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J.J. Thomson Discoverer of the Electron. Background Information Cathode Rays Form when high voltage is applied across electrodes in a partially evacuated.

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Presentation on theme: "J.J. Thomson Discoverer of the Electron. Background Information Cathode Rays Form when high voltage is applied across electrodes in a partially evacuated."— Presentation transcript:

1 J.J. Thomson Discoverer of the Electron

2 Background Information Cathode Rays Form when high voltage is applied across electrodes in a partially evacuated tube. Originate at the cathode (negative electrode) and move to the anode (positive electrode) Carry energy and can do work Travel in straight lines in the absence of an external field

3 Source of Electrical Potential Metal Plate Gas-filled glass tube Metal plate Stream of negative particles (electrons) A Cathode Ray Tube Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 58

4 Cathode Ray Experiment 1897 Experimentation Using a cathode ray tube, Thomson was able to deflect cathode rays with an electrical field. The rays bent towards the positive pole, indicating that they are negatively charged.

5 The Effect of an Obstruction on Cathode Rays Dorin, Demmin, Gabel, Chemistry The Study of Matter, 3 rd Edition, 1990, page 117 High voltage cathode source of high voltage yellow-green fluorescence shadow

6 Dorin, Demmin, Gabel, Chemistry The Study of Matter, 3 rd Edition, 1990, page 117 The Effect of an Electric Field on Cathode Rays High voltage cathode source of high voltage positive plate negative plate anode _ +

7 Cathode Ray Experiment Deflection region Drift region Displacement + - Anodes / collimators Cathode Volts

8 Thomson’s Calculations Cathode Ray Experiment Thomson used magnetic and electric fields to measure and calculate the ratio of the cathode ray’s mass to its charge. Magnetic deflection charge of ray particle magnetic field length of deflection region length of drift region mass of ray particle velocity of ray particle x x x x = Electric deflection charge of ray particle electric field length of deflection region length of drift region mass of ray particle velocity of ray particle x x x x = 2 magnetic deflection electric deflection magnetic field electric field x velocity =

9 Conclusions He compared the value with the mass/ charge ratio for the lightest charged particle. By comparison, Thomson estimated that the cathode ray particle weighed 1/1000 as much as hydrogen, the lightest atom. He concluded that atoms do contain subatomic particles - atoms are divisible into smaller particles. This conclusion contradicted Dalton’s postulate and was not widely accepted by fellow physicists and chemists of his day. Since any electrode material produces an identical ray, cathode ray particles are present in all types of matter - a universal negatively charged subatomic particle later named the electron

10 So what does J.J. Thomson have to do with mass spec? Just as J.J. Thomson used a magnetic field to affect charged particles, so does a mass spectrometer. The machine sorts ions according to their mass to charge ratio, something Thomson was able to calculate for the electron using the results of his cathode ray experiments. High voltage cathode source of high voltage positive plate negative plate anode _ +

11 What is mass spectrometry? Mass spectrometry is a technique used to separate a substance into ions based on their mass. Molecules are bombarded by high energy particles that cause them to lose one electron and carry a +1 charge. These ions undergo further fragmentation producing smaller positive ions. The spectrum produced plots intensity (abundance of ions) against the ions’ mass-to-charge ratio. Substances can be identified by their characteristic fragment ions represented on a mass spectrum

12 Mass spectrometers that break up molecules into fragments that can be characterized by electrical methods. [image link]image link Detector plate Least massive ions Ion-accelerating electric field Magnetic field Heating device to vaporize sample Positive ions Sample Electron beam accelerated Ion beam Most massive ions Slits

13 Mass Spectrophotometer electron beam magnetic field gas stream of ions of different masses lightest ions heaviest ions Dorin, Demmin, Gabel, Chemistry The Study of Matter 3rd Edition, page 138

14 Inlet - ensures that the sample enters the machine with minimal loss Source - sample components are ionized (the method by which this is done depends on the specific mass spectrometer being used.) Analyzer - accelerates ion and separates them Detector - records the charge induced when an ion passes by or hits a surface. Signal Processor - produces a mass spectrum, a record of the m/z's at which ions are present. *A vacuum must be used to maintain a low pressure. A low pressure reduces the collisions among the ions. Components of a Mass Spectrometer Inlet Signal processor SourceAnalyzer Detector Vacuum

15 The general operation of a mass spectrometer is: 1. create gas-phase ions 2. separate the ions based on their mass-to-charge ratio 3. measure the quantity of ions of each mass-to-charge ratio Electron Beam Ion Accelerating Array Molecular Source Magnetic Field Bends Path of Charged Particles Collector Exit Slit HoHo

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17 Mass Spectrometry - + Photographic plate Mass spectrum of mercury vapor Hill, Petrucci, General Chemistry An Integrated Approach  1999, page 320 Stream of positive ions

18 Mass Spectrum for Mercury Mass number Relative number of atoms Mass spectrum of mercury vapor The percent natural abundances for mercury isotopes are: Hg % Hg % Hg % Hg % Hg % Hg % Hg % Hg % Hg % Hg % Hg % Hg % Hg % Hg % (The photographic record has been converted to a scale of relative number of atoms)

19 The percent natural abundances for mercury isotopes are: Hg % Hg % Hg % Hg % Hg % Hg % Hg % Hg % Hg % Hg % Hg % Hg % Hg % Hg % ( )(196) + (0.1002)(198) + (0.1684)(199) + (0.2313)(200) + (0.1322)(201) + (0.2980)(202) + (0.0685)(204) = x = x x = amu Hg (% "A")(mass "A") + (% "B")(mass "B") + (% "C")(mass "C") + (% "D")(mass "D") + (% "E")(mass "E") + (% F)(mass F) + (% G)(mass G) = AAM ABCDEFGABCDEFG

20 Assume you have only two atoms of chlorine. One atom has a mass of 35 amu (Cl-35) The other atom has a mass of 36 amu (Cl-36) What is the average mass of these two isotopes? 35.5 amu Looking at the average atomic mass printed on the periodic table...approximately what percentage is Cl-35 and Cl-36? 55% Cl-35 and 45% Cl-36 is a good approximation Cl

21 Using our estimated % abundance data 55% Cl-35 and 45% Cl-36 calculate an average atomic mass for chlorine. Cl Average Atomic Mass = (% abundance of isotope "A")(mass "A") + (% "B")(mass "B") +... AAM = (% abundance of isotope Cl-35)(mass Cl-35) + (% abundance of Cl-36)(mass Cl-36) AAM = (0.55)(35 amu) + (0.45)(36 amu) AAM = (19.25 amu) + (16.2 amu) AAM = amu

22  An electric or magnetic field can deflect charged particles.  The particles have kinetic energy as they move through a magnetic field (KE=1/2mv 2 ).  The particles’ inertia depends on their mass.  A mass analyzer can steer certain masses to the detector based on their mass-to-charge ratios (m/z). by varying the electrical or magnetic field.  Typically ions in a mass spectrometer carry a +1 charge so the m/z ratio is equivalent to the ion’s mass. What’s mass got to do with it?

23 Ion focusing

24 What does a mass spectrum look like? Intensity or ion abundance is plotted on the y-axis. The m/z ratio is plotted on the x-axis. The base beak is from the ion that is the most abundant and is assigned an intensity of 100%. The molecular ion peak, M +, is the peak due to the parent ion (the original molecule minus one electron).

25 m/z % RELATIVE INTENSITY Mass spectrum of carbon dioxide, CO 2 molecular ion is seen at m/z C+C+ O+O+ CO + CO 2 + M+M+

26 Mass spectrums reflect the abundance of naturally occurring isotopes. Hydrogen Carbon Nitrogen Oxygen Sulfur Chlorine Bromine 1 H = % 2 H = 0.015% 12 C = 98.90% 13 C = 1.10% 14 N = 99.63% 15 N = 0.37% 16 O = % 17 O = 0.038% 18 O = 0.200% 32 S = 95.02% 33 S = 0.75% 34 S = 4.21% 36 S = 0.02% 35 Cl = 75.77% 37 Cl = 24.23% 79 Br = 50.69% 81 Br = 49.31% Natural Abundance of Common Elements

27 For example….Methane For carbon 1 in approximately 90 atoms are carbon-13 The rest are carbon-12 the isotope that is 98.9% abundant. So, for approximately 90 methane molecules…1 carbon is carbon-13

28 Where’s Waldo? C-13

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31 Why is the Mass Spectrometer an Important Analytical Instrument? Mass Spectrometers have been used in: 1) Forensics 2) Organic synthesis laboratories 3) The analysis of large biomolecules: proteins and nucleic acids 4) Drug Test 5) Determination of isotopic abundance 6) Identification of impurities in pharmaceutical products 7) Diagnosis of certain diseases.

32 ass_lehr.htmlhttp://www.infochembio.ethz.ch/links/en/spectrosc_m ass_lehr.html Electron-Intro.htmlhttp://dbhs.wvusd.k12.ca.us/AtomicStructure/Disc-of- Electron-Intro.html 72/Instructor_Resources/Chapter_12/47http://wps.prenhall.com/wps/media/objects/340/ /Instructor_Resources/Chapter_12/47 References


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