Chapter 7. Part I Dr. Stone Stan State

Transcription

1 Chapter 7 Part I Dr. Stone Stan State 1

2 2

3 Electromagnetic Radiation Perpendicular oscillating fields: Electric: PET scan: gamma rays X-rays Visible light Infrared (heat) Microwaves Magnetic MRI = magnetic resonance imaging, Radiowaves 3

4 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 = m/s. 4

5 What is the frequency of the yelloworange light (λ = 589 nm) produced by sodium-vapor streetlights? A x sec -1 B x 10 5 sec -1 C x 10-6 sec -1 D x sec -1 E. None of these c = λν ν = c λ The anwer is on the next slide. 5

6 ν = x 10 8 m/sec 589 x 10-9 m = 5.08 x sec -1 Frequency units can be Hertz (Hz) or sec -1 = 1/sec 6

7 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 = x J s (Planck s constant) 7

8 What is the energy of a photon of red light that has a wavelength of 685 nm? A x J B x J C x x J D x J E. None of these 8

9 E = hν E = hc λ 1 m λ = 10 nm -7 = 685 nm m J s m s -19 E = = J m 9

10 Answers to More practice (a) ultraviolet light ( λ= 100. nm), E = 1.98 x J (b) visible ( λ= 500. nm), E= 3.97 x J (c) infrared ( λ= 6.00 µm), E = 3.31 x J 10

11 Answers 1. Microwave 1.00 x 10 5 MHz, Mhz = 1 x 10 6 Hz a. wavelength = x 10-3 m b. Energy per photon = x J c. Energy per mole of photons = 3.98 x 10 1 J/mole (multiply answer in b by 6.02 x photons/mole) 2. UV light in the stratosphere Frequency = 1.50 x Hz Energy = 9.94 x J 3. Yellow diamond have nitrogen compounds that absorb purple light at a frequency of 7.23 x Hz. a. Wavelength = 4.15 x 10-7 m, 415 nm, 4150 Å b. Energy = 4.79 x J 11

12 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

13 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 J The answer is on the next slide. 13

14 The photo electric effect requires photons to be packets of energy. The packets can then strike a material surface and eject electrons. E = x J/sec x x 10 8 m/sec 250 x 10-9 m E = 7.94 x J KE electron = hν Φ KE = Photon E workfunction KE = 7.94 x J x J = 4.16 x J 14

15 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

16 The Hydrogen Spectrum Johannes Rydberg ( ) Revised Balmer s equation by changing wavelength to wavenumber (1/λ): 1 ( -2 1 ) 1 1 = 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

17 1 ( -2 1 ) 1 1 = 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 J 4 green 528 nm 3.76 x J 5 blue 471 nm 4.22 x J 6 violet 446 nm 4.45 x J 17

18 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

19 Use the cursor to get the pixel position

20 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

21 Practice: Hydrogen pixel position: 801 Equation of the line: y = x = y Solve for x 801 = x Try this yourself, before you go tto the solution on the next slide. 21

22 Solve for x: 801 = x Add to both sides: 1, = x Divide both sides by : / = x 645 = x 22

23 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( )/656= 1.7 % 23

24 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

25 The energy change for an electron to move from n= 3 to n = 2: ΔE = x J ( 1/4-1/9) ΔE = x J x x 10-1 ) ΔE = x J When an electron moves from n = 3 to n = 2 is that endergonic or exergonic? 25

26 - ΔE = exergonic Energy is released to the surroundings In this case the energy is in the form of a photon of visible light 26