# Lecture 49/7/05 Homework concerns Atomic Structure.

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Lecture 49/7/05 Homework concerns Atomic Structure

Cathode Ray Tubes Glass tube with most of the air removed and 2 electrodes Cathode ray goes between electrodes with applied voltage Run in straight lines Cause gases to glow Can heat metal Can be deflected by a magnetic field Attracted toward positively charged plates Gives off light when they strike a fluorescent screen

JJ Thomson (1897) Determined the charge to mass ratio for electrons 5.60 x 10 -9 g/Coulombs Used Cathode Ray Tube with both magnetic and electrical field

Robert Millikan (1911-1913) Measured the charge on an electron Oil drop experiment Found each to have a charge as a multiple of 1.60 x 10 -19 C Current value: -1.602176 x 10 -19 C Used to find mass of an electron 9.109382 x 10 -28 g

Fig. 2-2, p.42

Electrons (cont.) Later showed that cathode rays (electrons) were the same as β particles

Canal Rays Eugene Goldstein (1886) First evidence of fundamental positive particle Different gases gave different charge to mass ratios Ernest Rutherford called them ‘protons’

Neutrons James Chadwick (student of Rutherford) Observed radiation released when particles from radioactive Polonium struck Beryllium 1932 No charge Slightly heavier than proton

Plum-pudding model JJ Thomson Atom composed of a uniform sphere of positively charged matter with electrons embedded in the sphere

Rutherford’s Gold foil experiment

New model Most of the mass and all of the positive charge in the center

ATOMIC COMPOSITION Protons positive electrical charge mass = 1.672623 x 10 -24 g relative mass = 1.007 atomic mass units (amu) Electrons negative electrical charge relative mass = 0.0005 amu Neutrons no electrical charge mass = 1.009 amu

Atomic Number (z) Number of protons in nucleus Unique for every element

Mass Number, A C atom with 6 protons and 6 neutrons is the mass standard = 12 atomic mass units Mass Number (A)= # protons + # neutrons A = 5 p + 5 n = 10 amu A = 5 p + 5 n = 10 amu

ISOTOPES Atoms of the same element (same Z) but different mass number (A).

Isotopes Because of the existence of isotopes, the mass of a collection of atoms has an average value. Average mass = ATOMIC WEIGHT Boron is 20% 10 B and 80% 11 B. That is, 11 B is 80 percent abundant on earth. For boron atomic weight = 0.20 (10 amu) + 0.80 (11 amu) = 10.8 amu

Isotopes & Atomic Weight 6 Li = 7.5% abundant and 7 Li = 92.5% Atomic weight of Li = ______________ 28 Si = 92.23%, 29 Si = 4.67%, 30 Si = 3.10% Atomic weight of Si = ______________

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