Chapter 7 Part I Dr. Stone Stan State 1
2
Electromagnetic Radiation Perpendicular oscillating fields: Electric: PET scan: gamma rays X-rays Visible light Infrared (heat) Microwaves Magnetic MRI = magnetic resonance imaging, Radiowaves 3
c = λν λ is wavelength (in meters). ν is frequency (in hertz, or s -1 ). c is the speed of light in vacuum. c is a constant; in a vacuum c = 2.998 10 8 m/s. 4
What is the frequency of the yelloworange light (λ = 589 nm) produced by sodium-vapor streetlights? A. 1.96 x 10 14 sec -1 B. 5.09 x 10 5 sec -1 C. 1.96 x 10-6 sec -1 D. 5.09 x 10 14 sec -1 E. None of these c = λν ν = c λ The anwer is on the next slide. 5
ν = 2.998 x 10 8 m/sec 589 x 10-9 m = 5.08 x 10 14 sec -1 Frequency units can be Hertz (Hz) or sec -1 = 1/sec 6
Radiant energy is quantized. Having values restricted to wholenumber multiples of a specific base value Quantum the smallest discrete quantity of a particular form of energy Photon a quantum of electromagnetic radiation Energy of photon: E = hν h = 6.626 x 10 34 J s (Planck s constant) 7
What is the energy of a photon of red light that has a wavelength of 685 nm? A. 3.44 x 10 18 J B. 2.90 x 10-19 J C. 2.90 x x 10-28 J D. 1.36 x 10-22 J E. None of these 8
E = hν E = hc λ 1 m λ = 10 nm -7 = 685 nm 6.85 10 m 9-34 8 6.626 10 J s 2.998 10 m s -19 E = = 2.90 10 J -7 6.85 10 m 9
Answers to More practice (a) ultraviolet light ( λ= 100. nm), E = 1.98 x 10-18 J (b) visible ( λ= 500. nm), E= 3.97 x 10-19 J (c) infrared ( λ= 6.00 µm), E = 3.31 x 10-20 J 10
Answers 1. Microwave 1.00 x 10 5 MHz, Mhz = 1 x 10 6 Hz a. wavelength = 2.998 x 10-3 m b. Energy per photon = 6.626 x 10-23 J c. Energy per mole of photons = 3.98 x 10 1 J/mole (multiply answer in b by 6.02 x 10 23 photons/mole) 2. UV light in the stratosphere Frequency = 1.50 x 10 15 Hz Energy = 9.94 x 10-19 J 3. Yellow diamond have nitrogen compounds that absorb purple light at a frequency of 7.23 x 10 14 Hz. a. Wavelength = 4.15 x 10-7 m, 415 nm, 4150 Å b. Energy = 4.79 x 10-19 J 11
Photoelectric Effect Photoelectric effect: Phenomenon of light striking a metal surface and producing an electric current (flow of electrons) If radiation is below threshold energy, no electrons are released. 12
Photoelectric Effect (cont. 1) Explained by quantum theory: Photons of sufficient energy (hν) dislodge e from metal surface. Work function (Φ) amount of energy needed to dislodge an electron from the surface of a metal: Φ = hν 0 Where Φ = work function; ν 0 = threshold frequency Kinetic energy of ejected electrons: KE electron = hν Φ What is the Kinetic energy of an electron on a copper wire that is hit with a 450 nm photon? The threshold energy (work function) = 7.53 x 10-19 J The answer is on the next slide. 13
The photo electric effect requires photons to be packets of energy. The packets can then strike a material surface and eject electrons. E = 6.626 x 10-34 J/sec x 2.998 x 10 8 m/sec 250 x 10-9 m E = 7.94 x 10-19 J KE electron = hν Φ KE = Photon E workfunction KE = 7.94 x 10-19 J - 7.53 x 10-19 J = 4.16 x 10-20 J 14
Wave Particle Duality Light has properties of both a wave and a particle. Wave-like behavior of radiant energy: Wavelength Frequency Particle-like behavior of radiant energy: Photoelectric effect Quantized packets of energy 15
The Hydrogen Spectrum Johannes Rydberg (1854 1919) Revised Balmer s equation by changing wavelength to wavenumber (1/λ): 1 ( -2 1 ) 1 1 = 1.0097 10 nm 2 2 λ n1 n2 This constant will be on the exam. n 1 = starting electron energy level n 2 = ending electron energy level 16
1 ( -2 1 ) 1 1 = 1.0097 10 nm 2 2 λ n1 n2 E = hν E = hc λ n 2 Color λ (nm) E (J) 3 red 713 nm 2.79 x 10-19 J 4 green 528 nm 3.76 x 10-19 J 5 blue 471 nm 4.22 x 10-19 J 6 violet 446 nm 4.45 x 10-19 J 17
Get pixel positions for every line in your spectra. These values are on the x axis of the ImageJ plot. Make a plot of the pixel position vs the known wavelengths of He. The known wavelength will be on the x axis. The pixel position values will be on the y axis. 18
Use the cursor to get the pixel position. 114 267 19
Make a data table and then graph your data: X values Y values Use the equation of this line to calculate the wavelengths for each pixel position in the hydrogen spectrum 20
Practice: Hydrogen pixel position: 801 Equation of the line: y = 2.6864x -933.42 801 = y Solve for x 801 = 2.6864x 933.42 Try this yourself, before you go tto the solution on the next slide. 21
Solve for x: 801 = 2.6864x 933.42 Add 933.42 to both sides: 1,734.42 = 2.6864 x Divide both sides by 2.6864: 1734.42/2.6864 = x 645 = x 22
x = 645 nm, Compare this to the closest line in the Balmer Series The closest wavelength to 645 nm is 656 nm, the transition from n = 3 to n = 2 % error = 100%x(645-656)/656= 1.7 % 23
Use the Rydberg Equation for energy Calculate the energy of the transistion from n = 3 (initial) to n = 2 (final): Try this yourself, before you go tto the solution on the next slide. 24
The energy change for an electron to move from n= 3 to n = 2: ΔE = -2.179 x 10-18 J ( 1/4-1/9) ΔE = -2.179 x 10-18 J x 1.3888 x 10-1 ) ΔE = - 3.02 x 10-19 J When an electron moves from n = 3 to n = 2 is that endergonic or exergonic? 25
- ΔE = exergonic Energy is released to the surroundings In this case the energy is in the form of a photon of visible light 26