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Modern Physics
Dual Nature of Light
 Light exhibits wave phenomena as a light wave
is propagated by interchange of energy between
varying electric and magnetic fields (Maxwell)
 Light acts like particles composed of kinetic
energy and momentum when light interacts with
matter
 Both wave and particle
1. Wave Nature
 Light exhibits wave characteristics like:
Diffraction
Interference
Doppler Effect
Polarization
 Particles are not able to do any of those
characteristics
2. Particle Nature
 The Photoelectric Effect
o Light having a frequency above some
minimum value and incident on metals
causes electrons to be emitted from the
metal
o Albert Einstein explained the phenomenon
using quantum theory developed by Max
Planck
Red Light
electrons
-
Silver Metal
Intensity (brightness) of Incident Light
 Causes more electrons to be emitted but does not
affect their energy of emission
Frequency of Incident Light
 The higher the frequency, the greater the energy
of the emitted electron
 Blue Light vs. Red Light
Electric Eyes
Alarm Systems
Camera Light Meters
Elevator Doors
Assembly Line Counters Solar Cell
Quantum Theory (Max Planck)
 Matter absorbs and emits electromagnetic energy
in discrete amounts or packets
 Quantum – discrete amount or packet of energy
 Photon – basic unit of quanta, electronvolt (eV)
joules (j)
 Ephoton = hf h = Planck’s constant
The energy of a photon is directly
proportional the frequency of
the electromagnetic radiation
 A photon of visible light = 4eV – 6eV
E = hf = (6.6 x 10^-34j·s)(10^15Hz)
E = 6.63 x 10^-19j or 4eV
 Photon - massless particle of light but it carries
both energy and momentum
- it’s momentum suggests it has
mass-like qualities
Ephoton = hf = hc
λ
E is directly proportional to its frequency and
inversely proportional to its wavelength
The energy of a photon is 2.11eV
Determine the energy of the photon in joules
convert 2.11eV to joules
2.11eV 1.6 x 10^-19J = 3.38 x 10^-19J
1
1eV
Determine the frequency of the photon
E = hf
3.3 x 10^-19J = 6.63 x 10^-34J·s f
3.3 x 10^-19J
= f
6.63 x 10^-34J·s
5.0 x 10^14Hz
= f
What is the color associated with the photon?
Yellow!
10
9
8
7
6
V 5
e 4
E 3
K 2
1
0
-1
-2
-3
-4
-5
-6
-7
gold
2
3
copper
4
5
6
7
8
Frequency (x 10^15Hz)
Threshold Frequency
minimum frequency needed to emit e’s
gold – 2.0 x 10^15Hz
copper – 3.1 x 10^15Hz
Work Function
minimum KE needed to emit e’s
gold – 3eV
copper – 6eV
Slope of the line is Planck’s Constant
Photon 10eV
3eV e’s
7eV Metal
Metal work function 7eV
Photon- Particle Collisions
 Both energy and momentum are conserved
X- ray photon collides with an electron
-e
Photon
λ
ejected from atom
with KE
-e
Photon
lower f and longer λ given off by atom
Photon loses energy and momentum
Electron gains energy and momentum
Early Models of the Atom
Rutherford’s Model
 Scattering Experiments with Alpha particles on
gold leafs helped discover
o A net positive charge on the nucleus
o The atom is mostly space
 Trajectories of Alpha Particles
+
+
nucleus
+
HYPERBOLA PATH!
 Developed the Solar System Model of the atom
Bohr Model of the Hydrogen Atom
 An electron occupies specific orbit due a
quantum of energy
 Electrons in lower orbits (close to nucleus) have
less energy than electrons in higher orbits
 Electrons remain in orbit without losing energy
even though they are being accelerated towards
the nucleus by electrostatic Coulomb forces
 Electron can jump to higher orbit by absorbing a
quantum of energy in the form of a photon
Energy States of an Atom
 Stationary State, n = 1, ground state, lowest
energy level
 Excited State, electron in any other energy level
above ground state
 Ionization Potential, energy needed to remove
an electron from an atom
o 13.60 eV to remove an electron from H2
 See Energy Level Diagram (reference tables)
o Negative values, atom loses energy,
electron moves closer to the nucleus
o Positive values, atom absorbs energy,
electron moves farther from the nucleus
o –13.60eV, hydrogen is in the ground state,
electron is in the n=1 energy level, atom
lost energy
The Cloud Model
 Electrons are not confined to specific orbits
 Electrons are spread out in space in a form called
an electron cloud
 Each cloud is the densest region of finding an
electron there (mathematical concept)
Atomic Spectra
 Electrons returning to lower energy states give
off a series of frequencies of electromagnetic
radiation
Bright Line Spectrum
 Emission spectrum appears as a series of bright
lines against a dark background
 Electrons falling to lower energy levels
Absorption Spectrum
 Spectrum appears as a series of dark lines in a
white line spectrum
 Electrons jumping to higher energy levels
absorbing photons with energies corresponding
to differences in energy levels equal to emitting
frequencies
The Nucleus
 Contain nucleons
neutrons and protons
 Distance between protons = 10^-15m due to
repulsive Coulomb forces
+
10^-15M
+
 Gravitational force of attraction is too weak to
counterbalance the coulomb force
 Strong Nuclear Force – attractive force
between protons and neutrons that holds nucleus
together (stability)
o 100 times stronger than Coulomb force
o Strongest force
o Effective only within 10^-15m
o Diminishes at distances greater than
10^-15m,
Universal Mass Unit
 Atomic Mass Unit (u) 1/12 of the mass of the
atom carbon – 12 (6p, 6n, 6e)
 Proton = 1.0073u
1.67 x 10^-27kg
 Neutron = 1.0087u 1.67 x 10^-27kg
 Electron = 0.0005u 9.11 x 10^-31kg
 1u = 1.66 x 10^-27kg
Mass – Energy Relationship
 Mass and Energy are different forms of the
same thing
 E = mc²
E = energy
m = mass
c = 3.00 x 10^8m/s
 1kg mass converted to energy = 9.00 x 10^16J
= 5.62 x 10^35eV
Find the energy equivalent of 1u in megaelectronvolt
m = 1.66 x 10^-27kg
c = 3.0 x 10^8m/s
E= ?
1 eV = 1.6 x 10^-19J
10^6eV = 1MeV
E = mc²
E = (1.66 x 10^-27Kg)(3.00 x 10^8m/s)²
E = 1.49 x 10^-10J
1.49 x 10^-10J
1
1eV
1.6 x 10^-19J
E = 9.31 x 10^8eV
9.31 x 10^8eV
1
E = 931 MeV
1MeV
10^6eV
Nuclear Mass and Energy
 Mass-energy is conserved at all levels from
cosmic to subatomic
Molecular Example:
1kg Carbon combines with O2 to form CO2
 3.3 x 10^7J of energy is released from
4 x 10^-10kg of mass
 CO2 is slightly less in mass than Carbon and
Oxygen separately
Atomic Level Example
2 protons
2(1.0073u)
2 neutrons
2(1.0087u)
Total Mass
Helium Nucleus
2protons, 2 neutrons Total Mass
Nucleus Mass is 0.0304u less
2.0146u
2.0174u
4.0320u
4.0016u
When nucleons come together to form a nucleus,
energy is released and an equivalent amount of
matter is lost
To break up the nucleus and separate the nucleons,
work must be done against the strong nuclear force of
attraction
Find the energy equivalent of this mass difference in
electronvolts
1u = 931MeV
E = (0.0304u)(931MeV) = 28.3 MeV
u
Studying Atomic Nuclei
 Using Particle Accelerators
o Use electric and magnetic fields to increase
the KE of charged particles like electrons
and protons
o Projects them near the speed of light into
matter disrupting nuclei and releasing new
particles
o Ejected particles can give useful info about
the structure and forces within the nucleus