NUCLEAR CHEMISTRY. Atomic Structure Recall: Atoms – consist of a positively charged nucleus, which…

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Difference between the mass of an atom and the mass of its individual particles amu amu Mass Defect Nuclear Forces

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NUCLEAR CHEMISTRY

Atomic StructureRecall:Atoms – consist of a positively charged nucleus, which has protons and neutrons.

Isotope Symbol # protons # neutrons

Atomic Mass

Mo-101H-2C-14U-238Bi-210He-4

H21

C146

U23892

Bi21083

He42

Mo10142 42

1692832

59181461272

101214238210 4

Isotope – atoms of the same chemical element that have a different number of neutrons. Each isotope of a given element is designated by the total number of its protons plus its neutrons.

• Difference between the mass of an atom and the mass of its individual particles.

4.00260 amu 4.03298 amu

Mass DefectNuclear Forces

Nuclear Binding Energy• Energy released when a nucleus is

formed from nucleons.• High binding energy = stable nucleus.

E = mc2E: energy (J)m:mass defect (kg)c: speed of light

(3.00×108 m/s)

Nuclear Binding Energy

Unstable nuclides are radioactive and undergo radioactive decay.

The seven most widely recognized magic numbers as of 2007 are 2, 8, 20, 28, 50, 82, 126

He42

Types of Radiation• Alpha particle ()

– helium nucleus paper2+

• Beta particle (-)– electron

e0-1 1-

lead

• Positron (+)– positron e0

1 1+

• Gamma ()– high-energy photon 0 concrete

ChargeShielding

00

Penetrating Ability of Radiation

Nuclear Decay

He Th U 42

23490

23892

e Xe I 0-1

13154

13153

e Ar K 01

3818

3819

Pd e Ag 10646

0-1

10647

•Transmutation-One element becomes another.

More than 83 protons means that the nuclei is unstable (radioactive)

Alpha Emission

He Th U 42

23490

23892

parentnuclide

daughternuclide

alphaparticle

Numbers must balance!!

occurs when the nucleus has too many protons which cause excessive repulsion.

Beta Emission

e Xe I 0-1

13154

13153

electron

occurs when the neutron to proton ratio is too great.

Neutron to Proton

XAZPo210

84

Ex. Polonium-210 undergoes beta decay to produce this daughter nuclide

+

Beta Emission

XAZ = At210

85

e01-

e Ar K 01

3818

3819

positron

Positron EmissionOccurs when the neutron to proton ratio is too small.

XAZPo210

84

Ex. Polonium-210 undergoes positron emission to produce this daughter nuclide

+

Positron Emission

XAZ = Bi210

83

e01

Electron Capture

Pd e Ag 10646

0-1

10647

electron

occurs when the neutron to proton ratio in the nucleus is too small.

XAZPo210

84

Ex. Polonium-210 captures an electron to produce this daughter nuclide

+

Electron Capture

XAZ = Bi210

83

e01-

Gamma Emission

– Emission of high energy electromagnetic wave.

occurs when the nucleus is at too high an energy.

XAZPo210

84

Ex. Polonium-210 undergoes gamma decay to produce this daughter nuclide

+

Gamma Emission

XAZ =

00

Po21084

Types of RadiationTypes of Radiation

Nuclear Fission– splitting of heavier nuclei into lighter nuclei.

U23592 n1

0 Ba13756 Xe84

36 n10

energy

+ +15+ +

How much energy? E=mc2

_______= _______ x (_____ _ _____)2 c=3.0x108Energy mass speed of light

E=mc2 explains _____ _____ (total mass of nucleus is less than sum of individual

particles)mass defect

Nuclear Fusion - Energy released when two light nuclei combine or fuse

•However, a large amount of energy is required to start a fusion reaction:

o Need this energy to overcome ________ forces of protons.o Extremely high temperatures can provide start-up energy.

repulsion

More energy in fusing hydrogen that fission of uranium

Nuclear Fusion

A A Decay Decay SeriesSeries

A radioactive nucleus reaches a stable A radioactive nucleus reaches a stable state by a series of stepsstate by a series of steps

Copyright© by Houghton Mifflin Company. All rights reserved.

27

Half-life ConceptHalf-life Concept

Decay KineticsDecay KineticsDecay occurs by first order kinetics (the rate of decay is proportional to the number of nuclides present)

ktNN

0

lnN = number of nuclides remaining at time t

N0 = number of nuclides present initially

k = rate constant

t = elapsed time

Calculating Half-lifeCalculating Half-life

kkt 693.0)2ln(

2/1

t1/2 = Half-life (units dependent on rate constant, k)

Sample Half-LivesSample Half-Lives

Half-lifen

if mm )( 21

mf: final massmi: initial massn: # of half-lives

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