) c. Biological effects of radiation doses. (These are continually being revised See end of notes for

more information)) Effect Dose(rem) No observable effects 0 - 25 Decrease in white blood cell count 25 - 50 Nausea, no deaths 100 - 200 Death to 50% of irradiated subjects 450 100% death rate >800 d. One other factor that is important in determining the biological danger of radiation is its penetrating power. 1) α particles, while extremely ionizing are not very penetrating. The can be stopped by

clothing or a sheet of paper and do not penetrate past the dead outer layer of skin. Therefore, this type of radiation from an outside source is not too dangerous. However, if an α emitter is ingested or inhaled, it is very deadly.

2) β particles are more penetrating, they can pass through a piece of paper or light clothing but are stopped by a piece of wood. They can penetrate the dead outer layer of skin, but are stopped within the skin. They can cause skin damage so that it appears as if the skin were rburned. Ingested or inhaled, β emitters are very damaging.

3) γ rays are very penetrating, it requires about 3 cm of lead to stop γ rays. They will completely penetrate the body causing extensive cell damage. II. Nuclear Stability. A. Binding Energy. 1. Any nucleus is more stable than is its composite protons and neutrons. Therefore, the process:

Z protons + N neutrons ------> Z+NZ X (where Z+N = A = mass number )

is exothermic.

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a. The binding energy is the energy liberated in this process.

b. The binding energy per nucleon = binding energy

A . The most stable nuclei will have the highest values of the binding energy per nucleon. c. According to Einstein's special theory of relativity, when an amount of mass of M kg is destroyed, the amount of energy, E, liberated, in J, is given by: E = Mc2 where c is the speed of light ( 3.0x108 m/s ). In ordinary chemical reactions the energy changes

are so small that mass differences between reactants and products are beyond the detection range of our most sensitive balances, and the law of conservation of mass holds. However, for nuclear processes, such as the formation of nuclei, the mass changes are measurable.

d. The mass defect = mass lost when nuclei are formed from their composite protons and neutrons.

2. Example: The mass of a 73 Li nucleus is 7.016005 amu. Given that the mass of a proton is

1.007276 amu and that of a neutron is 1.008665 amu, calculate the mass defect, the binding

energy, and the binding energy per nucleon of 73 Li.

a. mass defect.

The process is: 3 11 H + 4

10 n ------>

73 Li

mass of protons = 3 (1.007276 amu) = 3.021828 amu mass of neutrons = 4 (1.008665 amu) = 4.034660 amu Total mass of protons & neutrons = 7.056488 amu

- mass of 73 Li nucleus = -7.016005 amu

mass defect = 0.040483 amu

When the 73 Li is formed from its protons and neutrons, this much mass is converted into energy .

b. Binding energy.

The mass defect in kg = 0.040483 amu

6.023x1026 amu/kg = 6.721x10-29 kg

The binding energy in J = (6.721x10-19 kg) ( 2.998x108 m/s)2 = 6.04x10-12 J

This is the energy liberated when one 73 Li nucleus is formed, it is equivalent to 3.64x1012 J/mol.

A more convenient unit of energy is the MeV (Million electron Volt). An electron volt is the energy change when 1 electron (charge = 1.602x10-19 C) is transferred across a potential energy difference of 1 volt (1 J/C). Therefore, 1 eV = 1.602x01-19 J 1 MeV = 1.602x10-13 J

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The binding energy in MeV = 6.04x10-12 J

1.602x10-13 J/MeV = 37.7 MeV

c. The binding energy per nucleon of 73 Li =

37.7 MeV7 = 5.38 MeV per nucleon.

A convenient conversion factor to remember is that the destruction of 1 amu of mass will yield 933.2 MeV. Therefore, the binding energy in MeV can be obtained by multiplying the mass defect in amu by 933.2. 3. The binding energy per nucleon has been calculated for most nuclei. A plot of binding energy per nucleon vs. the mass number (A) gives a display of the relative stabilities of the nuclei. Such a plot is shown below.

4. The most stable nuclei have mass numbers in the 50-60 range. The most stable nucleus is

5626 Fe with a

binding energy per nucleon of 8.6 MeV.

a. Note that 42 He, the alpha particle, has an unusually high binding energy per nucleon.

b. The stability of the lighter nuclei (A's less than about 40) increase rapidly with increasing mass number, while changes are much smaller for the heavier isotopes. B. Fission and Fusion. 1. Fission = the splitting apart of a heavy nucleus into lighter ones. a. Fission will be exothermic for isotopes with A's greater than 56.

b. There are certain isotopes, such as 235 92 U and

239 94 Pu, that can undergo sustained fission

chain reactions.

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When a 235 92 U captures a neutron, the resulting

236 92 U that is formed will split apart in

two lighter nuclei, one in the mass range of ≈ 100 and another in the mass range of

≈140. An example of this fission process is:

235 92 U +

10 n ---> [

236 92 U] ---->

9038 Sr +

143 54 Xe + 3

10 n

1) Note that 3 neutrons are also produced in the fission reaction.

2) If enough 235 92 U is located in the immediate vicinity, the probability of one of the product

neutrons encountering another 235 92 U will be large enough so that a chain reaction is set

up. Since there are more neutrons produced than consumed, the reaction will be a branching chain reaction and the rate would begin to increase without bounds leading to a nuclear explosion. 3) The critical mass = the minimum amount of fissionable material necessary to generate a

self-sustained chain reaction. In an atomic bomb a critical mass of 235 92 U or

239 94 Pu is

suddenly brought together. The actual amount will depend on the configuration and the

shape of the sample. The bombs dropped on Hiroshima and Nagasaki most probably

consisted of a few kilograms of 235 92 U ( used on Hiroshima) or

239 94 Pu (used on Nagasaki)

that were suddenly compressed into a small sphere; the explosions were equivalent to about

20 kilotons of TNT (1 kiloton of TNT will liberate 4 billion kJ of energy). 2. Fusion = merging of two nuclei into a heavier one. a. Fusion reactions are exothermic for the lighter isotopes. That is, for fusion reactions in which isotopes with A's < 56 are produced. b. Since the binding energy vs. A curve rises much more steeply for A's < 56 than it drops for A's > 56, more energy is obtained per gram of nuclear fuel for fusion reactions than for fission reactions. 1) Since fusion reactions have extremely large activation energies, very high temperatures, in excess of several million degrees, are required to initiate fusion reactions. 2) The energy of the sun and stars is furnished by the fusion of hydrogen into helium, and

ultimately, other elements.

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At temperatures less than 20 million degrees the conversion can be envisioned to take place through the following steps.

11 H +

11 H ---->

21 H +

01 e

21 H +

11 H ---->

32 He + γ

32 He +

32 H ---->

42 He + 2

11 H

The net reaction is the conversion of 4 11 H into a

42 He (and 2

01 e's). At much higher

temperatures further fusion processes occur producing carbon and heavier elements. c. Thermonuclear bombs derive most of their energy from a fusion reaction. A modern 20

megaton superbomb is a three stage fission-fusion-fission device. Its first stage (the detonator) consists of a few kilograms of 239Pu. The fission bomb detonator is surrounded by about 150 kg of LiD that can undergo fusion (6Li2D ----> 2 4He); this is the second stage. The LiD is surrounded by about 1000 pounds of 238U, that can undergo fission with fast neutrons, that acts as the third stage.

The total weight of a 20 megaton bomb is about 1.5 tons; its explosive power is roughly three times the total of all the bombs used in the Second World War.

C. Nuclear Reactors - Controlled Fission Reactions. 1. When a critical mass of 235U, or some other fissionable isotope, is brought together, a branching chain reaction will take place and an explosion will result. a. Chain branching occurs because in the fission reaction a neutron is required to induce fission and three neutron are produced in the reaction.

235 92 U +

10 n ---> [

236 92 U] ---->

9038 Sr +

143 54 Xe + 3

10 n

If some of the product neutrons are absorbed before they strike other 235U nuclei, then the rate of the chain reaction can be controlled and a sustained chain reaction can be achieved. b. In a Nuclear Reactor a sustained chain reaction takes place and a constant amount of energy is produced. ( See the accompanying diagram.) 2. Components. a. Fuel Rods- Contain uranium, usually in the form of uranium oxide (U3O8). 1) Natural uranium is only about 0.7% 235U, which is too low to support a chain reaction. 2) The uranium used must be enriched in 235U. One way in which this has been accomplished is through the diffusion, or centrification, of UF6(g).

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F is unusual in that it is one of the few naturally occurring substances in which only one isotope is found in nature. Therefore, molecules of UF6 will have slightly different masses depending on the isotope of uranium that is incorporated in the

Control

Rods

Fuel Rods

(U O )3 8

Heat

Exchanger

ModeratorShield

CORE

Steam to

Turbine

Schematic of Nuclear Reac tor

compound. A molecule of 238UF6 will be heavier than a molecule of 235UF6 by 3 amu. In accord with Graham's Law of Diffusion, the 235UF6 will diffuse fasted than a molecule of 238UF6 by an amount given by the expression,

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RateU-235RateU-238 =

M 238UF6

M 235UF6

= 352349 = 1.004

This is not much of a rate enhancement, but given long enough diffusion paths, the leading front of UF6(g) will be enriched in 235U. 3) Other methods, such as the use of centrifuges and mass spectrometers can also be used to enrich uranium. The enriched uranium is 3 - 4% uranium-235. b. Control Rods. Rods of Cd and B, elements with large neutron capture radii, that are inserted between the fuel rods to absorb neutrons and slowdown the chain reaction. When the control rods are fully inserted, the reactor is shut down. c. Moderators. These are substances that decrease the kinetic energy of the neutrons. Because the fission reaction is very exothermic, the emitted neutrons have very high kinetic energies. These fast neutrons are not easily captured by 235U nuclei. To improve the efficiency of neutron capture a moderator is used to slowdown the neutrons. 1) The moderator is a substance that surrounds the fuel rods that slow the neutrons and also carry off the heat generated by the reactor. The most common moderator is water. 2) Light Water Reactors. Use naturally occurring water, that is composed mainly of 1H2O. Light water reactors require the use of enriched uranium. Most U. S. reactors are light water reactors. 3) Heavy Water Reactors. Heavy water is D2O (D = deuterium = 2H). D2O is not as efficient in slowing neutrons as is light water. Therefore, the faster neutrons will travel greater distances and a 235U will have a higher probability of being struck. Essentially all the uranium-235 is reacted. Do not need enriched uranium to run heavy water reactors. 3. Breeder Reactors. Reactors that produce more fissionable fuel than they consume.

a. Uranium-238 can absorb fast neutrons to give 23994 Pu, which is fissionable.

23892 U +

10 n ------>

23992 U --------->

23993 Np +

0-1 e (t1/2 = 23.4 min)

23993 Np ------->

23994 Pu +

0-1 e (t1/2 = 2.35 da)

The half-life of plutonium-239 is 24,400 yr. b. Uranium-233 is also a fissionable isotope that can be produced from 232Th in a breeder reactor. The equations are,

23290 Th +

10 n --------->

23392 U + 2

0-1 e

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c. In a breeder reactor the core is surrounded by a blanket of uranium-238, so that some of the emitted neutrons will form plutonium-239. If, on the average, for every fission process more than one neutron is absorbed by a uranium-238, then the amount of plutonium-239 produced will be greater than the amount of nuclear fuel consumed. The doubling time = time required for the fuel to double. For fast breeders doubling times are 7 - 10 yrs. d. There are a number of problems associated with breeder reactors, the neutrons used are fast neutrons so that a moderator, such as water, is not used. Therefore other substances, such as liquid sodium, are used as coolants. These are very corrosive.

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Table of exposure levels and symptoms

Dose-equivalents are presently stated in sieverts (Sv). 1 Sv = 100 REM

0.05–0.2 Sv (5–20 REM)

No symptoms. Possible potential for cancer and mutation of genetic material. A few researchers

contend that low dose radiation may be beneficial.

50 mSv is the yearly federal limit for radiation workers in the United States.

In the UK the yearly limit for a classified radiation worker is 20 mSv.

In Canada, the single-year maximum is 50 mSv, but the maximum 5-year dose is only 100 mSv.

Company limits are usually stricter so as not to violate federal limits.

0.2–0.5 Sv (20–50 REM)

No noticeable symptoms. Red blood cell count decreases temporarily.

0.5–1 Sv (50–100 REM)

Mild radiation sickness with headache and increased risk of infection due to disruption of

immunity cells. Temporary male sterility is possible.

1–2 Sv (100–200 REM)

Light radiation poisoning, 10% fatality after 30 days (LD 10/30). Typical symptoms include

mild to moderate nausea (50% probability at 2 Sv), with occasional vomiting, beginning 3 to 6

hours after irradiation and lasting for up to one day. This is followed by a 10 to 14 day latent

phase, after which light symptoms like general illness, and fatigue appear (50% probability at 2

Sv). The immune system is depressed, with convalescence extended and increased risk of

infection. Temporary male sterility is common. Spontaneous abortion or stillbirth will occur in

pregnant women.

2–3 Sv (200–300 REM)

Severe radiation poisoning, 35% fatality after 30 days (LD 35/30). Nausea is common (100% at 3 Sv),

with 50% risk of vomiting at 2.8 Sv. Symptoms onset at 1 to 6 hours after irradiation and last for 1 to 2

days. After that, there is a 7 to 14 day latent phase, after which the following symptoms appear: loss

of hair all over the body (50% probability at 3 Sv), fatigue and general illness. There is a massive loss of

leukocytes (white blood cells), greatly increasing the risk of infection. Permanent female sterility is

possible. Convalescence takes one to several months.

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3–4 Sv (300–400 REM)

Severe radiation poisoning, 50% fatality after 30 days (LD 50/30). Other symptoms are similar

to the 2–3 Sv dose, with uncontrollable bleeding in the mouth, under the skin and in the kidneys

(50% probability at 4 Sv) after the latent phase.

4–6 Sv (400–600 REM)

Acute radiation poisoning, 60% fatality after 30 days (LD 60/30). Fatality increases from 60% at

4.5 Sv to 90% at 6 Sv (unless there is intense medical care). Symptoms start half an hour to two

hours after irradiation and last for up to 2 days. After that, there is a 7 to 14 day latent phase,

after which generally the same symptoms appear as with 3-4 Sv irradiation, with increased

intensity. Female sterility is common at this point. Convalescence takes several months to a

year. The primary causes of death (in general 2 to 12 weeks after irradiation) are infections and

internal bleeding.

6–10 Sv (600–1,000 REM)

Acute radiation poisoning, near 100% fatality after 14 days (LD 100/14). Survival depends on

intense medical care. Bone marrow is nearly or completely destroyed, so a bone marrow

transplant is required. Gastric and intestinal tissue are severely damaged. Symptoms start 15 to

30 minutes after irradiation and last for up to 2 days. Subsequently, there is a 5 to 10 day latent

phase, after which the person dies of infection or internal bleeding. Recovery would take several

years and probably would never be complete.

Devair Alves Ferreira received a dose of approximately 7.0 Sv (700 REM) during the Goiânia

accident and lived partially due to his fractionated exposure.

10–50 Sv (1,000–5,000 REM)

Acute radiation poisoning, 100% fatality after 7 days (LD 100/7). An exposure this high leads to

spontaneous symptoms after 5 to 30 minutes. After powerful fatigue and immediate nausea caused by

direct activation of chemical receptors in the brain by the irradiation, there is a period of several days

of comparative well-being, called the latent (or "walking ghost") phase. After that, cell death in the

gastric and intestinal tissue, causing massive diarrhea, intestinal bleeding and loss of water, leads to

water-electrolyte imbalance. Death sets in with delirium and coma due to breakdown of circulation.

Death is currently inevitable; the only treatment that can be offered is pain therapy. Louis Slotin was

exposed to approximately 21 Sv in a criticality accident on 21 May 1946, and died nine days later on

30 May.

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50–80 Sv (5,000–8,000 REM)

Immediate disorientation and coma in seconds or minutes. Death occurs after a few hours by

total collapse of nervous system.

More than 80 Sv (>8,000 REM)

U.S. military forces expect immediate death. A worker receiving 100 Sv (10,000 REM) in an accident at Wood River, Rhode Island, USA on 24 July 1964 survived for 49 hours after exposure, and an operator receiving 120 Sv (12,000 REM) to his upper body in an accident at Los Alamos, New Mexico, USA on 30 December 1958 survived for 36 hours.

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Problems

1. Complete and balance the following nuclear equations.

a. 23491 Pa ------>

0-1 e + b.

21484 Po ------>

21082 Pb +

c. 125 B ------->

126 C + d.

147 N +

42 He ------>

11 H +

e. 23892 U +

42 He ------->

0-1 e + f.

25098 Cf +

115 B ------> 4

10 n +

g. 85 B ----->

84 Be + h.

15

30 P -------> 3014 Si +

2. Calculate the binding energy per nucleon, in eV and kJ/mol, of the 115 B nucleus, given that the

mass of a 115 B nucleus is 11.00656 amu.

3. complete the notations for the following processes.

a. 24 Mg(

21 H,

42 He)_____ d.

126 C(

21 H,

10 n)____ g.

5927 Co(

10 n,

42 He)______

b. 2612 Mg(

21 H,

11 H)_____ e.

13052 Te(

21 H,2

10 n)_____

c. 4018 Ar(

42 He,

11 H)______ f.

5525 Mn(

10 n,

00 γ)_______

4. At the centers of very heavy stars, two 16O atoms can fuse to produce an atom of 32S. Calculate the energy released in kJ/mol. (mass 32S = 31.9721 amu, 16O = 15.9949 amu ) 5. Through a series of decay steps in the 4n+2 series, 238U is converted to 214Pb. How many alpha particles and beta particles are lost in this process? 6. The binding energy of 2H is 1.7x10-16kJ per nucleon and that of 4He is 1.1x10-15kJ per nucleon. Calculate the energy released when 2 moles (4.0g) of 2H is fused to form of 4He. Given that the heat of combustion of CH4 is 890 kJ/mol, how many grams of methane would have to be burned in order to produce this much energy? 7. Given that the binding energy per nucleon of 12C = 1.233x10-15 kJ, calculate the mass of the 12C in amu. From this mass show that the mass of the 12C isotope is 12.000 amu. 8. If a 235U, after the absorption of a slow neutron, undergoes fusion to give 142Ba and 93Kr, write the balanced nuclear reaction. 9. Given that 13B undergoes beta decay to give 13C and the isotopic mass of 13B = 13.0178 amu and that of 13C = 13.0034 amu, calculate the energy change for the radioactive decay process. (Note that these masses are not nuclear masses)

10. Assume that you have prepared 100 g of a compound containing a radioactive element whose half-life is 20 min. How many grams of the radioactive compound would remain 2.0 hrs after preparation?

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11. 50.0 grams of a radioactive isotope was found to decay at a rate such that only 6.75 g of the isotope remains after 30 hrs. What is the half-life of the isotope? 12. A wooden artifact from an Egyptian tomb was dated using 14C. Suppose that the artifact gave

8.3 counts per min per g of C. Given that fresh wood gives 15.3 counts per min per g of C and the half-life of 14C is 5730 yrs, calculate the age of the artifact.

13. Potassium-40 undergoes K capture to form 40Ar with a half-life of 1.3x109 yr. Suppose that analysis of a moon rock gave an potassium-40/argon-40 molar ratio of 0.816, calculate the age of the rock. 14. The age of the earth is about 2 billion years. Using the half-lives given in this chapter, calculate the maximum ratio of 206Pb to 238U you would expect in the oldest possible rock sample.