Rutherford’s model : Solar system Taught us where the subatomic parts of the atom are located

Problem with solar system model – If electrons travel in an orbit, why does the atom not collapse on itself?

Light and Sound In 1905 Einstein derived an equation relating mass and energy. You should be familiar with this equation: E = mc2 This equation has been changed a bit since, but a relationship has now, for the first time in history, been established between matter and energy, and between physics and chemistry.

Light and Sound Because Einstein was able to prove a relationship between matter and energy, we today can understand more about matter by learning all about energy. We can see this relationship between energy and matter specifically when we look at some of the unusual properties of the wave nature of energy.

The nature of light has been debated for thousands of years. In the 1600's, Newton argued that light was a stream of particles. Huygens countered that it was a wave. Both had good arguments, but neither could prove their case. The Nature of Light: Wave or Particle? wave! particle!

Young's Double Slit Experiment In 1801, Thomas Young settled the argument (apparently) with his Double Slit Experiment. When we look at the results of Young's experiment we can see one of the unusual properties of energy that we were talking about. Let's start by reviewing light

Young showed that light is a wave. Maxwell showed that electromagnetic waves exist and travel at the speed of light. Light was shown to be an electromagnetic wave. The frequency of an electromagnetic wave is related to its wavelength. For electromagnetic waves (including light), in a vacuum: Light is an Electromagnetic Wave c = λ f

The electric and magnetic waves are perpendicular to each other, and to the direction of propagation. Electromagnetic Waves

If you look at white light through a prism... This is what you get.

I. White Light A Prism breaks white light up into individual wavelengths –Light is made up of many wavelengths –Light is a form of radiation called electromagnetic radiation

II. Electromagnetic spectrum Light is one type of wave that comes from the sun electromagnetic radiation are all the different types of energy in the form of waves

Summary Light, color, energy, and waves are all the same phenomena represented by different variables.

III. Properties of waves: Wavelength ( ) -the distance between corresponding points on adjacent waves. frequency (v)- number of waves that pass a given point in a second. Units hertz (Hz) = 1/s Speed of light (c) = light has a constant speed of 3 x 10 8 meters per second -

Rules #1: Wavelength ( ) –are inversely related to frequency (v)- As wavelength ( ) increases frequency (v) decreases-

Rule 1 formula Frequency increases wavelength decreases = c

Units for waves c = c:speed of light (3.00  10 8 m/s) :wavelength (m, nm, etc.) :frequency (Hz) 1m = 1.0 X 10 9 nm

Fo r all waves: velocity = wavelength x frequency v = λf T herefore for light: c = λf Electromagnetic Radiation All electromagnetic radiation travels at the same velocity: the speed of light (c) c = 3.00 x 108 m/s.

5 All electromagnetic waves travel through a vacuum at Athe same speed. B speeds that are proportional to their frequency. C speeds that are inversely proportional to their frequency. D none of the given answers

6 In a vacuum, the velocity of all electromagnetic waves A is zero. Bis 3.0 × 108 m/s. Cdepends on the frequency. D depends on their amplitude.

7 For a wave, the frequency times the wavelength is the wave's A velpcity B amplitude. Cintensity. D power.

8 Electromagnetic radiation travels through vacuum at a speed of A186,000 m/s B 125 m/s C 3.00 x 10 8 m/s D It depends on wavelength

9 The wavelength of light that has a frequency of 1.20 x 1013 Hz is A 25 m B 2.5 x 10-5 m C m D 2.5 m c = λf c = 3.00 x 10 8 m/s

10 What is the frequency of light whose wavelength is 600 nm? A 5.0 x 1014 Hz B 1.0 x 1015 Hz C 1.5 x 1015 Hz D 2.0 x 1015 Hz c = λf c = 3.00 x 10 8 m/s 1m = 1.0 X 10 9 nm

Rules #2: Energy (e) –is directly related to frequency (v)- As energy (e) increases frequency (v) increases-

LOWENERGYLOWENERGY HIGHENERGYHIGHENERGY ROYG.BIV redorangeyellowgreenblueindigoviolet Rule 2- Example Energy directly relates to frequency

E:energy (J, joules) h:Planck’s constant (6.63  J·s) :frequency (Hz) E = h The energy of a photon is directly proportional to its frequency.

Radiation

Relationships Direct –Energy and Frequency Inverse –Wavelength and Frequency –Wavelength and Energy

Variables Variablerepresentsunit C H E

Constants Variablequantity C H 1m = 1.0 X 10 9 nm

4 Steps in solving wave formulas 1. write givens 2. determine equation 3. Convert units (if needed) 4. Solve

Wavelength Formula EX: Find the frequency of a photon with a wavelength of 434 nm.

Formula GIVEN: = ? = 434 nm = 4.34  m c = 3.00  10 8 m/s WORK : = c EX: Find the frequency of a photon with a wavelength of 434 nm. 434 nm X 10 9 nm

Formula GIVEN: = ? = = 4.34  m c = 3.00  10 8 m/s WORK : = 3.00  10 8 m/s = 6.91  Hz EX: Find the frequency of a photon with a wavelength of 434 nm. Step 3 do the work m

Energy formula GIVEN: E = ? = 4.57  Hz h = 6.63  J·s WORK : E = h E = ( 6.63  J·s ) x ( 4.57  Hz ) E = 3.03  J EX: Find the energy of a red photon with a frequency of 4.57  Hz.

Combined formula GIVEN: = ? E = 6.2  joules WORK : = c EX: Find the of photon with an energy of 6.2  Joules C= 3.00  10 8 m/s

Combined formula GIVEN: = ? E = 6.2  joules h =  J·s C= 3.00  10 8 m/s WORK : step 1 – formula for = c EX: Find the of photon with a energy of 6.2  Joules

Combined formula GIVEN: = ? E = 6.2  Hz h = 6.63  J·s C= 3.00  10 8 m/s WORK : step 2 – formula for = E h EX: Find the of photon with a frequency of 6.2  Hz.

Combined formula GIVEN: = ? E = 6.2  Hz h =  J·s C= 3.00  10 8 m/s WORK : step 3 – Solve for and plug in to first formula EX: Find the of photon with a frequency of 6.2  Hz.

Homework CHAPTER ALL 2.23 A,B,C 2.27 A ALL CHAPTER ALL ALL 6.35 A 6.37 AB 6.51 ALL 6.55 ALL 6.71 AB 6.90 ALL