Atomic Structure

Law of Conservation of Mass (Lavoisier) The total mass remains constant in a chemical reaction.

Law of Definite Proportions (Proust) All samples of a compound have the same composition. All samples have the same proportion by mass of the elements present.

Law of Multiple Proportions (Dalton) When two or more different compounds of the same two elements are compared, the masses of one element that combine with a fixed mass of the second element are in small whole number ratios. http://cwx.prenhall.com/petrucci/medialib/media_portfolio/text_images/003_MULTIPLEPROP.MOV

Important Scientists Dalton Thomson Millikan

Scientists Rutherford Planck

Scientists Einstein Bohr DeBroglie Schrodinger

Mass Spectrometry

Mass Spectrometer

Mass Spectrum of Mercury

Calculating Average Atomic Mass Ex: A sample of naturally occurring carbon contains 98.892 % carbon-12 and 1.108 % carbon-13. What is the average atomic mass of carbon?

Wave Properties

c =  c = 3.00 x 108 m/s Example: Calculate the frequency, in Hertz, of an x-ray with a wavelength of 8.21 nm.

Electromagnetic Spectrum

Sodium, potassium, and lithium flame tests http://cwx.prenhall.com/petrucci/medialib/media_portfolio/text_images/039_FlameTestsMet.MOV

Planck’s Quantum Hypothesis Energy can only be absorbed or emitted as a quantum. E = h E = energy (J) h = Planck’s constant 6.63 x 10-34 Js  = frequency (s-1)

The Photoelectric Effect http://cwx.prenhall.com/petrucci/medialib/media_portfolio/text_images/040_PhotoelectEff.MOV

Photoelectric Effect Simulation

PHET simulation: http://phet.colorado.edu/en/simulation/photoelectric Einstein won the Nobel prize in Physics in 1921 for his explanation of the photoelectric effect (not for his theory of relativity)

Key Points to Photoelectric Effect Light with higher energy (higher frequency) is able to ionize atoms (remove electrons) “Before Einstein's explanation, the photoelectric effect was a real mystery. Scientists couldn't really understand why low-frequency high-intensity light (lots of red light) would not cause electrons to be emitted, while higher-frequency low-intensity light (a faint amount of violet) would. Knowing that light is made up of photons, it's easy to explain now. It's not the total amount of energy (i.e., the intensity) that's important, but the energy per photon.” http://physics.bu.edu/~duffy/py106/PhotoelectricEffect.html

Energy of Photons Energy of a photon E = h  Energy of one mole of photons E = 6.022 x 1023 h 

Examples Calculate the energy, in joules, of a photon of violet light that has a frequency of 6.15 x 1014 s-1. Calculate the energy, in joules per photon, of ultraviolet light with a wavelength of 235 nm. A laser produces red light of wavelength 632.8 nm. Calculate the energy, in kJ of one mole of photons of this light.

3. Calculate the frequency of the radiation released by the transition of an electron in the H atom from the n=5 energy level to the n=3 energy level.

Bohr’s Model

Bohr’s Model En = -2.178 x 10-18 J n2 Calculate the energy of an electron on the second energy of a H atom. Calculate the energy change that occurs, in J, when an electron falls from the n=5 to the n=3 energy level.

Photoelectron Spectroscopy (PES) ***Key point : PES provides evidence for the shell model of the atom***

Photoelectron Spectroscopy PES apparatus: iramis.cea.fr

PES Data Number of electrons Energy to remove an electron Electrons generally farther from the nucleus Electrons generally closer to the nucleus Each peak represents the electrons in a single sublevel in the atom The bigger the peak – the more electrons Number of electrons Energy to remove an electron (binding energy) (increases to the left!) 

Hydrogen vs. Helium Helium Hydrogen #e- #e-  energy  energy 1 electron in 1s 2 electrons in 1s The helium peak is twice as tall because there are twice as many electrons in the 1s sublevel

Hydrogen vs. Helium Helium Hydrogen #e- #e-  energy  energy 1 electron in 1s 2 electrons in 1s The helium peak is farther to the left (higher energy) thus more energy is needed to remove the 1s electrons in helium. They must be held more tightly because there is a higher effective nuclear charge. (Helium has 2 protons pulling on 1s but hydrogen only has 1)

Oxygen (1s22s22p4) Energy to remove an electron (binding energy) Number of electrons 2 electrons in 1s 2 electrons in 2s 4 electrons in 2p Energy to remove an electron (binding energy) (increases to the left!) 

Scandium (1s22s22p63s23p64s23d1) Energy to remove an electron Number of electrons *Notice that it takes more energy to remove an electron from 3d than from 4s. This is because as electrons are added to 3d they shield 4s thus it’s easier (takes less energy) to remove 4s electrons compared to 3d electrons. Remember when transition metals make positive ions - it’s the s electrons that are lost first! 2 in1s 2 in 3s 2 in 4s 1 in 3d 2 in 2s 6 in 2p 6 in 3p Energy to remove an electron (binding energy) (increases to the left!) 

Key Points to Reading a PES Size of peaks indicates # of electrons Go in energy level or shell order (all the 3rd energy level data is together) Sometimes 1s is omitted because it’s so high in energy that it’s off the graph

Example 1: A Energy  Number of electrons Identify the element whose PES data is shown to the right. Sodium Why is one peak much larger Than the others? This peak represents 6 electrons In the 2p sublevel the other Peaks represent only 1 or 2 electrons In which sublevel are the electrons Represented by peak A 3s A Energy 

Example 2: Oxygen Nitrogen #e- #e-  energy  energy The PES data above shows only the peak for the 1s electrons. Why is the peak for Oxygen farther to the left? It takes less energy to remove a 1s electron from nitrogen because it has a lower effective nuclear charge (less protons) than oxygen

Example 3: Number of electrons Draw the expected PES Spectrum for the element boron Energy 

What element is this? Explain.

Identify all of the peaks in the PES below. What is missing Identify all of the peaks in the PES below. What is missing? What neutral element is this data from?

Write an electron configuration and draw a PES spectra you might see for aluminum