Unit 4- Atomic Theories and Nuclear

ChemistryChapters

3 & 22

J.J. Thomson

Kishan AlluriChristina Costeas

Sandy JiangPatrick Morgan

Ben Wong

Cathode Ray Experiment http://www.youtube.com/watch?v=XU8nMK

kzbT8

Background Born in Manchester, England in December

1856 Studied physics and mathematics Won a nobel prize in 1906

Experiment Observations(on page 71)1. An object placed between the cathode

and the opposite end of the tube cast a shadow on the glass.

2. A paddle wheel placed on rails between the electrodes rolled along the rails from the cathode toward the anode.

3. Cathode rays were deflected by a magnetic field in the same manner as a wire carrying electric current, which was known to have a negative charge.

4. The rays were deflected away from a negatively charged object.

Discovery of Electrons-1897 Experiments supported hypothesis that

the particles that compose cathode rays are negatively charged

Measured the ratio of cathode-ray particles to their mass—found it was always the same

Concluded that all cathode rays are composed of identical negatively charged particles called electrons

Experiments revealed the electron has a very large charge for its tiny mass

RobertMillikan

By: Katrina Leung,Jesse Stathis, Jenna Taormina,

Danielle Zhang, Matt Odea

Millikan’s Atomic Theory Electrons are really small Mass of electron 1/2000 the mass of a

hydrogen atom Electron has a mass of 9.109 * 10^-31kg or

1/1837 the mass of a hydrogen atom Electrons carry a negative charge

Video on Millikan’s Oil Experiment

http://www.youtube.com/watch?v=XMfYHag7Liw

New Inferences aboutAtomic Structure

1. Because atoms are electrically neutral, they must contain a positive charge to balance the negative electrons.

2. Because electrons have so much less mass than atoms, atoms must contain other particles that account for most of their mass.

Lord Rutherford of Nelson

As flawlessly explained by: Adam, Jane, Justin, Jeremy,

rob

Rutherfordian History

Born 1871 in New Zealand

In early work, discovered radioactive half-life

Had an element named after him-rutherfordium

Became known as the father of nuclear physics

Discovery of the Atomic Nucleus Assistants Geiger and Marsden

bombarded a thin piece of gold foil with a narrow beam of alpha particles.

Some of the particles were redirected by the gold foil back towards their source.

Rutherford thus concluded that the force must be caused by a very densely packed bundle of matter with a positive charge, which he called the nucleus.

The Experiment

Rutherford had discovered that the volume of the nucleus was very small compared to the total volume of the atom, suggesting that there was a lot of empty space.

James ChadwickBy Monica, Santosh, Jennie, Antonella, and

Drew

James Chadwick’s Discovery of the Neutron In 1932, Chadwick discovered a previously

unknown particle in the atomic nucleus.3 This particle became known as the neutron because of its lack of electric charge. Chadwick's discovery was crucial for the fission of uranium 235.

A. Mass Number

mass # = protons + neutrons

¨ always a whole number

¨ NOT on the Periodic Table!

© Addison-Wesley Publishing Company, Inc.

B. Isotopes

Atoms of the same element with different mass numbers.

C126

Mass #

Atomic #

¨ Nuclear symbol:

¨ Hyphen notation: carbon-12

B. Isotopes

© Addison-Wesley Publishing Company, Inc.

B. Isotopes

Chlorine-37◦ atomic #: 17

◦ mass #: 37

◦ # of protons:17

◦ # of electrons:17

◦ # of neutrons:20 Cl3717

Chlorine-35◦ atomic #: 17

◦ mass #: 35

◦ # of protons:17

◦ # of electrons:17

◦ # of neutrons:18

B. Isotopes

Cl3517

C. Johannesson

C. Relative Atomic Mass

12C atom = 1.992 × 10-23 g

¨ 1 p = 1.007276 amu

1 n = 1.008665 amu

1 e- = 0.0005486 amu

© Addison-Wesley Publishing Company, Inc.

¨ atomic mass unit (amu)

¨ 1 amu = 1/12 the mass of a 12C atom

D. Average Atomic Mass

weighted average of all isotopes on the Periodic Table round to 2 decimal places

100

(%)(mass(mass)(%) )

Avg.AtomicMass

Avg.AtomicMass

D. Average Atomic Mass

EX: Calculate the avg. atomic mass of oxygen if its abundance in nature is 99.76% 16O, 0.04% 17O, and 0.20% 18O.

100

(18)(0.20)(17)(0.04))(16)(99.76 16.00amu

Avg.AtomicMass

D. Average Atomic Mass

EX: Find chlorine’s average atomic mass if approximately 8 of every 10 atoms are chlorine-35 and 2 are chlorine-37.

100

(37)(2)(35)(8)35.40 amu

New Terms for Nuclear Chemistry Atom = nuclide Nucleus = nucleon

A. Mass Defect Difference between the mass of an

atom and the mass of its individual particles.

4.00260 amu 4.03298 amu

Mass of subatomic particles: Protons - 1.007276 u Electrons – 0.0005486 u Neutrons – 1.008665 u

Ex. Helium-4 has an atomic mass of 4.002602Calculate the mass defect:2 Protons = (2x1.007276)= 2.014552u2 Electrons = (2x0.0005486) = 0.001097u2 Neutrons = (2x1.008665)= 2.017330u Total = 4.032979 uMass defect = 4.032979 - 4.002602=

0.030377

What causes the loss of mass? It is nuclear binding: (Energy released when

a nucleus is formed from nucleons)

According to E=mc2:◦ Mass can be converted into energy and energy to

mass The mass defect is caused by the

conversion of mass to energy upon formation of the nucleus.

Calculating the Binding Energy: E=mc2

◦ E-energy (J, Joules)◦ m- mass (kg)◦ c- speed of light (3.00x108 m/s)

◦ J= kg x m2

s2

Steps:1. Calculate the mass defect in (u).2. Convert to kg using 1u=1.6605X10-27kg3. Plug in to energy equation

Practice:1. Calculate the binding energy of Sulfur-32. The measured atomic mass is 31.972070 u.

(4.36 x 10-11 J)

2. Calculate the nuclear binding energy per mole of Oxygen-16. The measured atomic mass of oxygen is 15.994915u. (1.23 x 1013 J)

3. Calculate the binding energy per nucleon of a Manganese-55 atom. It’s measured atomic mass is 54.938047 u.

(1.41 x 10-12J)

B. Nuclear Binding Energy

Energy released when a nucleus is formed from nucleons.

High binding energy = stable nucleus.

E = mc2

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

(3.00×108 m/s)

Calculate the binding energies of the following two nuclei, and indicate which nucleus releases more energy when formed.

A. Potassium–35, Atomic mass 34.988011

B. Sodium – 23, Atomic Mass 22.989767

Do now:

A. Potassium–35, Atomic mass 34.9880114.47 X 10-11 J

B. Sodium – 23, Atomic Mass 22.9897672.99 x 10 -11J

Potassium releases more energy

Do now:

B. Nuclear Binding Energy

Unstable nuclides are radioactive and undergo radioactive decay.

Binding energy per nucleon- Binding energy per nucleon, is the binding energy of the

nucleus divided by its number of nucleons.

- The higher the binding energy per nucleon, the more tightly the nucleons are held together.

- Elements with intermediate atomic masses have the greatest binding energies per nucleon and are therefore the most stable.

The Band of Stability

What observations can you make?

Atoms having low atomic numbers, are the most stable. ◦ N:P = 1:1

As atomic number increases the N:P increases (1.5:1)

Band of Stability

Why does this happen?

- Explained by the relationship btwn nuclear force and electrostatic forces btwn protons.

Protons in the nucleus repel all other protons through electrostatic repulsion.

As #p increase, the repulsive electrostatic force between protons increase faster than the nuclear force

More neutrons are required to increase the nuclear force to stabilize the nucleus.

Band of Stability

Beyond Bismuth (#83), the repulsive forces of protons are so great that no stable nuclide exists.

Stable nuclei tend to have even number of nucleons

Band of Stability

A. Types of Radiation Alpha particle ()

◦ helium nucleus ◦ 2 protons and 2 neutrons bound together and

emitted from the nucleus◦ Restricted nearly to heavy atomic nuclei◦ Mostly because protons numbers need to be

reduced to stabilize the nuclei

A. Types of Radiation Beta (-)

◦ Is an electron emitted from the nucleus during some kinds of radioactive decay.

◦ Usually for the nuclides above the band of stability, to decrease the number of neutrons.

◦ Electron is emitted from the nucleus as a beta particle.

A. Types of Radiation Positron particle (+)

◦ Particle has the same mass as an electron but has a positive charge and emitted from the nucleus

◦ Proton is converted into a neutron

He42

A. Types of Radiation Alpha particle ()

◦ helium nucleus

2+

Beta particle (-) electron e0

-1 1-

Positron (+) positron e0

11+

Gamma () high-energy photon 0

B. Nuclear Decay

Alpha Emission

He Th U 42

23490

23892

parentnuclide

daughternuclide

alphaparticle

Numbers must balance!!

B. Nuclear Decay

Beta Emission

e Xe I 0-1

13154

13153

electronPositron Emission

e Ar K 01

3818

3819

positron

B. Nuclear Decay

Electron Capture

Pd e Ag 10646

0-1

10647

electronGamma Emission Usually follows other types of

decay.

Transmutation One element becomes another.

Complete the following:

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

Complete the following:

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

Complete the following:

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

Complete the following:

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

Complete the following:

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

B. Nuclear Decay

Why nuclides decay…◦ need stable ratio of neutrons to protons

DECAY SERIES TRANSPARENCY

Characteristics of stable nuclei: When number of protons is plotted against

number of neutrons, a belt-like graph is obtained.

Atoms of low atomic numbers have stable nuclei with a ratio of 1:1, p:n

As atomic number increases, the p:n increases

What happens when p:n increase? Explained by nuclear and electrostatic

forces between protons.◦ Protons in the nucleus repel each other bc of

electrostatic repulsion◦ However Nuclear forces allow protons to attract to

other protons in close proximity As #p increase the repulsive electrostatic

force between protons increase faster than the nuclear force. Therefore creating an unstable nucleus.◦ More neutrons are required to stabilize this force.

Radioactive Decay The spontaneous disintegration of a nucleus

into a slightly lighter nucleus, accompanied by the emission of particles, electromagnetic radiation or both.

Leads to more stable nucleons.

C. Half-life

Half-life (t½)◦ Time required for half the atoms of a

radioactive nuclide to decay.◦ Shorter half-life = less stable.

C. Half-life

nif mm )( 2

1

mf: final massmi: initial massn: # of half-livesn: t/T t = time elapsed

T = Length of half life

C. Half-life Fluorine-21 has a half-life of 5.0 seconds. If you

start with 25 g of fluorine-21, how many grams would remain after 60.0 s?

GIVEN:

t½ = 5.0 s

mi = 25 g

mf = ?

total time = 60.0 s

n = 60.0s ÷ 5.0s =12

WORK:

mf = mi (½)n

mf = (25 g)(0.5)12

mf = 0.0061 g

Add in new equation

More Half-life problems: (wkst) The half-life of carbon-14 is 5715 years.

How long will it be until only half of the carbon-14 in a sample remains? (5715y)

The EPA is concerned about the levels of radon gas in homes. The half-life of radon-222 isotope is 3.8days. If a sample of gas taken from a basement contained 4.38ug of radon-222, how much will remain after 15.2 days? (0.274ug)

Uranium-238 decays through alpha decay with a half life of 4.46x109years. How long would it take for 7/8 of a sample of uranium-238 to decay? (three half-lives or 1.34 x 1010 years)

Decay Series Parent Nuclide:

The heaviest nuclide of each decay series

Daughter Nuclide:

The nuclide produced by decay of the parent nuclide

A. F ission

splitting a nucleus into two or more smaller nuclei

1 g of 235U = 3 tons of coal

U23592

A. F ission

chain reaction - self-propagating reaction

critical mass - mass required to sustain a chain reaction

B. Fusion combining of two nuclei to form one nucleus of larger

mass thermonuclear reaction – requires temp of

40,000,000 K to sustain 1 g of fusion fuel =

20 tons of coal occurs naturally in

stars

HH 31

21

C. Fission vs. Fusion

235U is limited danger of

meltdown toxic waste thermal pollution

fuel is abundant no danger of

meltdown no toxic waste not yet sustainable

FISSION FUSION

A. Nuclear Power Fission Reactors

Cooling Tower

A. Nuclear Power Fission Reactors

A. Nuclear Power Fusion Reactors (not yet sustainable)

A. Nuclear Power

Fusion Reactors (not yet sustainable)

Tokamak Fusion Test Reactor

Princeton University

National Spherical Torus Experiment

B. Synthetic Elements

Transuranium Elements◦ elements with atomic #s above 92◦ synthetically produced in nuclear reactors and

accelerators◦ most decay very rapidly

Pu He U 24294

42

23892

C. Radioactive Dating half-life measurements of radioactive

elements are used to determine the age of an object

decay rate indicates amount of radioactive material

EX: 14C - up to 40,000 years238U and 40K - over 300,000 years

D. Nuclear Medicine Radioisotope Tracers

◦ absorbed by specific organs and used to diagnose diseases

Radiation Treatment◦ larger doses are used

to kill cancerous cells in targeted organs

◦ internal or external radiation source

Radiation treatment using

-rays from cobalt-60.

E. Nuclear Weapons

Atomic Bomb◦ chemical explosion is used to form a critical mass of

235U or 239Pu◦ fission develops into an uncontrolled chain reaction

Hydrogen Bomb◦ chemical explosion fission fusion◦ fusion increases the fission rate◦ more powerful than the atomic bomb

F. Others Food Irradiation

◦ radiation is used to kill bacteria

Radioactive Tracers◦ explore chemical pathways◦ trace water flow◦ study plant growth, photosynthesis

Consumer Products◦ ionizing smoke detectors - 241Am