NUCLEAR CHEMISTRY

2F-1 (of 15)

NUCLEONS – The particles found in the nucleus

Protons (+)Neutrons (0)

ATOMIC NUMBER (Z) – The number of protons in the nucleus, also equal to the charge of the nucleus

MASS NUMBER (A) – The number of nucleons in the nucleus, or protons plus neutrons in the nucleus

and are isotopes

NUCLIDE – An atom with a specific number of protons and neutrons

ISOTOPES – A set of nuclides with the same number of protons

Hg80

196A Z

Protons: 80 Neutrons: 196 - 80 = 116

Hg80

196 Hg80

198

2F-2 (of 15)

NUCLEAR REACTIONS

Reactions that produce new atoms

TRANSMUTATION – When an atom of one element is changed into an atom of another element

14 N + 7

O + 17

8

Artificial elements are made by bombarding large nuclei with smaller ones

238 U +92

1 H 1

238 Np +93

He → 4

2

12 n 0

2 H → 1

2F-3 (of 15)

In all nuclear reactions (1) the mass number is conserved and (2) atomic number (or charge) is conserved

STABILITY SERIES

Nuclides are stable when their nuclei have enough neutrons to minimize proton-proton repulsion

(a) For Z < 20

Stable nuclei have n:p ratio of 1:1

(b) For Z > 20

As Z increases, stable nuclei have n:p ratio that increases from 1:1 to eventually 1.5:1

STABLE NUCLIDES (STABLE ISOTOPES) – Atoms with nuclei that last forever

RADIOACTIVE NUCLIDES (RADIOISOTOPES) – Atoms with nuclei that eventually break down to more stable nuclei

2F-4 (of 15)

Nuclides to the left of the line of stability are unstable because they are neutron poor

Nuclides to the right of the line of stability are unstable because they are neutron rich

Nuclides beyond the line of stability (with Z > 83) are unstable because they have too many total protons

Stable 16O atom:8 n, 8 p

(1:1 ratio)

Stable 200Hg atom:120 n, 80 p1.5:1 ratio

2F-5 (of 15)

Most of the stable nuclides have even numbers of protons and neutrons

Neutrons

Protons

Even Odd

Even

Odd

166

57

53

8

2F-6 (of 15)

There are 284 known stable nuclides

NUCLEAR DECAY – The process in which a radioactive nuclide turns into a more stable nuclide

The type of decay depends on whether the radioactive nuclide has too many total protons, if it is neutron rich, or if it is neutron poor

2F-7 (of 15)

α’s are emitted from radioisotopes beyond the line of stability, those with too many total protons (Z > 83 or A > 200)

(1) ALPHA DECAY (α) – The release of a helium-4 nucleus (4He2+) from a radioactive nucleus to become more stable

U →92

238 Th90

234 4 α +2

A and Z are always conserved in nuclear changes

Alpha particles can be stopped by the outermost layer of skin

2F-8 (of 15)

β-’s are emitted from radioisotopes that are to the right of the line of stability, those that are neutron rich

Essentially a neutron decays into a proton and an electron

(2) BETA MINUS DECAY (β-) – The release of an electron from a radioactive nucleus to become more stable

C → 6

14 N 7

14 0 β- +-1

Beta particles penetrate about 1 cm into the body

2F-9 (of 15)

EC occurs in radioisotopes to the left of the line of stability, those that are neutron poor

Essentially an electron and a proton turn into a neutron

(3) ELECTRON CAPTURE (EC) – An electron is captured by the nucleus to become more stable

Be + 4

7 Li 3

7 0 e- →-1

2F-10 (of 15)

β+’s are just like electrons, but with a positive charge

An electron is matter, but a β+ is ANTIMATTER

When a β+ and β- meet, they are ANNIHILATED, meaning all of their mass is converted into energy

A β+/β- annihilation forms 2 equal energy EM radiation photons

(4) POSITRON DECAY (β+) – The release of an electron with a positive charge from a nucleus to become more stable

2F-11 (of 15)

β+’s are emitted from radioisotopes to the left of the line of stability, those that are proton rich

Essentially a proton decays into a neutron and an antimatter electron

C → 6

11 B 5

11 0 β+ + 1

(4) POSITRON DECAY (β+) – The release of an electron with a positive charge from a nucleus to become more stable

2F-12 (of 15)

γ’s are emitted along with other forms of decay, or when an excited nucleus releases energy

(5) GAMMA DECAY (γ) – The release of any high energy photon of electromagnetic radiation

Ho → 67

163m Ho67

163 γ + 0

0

Gamma rays are deeply penetrating

2F-13 (of 15)

K →19

40 40 Ca20

0 β- +-1

+ γ 0

0

2F-14 (of 15)

Several neutrons, and lots of energy are released when nuclei fission

(6) SPONTANEOUS FISSION – When a large nucleus (Z > 80) breaks into two, approximately equal halves

U →92

239 Cd + 48

120 Ru +44

116 3 n 0

1

U →92

239 Ag + 47

119 Rh +45

118 2 n 0

1

Daughter Products

usually very radioactive, and always different

2F-15 (of 15)

THE RATE OF NUCLEAR DECAY

Each radioisotope undergoes nuclear decay at its own unique rate

HALF-LIFE (t1/2) – The time required for half of the radioisotopes in a sample to decay

The shorter the half-life, the more unstable the radioisotope

Half-life for 125I = 60 days

At 0 days:

At 60 days:

At 120 days:

At 180 days:

At 240 days:

16 125I atoms

8 125I atoms

4 125I atoms

2 125I atoms

1 125I atom

2G-1 (of 17)

Half-lives range from

1 x 10-21 seconds for 18Na

5 x 1015 years for 142Ce

Common half-lives

5,730 years for 14C

4.5 x 109 years for 238U

2G-2 (of 17)

THE DECAY EQUATION

n = n0e-kt

n0 = at time 0, number of atoms of a radioisotope (or g or disintegrations/time)

k = decay constant of a radioisotope (disintegrations atom-1 time-1)

t = time of decay

n = at time t, number of atoms of a radioisotope (or g or disintegrations/time)

2G-3 (of 17)

Half-life (t1/2) is the time needed so that ½ of n0 disintegrates

n0 = n0e-kt1/2

___

2

ln (1/2) = -kt1/2

ln 2 = t1/2

_____

k

1 = e-kt1/2

___

2

ln 2 = kt1/2

ln 2 = k _____

t1/2

or

n = n0e-kt

2G-4 (of 17)

THE DECAY EQUATION

n = n0e-kt n = n0e

- (ln2/t1/2)t

2G-5 (of 17)

Calculate the mass of 110Ag remaining after 2.00 minutes if you start with 1.00 g 110Ag and its half-life is 24 seconds.

= (1.00 g)e-(ln2/24 s)(120. s)

= 0.031 g

n = n0e-(ln2/t1/2)t

n and no can be anything proportional to the number of the radioactive atoms:

(1) grams, (2) moles, (3) disintegrations per time, (4) percentages, or of course (5) atoms

2G-6 (of 17)

Starting with 2.00 g of a radioisotope, after 1.00 hour only 0.63 g remain. Calculate the half-life.

n = n0e-(ln2/t1/2)t

n = e-(ln2/t1/2)t

___

n0

ln (n/n0) = -(ln2/t1/2)t

t1/2 =

-(ln2)t__________

ln (n/n0)

= 0.60 h = (ln2)t __________

ln (n0/n)

= (ln2)(1.00 h) _____________________

ln (2.00 g/0.63 g)

2G-7 (of 17)

CARBON DATING

In the atmosphere

14 N + 7

C + 14

6 n → 1

0 H 1

1

C + O2 → CO2 14

6

14 6

The carbon in all living organisms has the same percentage of 14C that the atmosphere has

15.3 dist. min-1 g-1 of carbon

When an organism dies, it stops taking in 14C, so the percentage starts dropping

2G-8 (of 17)

An axe with an elk antler sleve produces 4.8 cpm g-1 of carbon. How old is the axe?

n = n0e-(ln2/t1/2)t

n = e-(ln2/t1/2)t

___

n0

ln (n/n0) = -(ln2/t1/2)t

= 9,600 y t1/2 ln (n0/n) = t ______________

ln 2

= (5,730 y) ln (15.3 cpm g-1/4.8 cpm g-1) _______________________________________________

ln 2

2G-9 (of 17)

Much older objects can be dated with radioisotopes of longer half-lives

238U decays to 206Pb, so a material containing uranium can be dated by measuring the amount of 206Pb compared to 238U

2G-10 (of 17)

A rock weighing 4.267 g contains 1.023 g 238U and 0.112 g 206Pb. Calculate the age of the rock.

= 7.7 x 108 y t1/2 ln (n0/n) = t ______________

ln 2

= (4.5 x 109 y) ln (1.152 g/1.023 g) _______________________________________

ln 2

t1/2 = 4.5 x 109 y

n = 1.023 g

n0 = the original mass of 238U

0.112 g 206Pb

x mol 206Pb ______________

206 g 206Pb

x 1 mol 238U ______________

1 mol 206Pb

x 238 g 238U _____________

mol 238U

= 0.129 g + 1.023 g = 1.152 g

2G-11 (of 17)

STABILITY OF NUCLEI

Mass of proton + electron :

Mass of neutron :

1.007825 amu

1.008665 amu

Calculate the mass of a 23Na atom.

11 p+ + e-

12 n

= 11.086075 amu

= 12.103980 amu

11 x 1.007825 amu

12 x 1.008665 amu

= 23.190055 amu

Mass spectrometer data

Mass 23Na : 22.989773 amu

2G-12 (of 17)

23.190055 amu – 22.989773 amu = 0.200282 amu

BINDING ENERGY – The mass of an atom that has been converted into energy to hold the nucleus together

Through E = mc2 mass units can be converted into energy units

Mass loss of a 23Na atom:

1.000 amu = 1.492 x 10-10 J= 9.315 x 108 eV= 931.5 MeV

(Joule)(Electron Volt)(Million Electron Volt)

2G-13 (of 17)

0.200282 amu

x 931.5 MeV ______________

1.000 amu

= 186.6 MeV

This is the BINDING ENERGY of the 23Na nucleus

The stability of a nucleus is measured by its BINDING ENERGY PER NUCLEON

186.6 MeV________________

23 nucleons

= 8.113 MeV/nucleon

2G-14 (of 17)

Calculate the binding energy per nucleon for 56Fe if it has a mass of 55.934930 amu.

26 p+ + e-

30 n

= 26.203450 amu

= 30.259950 amu

26 x 1.007825 amu

30 x 1.008665 amu

= 56.463400 amu

56.463400 amu – 55.934930 amu = 0.528470 amu

x 1 _______________

56 nucleons

0.528470 amu

x 931.5 MeV ______________

1.000 amu

= 8.791 MeV/nucleon

2G-15 (of 17)

56Fe is the most stable atom

When large atoms break down they release energy

When small atoms combine they release energy

2G-16 (of 17)

FUSION – The combining of small nuclei to produce large nuclei

Fusion occurs in stars

4 H → 1

1 He 2

4

Very high temperatures or pressure are needed to overcome the repulsion of the positive hydrogen nuclei

Fusion releases much more energy than fission

Stars can fuse atoms to create even atomic numbered elements all the way up to 56Fe

2G-17 (of 17)

NUCLEAR REACTORS

235U is used as a fuel

U + 92

235 U92

236 n → 0

1

236U decays by spontaneous fission

CHAIN REACTION – When at least one neutron per fission produces a new 236U

Nuclear reactions release over 100 times more energy than chemical reactions

2G’-1 (of 4)

Not enough neutrons are

captured for a chain reaction

CRITICAL MASS – The minimum amount of 235U needed to support a chain reaction

Enough neutrons are captured to just maintain a chain reaction

So many neutrons are captured the chain reaction is

an explosion

2G’-2 (of 4)

Water – Acts as a MODERATOR to slow down the neutrons, as a COOLANT to keep the reactor core from overheating, and as PROTECTION because it absorbs radiation

Cd or B Control Rods – Absorb neutrons to control the rate of the chain reaction

Fuel Elements – Metal casings containing 235U

Reactor Core

2G’-3 (of 4)

San Onofre Nuclear Generating Station

Heat from the nuclear fission boils water, and steam turns a turbine, which produces electricity

Used up full elements contain radioactive daughter products, which must be disposed of safely

2G’-4 (of 4)