CATHODE RAYS

To produce a supply of electrons ,

hot cathodes

are used. By heating a metal such as tungsten to a high temperature, electrons can be ejected. This is called

thermoionic emission.

Before they were known to be tiny particles carrying a small charge

e

, beams of

electrons

were called

cathode rays

because they came from the cathode inside the tube.

Properties of Cathode Rays: - They travel from the cathode in a straight line, - They cause certain surfaces to fluoresce.

- They posses kinetic energy.

- They can be deflected by an electric field.

- They can be deflected by a magnetic field.

- They produce X-rays on striking matter.

DISCOVERY OF THE ELECTRON Thomson in 1897 performed an experiment in which he measured the speed and the charge-to-mass ratio (e/m) of cathode rays. The value (

e/m )

was always the same. Cathode rays soon came to be recognized as beam of particles, which we now call electrons.

Magnetic force = evB Electric force = eE

14.1

Combine the equations that describe the deflection of an electron by a magnetic field and by an electric field to give the charge/mass ratio in Thomson’s experiment.

Deflection by B

F magnetic = F centripetal

Bev



mv

2

r e m



v Br

Deflection by E

F magnetic = F electric

Bev = eE e m



v



E B e m



E

2

B r

  19  31

= 1.75x10

11 C/kg

MILLIKAN’S OIL DROP EXPERIMENT Robert Millikan was awarded the Nobel Prize in 1911 for determining the charge on the electron . The experiment was simple enough. Oil was sprayed out of an atomizer. The small oil drops would pick up some unknown number of electrons due to friction between the oil and the nozzle. These oil drops were sprayed between two oppositely charged parallel plates. The electric field E between the plates was adjusted until the drop was suspended in midair. The downward pull of gravity then just balanced by the upward force due to the was electric field.

qE = mg

The mass of the droplet was determined by measuring its terminal velocity in the absence of the electric field. The experimental results showed that any charge seems to be an integral multiple of 1.6 x 10 -19 C means that electric charge is quantized.

ROBERT MILLIKAN

14.2

An electron emitted from a hot cathode is accelerated by a voltage of 1000 V.

a.

Calculate the KE and the speed acquired by the electron.

V = 1000 V q = 1.6x10

-19 C m = 9.11x10

-31 kg

KE = W = qV

KE = (1.6x10

= 1.6x10

-16

KE

 1 2

mv

2

-19 J ) 1000

v

 2

KE



m

 16

)

 31

= 1.8x10

7 m/s

b.

The electron enters a uniform magnetic field of 1x10 -3 radius of the path.

T. Find the B = 1x10 -3 T

Bev



mv

2

r r



mv Be



9 11 10

 31

( .

x

7

)

 3  19

)

= 0.10 m

c.

If the electric field plates are 2x10 -2 between them?

m apart, what is the voltage d = 2x10 -2 m

F M = F E eBv = eE E = Bv

=1x10 -3 (1.8x10

7 ) = 1.8x10

4 V/m

V = Ed

= 1.8x10

4 = 360 V (2x10 -2 )

QUANTUM THEORY Planck postulated that electromagnetic energy is absorbed or emitted in discrete packets or

quanta

. The energy of such photons is proportional to the frequency of the radiation:

h h E = h f

is Planck’s constant

= 6.63 xl0 -34 J/Hz

The energy is usually given in electronvolts: 1

eV

= 1.6x10

-19 J

Max Planck (1858-1947) 1918 Nobel Prize

PHOTOELECTRIC EFFECT In

photoelectric emission

, electrons are ejected from metal surfaces when electromagnetic radiation (

photons

) of high enough frequency falls on them.

Example: zinc exposed to UV (ultraviolet) rays and X rays; sodium exposed to X-rays UV rays and all colors except orange and red.

Einstein used Planck’s equation to explain the photoelectric effect . When light strikes a metallic surface its energy is transferred to a single electron, the electron might be expected to leave the metal with the same amount of energy. However, at least an amount of energy is needed to remove the electron from the metal. This is called the

work function

of the surface and is given by:

f o W o = h f o

is called the

threshold frequency.

Photoelectric Emission: 1.

Electrons are emitted only when the frequency of the light is above some intense the light.

threshold value (

f o

), no matter how 2. The maximum kinetic energy of the emitted electrons depends on the frequency of the light.

3. The photoelectrons are emitted almost at once when the light strikes.

4. The minimum energy required to release an electron from a certain metal is called the work function of the metal

W o

.

5.

The number of photoelectrons emitted is directly proportional to the intensity of the light. The number of photons emitted per second =

P hf

where P is the power of the source.

The stopping potential

V stop

is the voltage needed to stop the emission of electrons:

V stop



K

max

q

Blue light will eject electrons but red light will not.

The

kinetic energy maximum

that the ejected electrons have is given by:

K max Effect = h f - h f o Einstein’s Photoelectric

A graph of the maximum kinetic energy of the emitted electrons vs. the frequency of the light can be used to find the following:

KE MAX W o f f o

slope of the line : Planck's constant x- intercept : threshold frequency y-intercept : work function

14.3

A surface has a threshold wavelength of 0.65 μm. Calculate: a.

The threshold frequency

λ o

= 0.65x10

c = 3x10 8 -6 m/s m

f o



c



o



x

8  6

= 4.6x10

14 Hz b.

The work function in electronvolts

W o = h f o

= 6.63x10

-34 (4.6x10

14 ) = 3.06x10

-19 J = 3.06x10

-19 /1.6x10

-19 = 1.9 eV

c.

The maximum speed of the electrons emitted by violet light of wavelength 0.4 μm λ = 0.4x10

-6 m

E



hf



h c

   34

(

x

8

) 0 4 10

 6

)

= 5x10 -19 J

K max = hf - W o =

= 5x10 -19 - 3.06x10

-19 = 2x10 -19 J

v



2

K m



K

 2 1

mv

2  19 )  31

= 6.5x10

5 m/s

14.4

Light of wavelength 650 nm is required to cause electrons to be ejected from the surface of a metal. What is the K of the ejected electrons if the surface is bombarded with light of 450 nm?

K



h c

 

h c



o

λ o

= 650x10 λ = 450x10 -9 -9 m m

 

hc

   1   1

o

    34  8  

1

 9

(450 10 )



1

 9  

= 1.36x10

-19 J

PHOTON INTERACTIONS Photons of light have energy and momentum but no mass . Thus, to determine the momentum of a photon, begin by considering the momentum of a particle with mass:

p = mv

For the mass of the photon, substitute its energy equivalent from

E = mc 2

,

m



E c

2

p



mv



E c

2

c



E c



E



hf

therefore:

p



E c



f hf

 

h



The momentum of a photon, then, is proportional to its frequency and inversely proportional to its wavelength. Since photons have momentum, their momentum is conserved in collisions.

The Compton Effect uses this concept to explain the scattering that occurs when a photon collides with an electron at rest. As expected, the electron's increase in kinetic energy is matched by a decrease in kinetic energy of the photon. A decrease in KE of the photon would mean a decrease in frequency and an increase in its wavelength after the collision.

14.5

A photon of wavelength 0.47 nm strikes an electron at rest. After the collision, the photon has a wavelength of 0.50 nm. Calculate: a.

the frequency of the photon before the collision,

λ o λ f

= 0.47x10

-9 = 0.50x10

-9 m m

f



c



o

  8   9

= 6.38x10

17 Hz b.

the frequency of the photon after the collision

f



c



f

  8   9

= 6.0x10

17 Hz

c.

the speed of the electron after the collision.

v e



K electron (photon)

1

m v e e

2 2

= K o(photon) - K f



hf o



hf f

2 

o



f f

 

m

2 (6.63 10

 34

= 7.6x10

6 m/s

  17  31

WAVE NATURE OF MATTER Just as light exhibits properties of both particles and waves, particles such as electrons, protons, and neutrons also exhibit wave properties.

Thus the

wave-particle theory

extends to matter as well as light.

In 1923, Louis de Broglie suggested that the

wavelength

of a particle of mass

m

traveling at speed

v

is given by:

 

h mv

Units: meters (m)

Louis de Broglie (1892-1987)

In 1927, two Americans, Davisson and Germer, produced diffraction patterns by scattering electron beams from the surface of a metal crystal. The calculated wavelength of the electron waves agreed with de Broglie’s prediction. It was later shown that protons, and neutrons as well as other particles exhibit wave properties as well as particle properties.

14.6

What is the de Broglie wavelength of an electron that has a kinetic energy of 100 eV?

K = 100 eV = (100 eV)(1.6x10

-19 J/eV) = 1.6x10

-17 J m = 9.11x10

-31 kg

K

 2 1

mv

2  

h mv



v



2

K m

  17 ) 9 11 10  31

= 5.93x10

6 m/s

6 63 10

 34  31

( .

x

6

)

= 1.23x10

-10 m

PAIR PRODUCTION A photon can actually create matter, such as the production of an electron and a positron . A positron has the same mass as an electron, but the opposite charge . In this process the photon disappears when creating the pair. This is an example of rest mass being created from pure energy according to Einstein's equation: E = mc 2 .

Notice that a photon cannot create an electron alone since electric charge would not then be conserved. The inverse of pair production also occurs: if an electron collides with a positron, the two annihilate each other and their energy, including their mass, appears as electromagnetic energy of photons.

Electron-positron annihilation is the basis for the type of medical imaging known as PET.

27.7 a.

What is the minimum energy of a photon that can produce an electron-positron pair?

m = 9.11x10

-31 kg

E = mc 2

= 2 (9.11x10

-31 )(3x10 8 ) 2 = 1.64x10

-13 J = 1.02 MeV b.

What is the photon's wavelength?

E



hc

  

hc E

  34  13  8

= 1.2x10

-12 m

STRUCTURE AND PROPERTIES OF THE NUCLEUS A nucleus can be considered to be made up of two types of particles: protons and neutrons . These particles also have wave properties.

Proton: The nucleus of the simplest atom, hydrogen. Neutron: A particle found in the nucleus that is electrically neutral and that has a mass almost identical to the proton. Nucleons: The term that refers to the two constituent particles of a nucleus (protons and neutrons).

Isotopes: Nuclei that contain the same number of protons but different numbers of neutrons.

Thomson’s Model for the Atom

J.J. Thomson’s plum pudding model consists of a sphere of positive charge with electrons embedded inside.

This model would explain that most of the mass was positive charge and that the atom was electrically neutral.

Positive pudding Electron

Thompson’s plum pudding

The size of the atom (



10 -10 m) prevented direct confirmation.

Rutherford’s Experiment

The Thomson model was abandoned in 1911 when Rutherford bombarded a thin metal foil with a stream of positively charged alpha particles.

Rutherford Scattering Exp.

Alpha source

Most particles pass right through the foil, but a few are scattered in a backward direction.

Gold foil Screen

The Nucleus of an Atom

If electrons were distributed uniformly, particles would pass straight through an atom. Rutherford proposed an atom that is open space with positive charge concentrated in a very dense nucleus.

Alpha scattering + -

Gold foil Screen

Electrons must orbit at a distance in order not to be attracted into the nucleus of atom.

The Bohr Atom

Energy levels,

n

+ The Bohr atom Bohr’s postulate : When an electron changes from one orbit to another, it gains or loses energy equal to the difference in energy between initial and final levels.

Bohr’s Atom and Radiation

Emission

When an electron drops to a lower level, radiation is emitted; when radiation is absorbed, the electron moves to a higher level.

Absorption

Energy: hf = E

f

- E

i

By combining the idea of energy levels with classical theory, Bohr was able to predict the radius of the hydrogen atom.

ENERGY LEVEL DIAGRAMS Radiation is absorbed or released when an atom changes from one stationary state to another. The energy of the emitted or absorbed photon is said to be quantized and is equal to the difference in the energy between these two states:

E f



E o

The transition on an energy level diagram is shown below.

Absorption of a photon to promote the electron to a higher energy level .

n F n o

n = 5 n = 4 n = 3 n = 2 n = 1

The transition on an energy level diagram is shown below.

Emission of a photon when the electron falls from a higher to a lower energy level.

n o n F

n =



n = 6 n = 5 n = 4 n = 3 n = 2 n = 1

Atomic Spectra

In an emission spectrum, light is separated into characteristic wavelengths.

Emission Spectrum

Gas  1

Absorption Spectrum

 2

In an absorption spectrum, a gas absorbs certain wavelengths, which identify the element.

14.8

A hypothetical atom has 4 energy levels: 0 eV, 1 eV, 3 eV and 6 eV.

a.

Draw an energy level diagram for this atom indicating the quantum numbers and the energies associated with them.

b.

Use arrows to show all of the possible transitions between energy levels.

The 12 possible transitions are indicated by the arrows showing either absorption or emission of a photon.

c.

For which transition is the associated photon energy highest? The highest possible photon energy is 6 eV, corresponding to a transition between the n = 1 and n = 4 levels.

d.

For which transition is the associated photon energy lowest?

The lowest photon energy is 1 eV, corresponding to a transition between the n = 1 and n = 2 levels.

e.

For which transition is the associated photon wavelength longest?

The longest wavelength corresponds to the lowest energy since

 

hc E

The transition between n = 1 and n = 2 corresponds to the longest wavelength.

f.

For which transition is the associated photon wavelength shortest?

The transition between n = 1 and n = 4 (highest energy) corresponds to the shortest wavelength.

g.

A photon incident on the hypothetical atom causes the electron to make a transition from the n = 2 orbital to the n = 4 orbital. What is the wavelength of the photon?

E

4 

E

2  

5 eV

 

hc



E

5 eV



248 nm

h.

How many wavelengths of emitted radiation are possible when the electron returns to the n = 2 state?

i.

An electron moving with a speed of 1.25x10

6 m/s collides with the hypothetical atom. Is the energy provided by the electron enough to excite the atom to the n = 3 state? Is it enough for the atom to reach the n = 4 state?

KE e

 1 2

mv

2  1 2  31  6

KE e

 7.12 10  19 J  19 J/eV  4.45 eV 2   19 J

The electron has enough energy to excite the atom to the n = 3 state but not to the n = 4 since that will require 6 eV.

Atomic Number: The number of protons in a nucleus (designated by the letter

Z ).

Atomic Mass Number: The total number of protons and neutrons (designated by the letter

A ).

Z = Atomic number N = Number of neutrons A = Atomic mass

A = Z + N

BINDING ENERGY

The

total mass

of a stable nucleus is always

less

than the

sum

of the masses of its constituent particles.

Mass Defect:

The difference in the mass of a nucleus and the sum of the masses of its constituent particles.

Nuclear Binding Energy:

The amount of energy that must be put into a nucleus to break it into its constituent particles. It is the energy equivalent of the mass defect found by using:

E = m c 2

The most convenient unit of energy to use is the

electronvolt (eV)

E = 931MeV

E = mc

2

One

atomic mass unit

(amu) = 1.6605 x 10 -27 kg ------------------------------------------------------------------- E = mc

2

E = (1.6605 x 10 = 1.49 x 10 -10 -27 J kg) (3 x 10 8 m/s) 2 1.49 x 10 -10 J / 1.6 x 10 -19 J /eV = 9.31 x 10 8 = 931 x 10 6 eV eV = 931 MeV

one amu = 931 MeV

The Binding Energy of Helium

The larger the

binding energy

of a nucleus, the more stable it is.

14.9

The mass of a proton is: 1.007825 u. The mass of a neutron is 1.008665 u. The mass of the nucleus of the radioactive hydrogen isotope tritium H-3 is 3.016049 u.

a.

What is the nuclear mass defect of this isotope?

1 3

H

1 proton + 2 neutrons 1(1.007825) + 2(1.008665) = 3.025155 u Mass defect = 3.025155 - 3.016049 = 0.009106 u b.

What is the binding energy of tritium? E = 0.009106 (931) = 8.47 MeV

TRANSMUTATION Transmutation is the changing of one element into another via radioactive decay.

Transmuting Uranium into Neptunium

RADIOACTIVE DECAY Alpha

particles are 8,000 times as heavy as beta particles.

Paper or clothing will block alpha particles.

Beta

particles require a few sheets of aluminum foil.

Gamma

radiation is extremely dangerous, a thousand times more potent than x-rays.

Alpha Particle The particle is emitted in

alpha decay

. It is essentially a helium nucleus . It contains two protons and two neutrons. It has a charge of q = -2 and a mass of A = 4. When a nuclei decays by emitting an alpha particle, the number of protons is reduced by two and its mass is reduced by four . Alpha particles are emitted by very large nuclei where the strong nuclear force is insufficient to hold the nuclei together. It is abbreviated: α or

2 4

He

Alpha Particle Emission

Distribution of Energy in Alpha Emission

Beta Particle The particle is emitted in

beta decay

. Beta particles are negative electrons or

positrons

emitted by the nucleus. It is not an orbital electron, but one created in the nucleus by the decay of a neutron into a proton and an electron . Beta particles are emitted by nuclei that have too many neutrons relative to the number of protons.

0 1

n

1 1

p

 1 0

e

and

1 1

p

  0 1

n

1 0

e

Another particle called the

neutrino

beta decay. It is abbreviated



The weak nuclear force is also emitted in is involved in the production of a beta particle in the nucleus.

Beta Particle (Electron) Emission

Beta Particle (Positron) Emission by Oxygen-15 A

positron

has the same mass as the electron, but opposite charge.

Gamma Radiation Radiation emitted in gamma decay.

Gamma radiation is composed of high-energy photons.

It is emitted by excited state nuclei. Gamma radiation has no charge and no mass. It is abbreviated :



14.10

Write the nuclear equation for the transmutation of radium 226 into radon-222 by the emission of an alpha particle.

226 88

Ra

 222

Rn

86  2 4

He

14.11 Write the nuclear equation for the transmutation of lead-209 into bismuth-209 by the emission of a beta particle.

209 82

Pb

 209

Bi

83   0 1

e

NUCLEAR ENERGY In 1934, Enrico Fermi and Emilio Segré bombarded uranium with neutrons, producing new radioactive isotopes.

In 1939, German scientists Otto Hahn and Fritz Strassmann found that barium was produced by bombarding uranium with neutrons. Lisa Meitner and Otto Frisch proposed that the neutrons caused the uranium to divide into two smaller nuclei, accompanied by a tremendous release of energy.

Fission: A division daughter nuclei.

of a nucleus into two or more smaller

Fusion: Two or more nuclei nucleus. The sun combine to form a larger produces its energy by nuclear fusion.

Fusion

is the opposite of

fission

. Deuterium must be moving extremely fast to fuse.

A Nuclear Reactor

The Reactor Vessel The water in the reactor vessel has three purposes: - The water, being composed of relatively light molecules, acts as a moderator.

- Water also acts to remove heat from fuel rods which otherwise would melt.

- The heated water, converted to steam, is then converted into electrical energy.

Chain Reaction: Neutrons produced by the fission of one nucleus induce the fission of other nuclei.

A chain reaction occurs if more than one neutron goes on to cause another fission.