Catalyst 11/2/10 No New Objective Please have out chart to be checked

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Tri al Energy per photon (eV)* Relative intensity of light Electrical current (mA)† Max kinetic energy of e- if ejected from metal (eV) 1 2 3 4 5 6 7 2.0 2.0 2.0 4.0 4.0 4.0 6.0 low medium high low medium high low 0 0 0 29 43 60 27 0.0 0.0 0.0 0.9 0.9 0.9 2.9 Catalyst 11/2/10 No New Objective Please have out chart to be checked 1. Which of the following statements best explains the results of Trials 1–3 ? A. The light was too intense to eject electrons from the metal in the photocell. B. The light was too intense to eject photons from the metal in the photocell. C.The energy per electron was too high to eject photons from the metal in the photocell. D. The energy per photon was too low to eject electrons from the metal in the photocell. 2. When 8.0 eV photons were shone on the photocell, electrons ejected from the metal in the photocell had a maximum kinetic energy of 4.9 eV. Based on this information and Table 1, the relative intensity of the light shone on the photocell: A.was high. B.was medium. C.was low. D.cannot be determined.

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Catalyst 11/2/10 No New Objective Please have out chart to be checked 1. Which of the following statements best explains the results of Trials 1–3 ? A. The light was too intense to eject electrons from the metal in the photocell. - PowerPoint PPT Presentation

Transcript of Catalyst 11/2/10 No New Objective Please have out chart to be checked

Atomic Theory

TrialEnergy per photon (eV)*Relative intensity of lightElectrical current (mA)Max kinetic energy of e- if ejectedfrom metal (eV)123456789 2.02.02.04.04.04.06.06.06.0 lowmediumhighlowmediumhighlowmediumhigh 000294360274055 0.00.00.00.90.90.92.92.92.9*eV = electron volts mA = milliampsCatalyst 11/2/10 No New Objective Please have out chart to be checked1. Which of the following statements best explains the results of Trials13?A. The light was too intense to eject electrons from the metal in the photocell.B. The light was too intense to eject photons from the metal in the photocell.C.The energy per electron was too high to eject photons from the metal in the photocell.D. The energy per photon was too low to eject electrons from the metal in the photocell.2. When 8.0 eV photons were shone on the photocell, electrons ejected from the metal in the photocell had a maximum kinetic energy of 4.9eV. Based on this information and Table1, the relative intensity of the light shone on the photocell:A.was high.B.was medium.C.was low.D.cannot be determined.

AgendaSubatomic Particle Review (15-30)Ion and Isotope Notes/Practice (30)Homework: 1, 2, 5 & 7: Pick 10 additional elements to add to chart4 & 5 atoms, ions, and isotopes worksheetName

1. 2.3....20.Symbol

atomic mass (A)

atomic number(Z)

protons(=Z)

neutrons(=A-Z)

electrons(=Z if neutral )

Homework:SymbolAtomic MassAtomic NumberHHydrogen11Protons: 1Neutrons: 0Electrons: 1NaSodium2311Protons: 11Neutrons: 12Electrons: 11RheniumRe18675Protons: 75Neutrons: 111Electrons: 75Try these:Name/ Atomic # Mass P N ESymbol

NiobiumInArsenicLeadAuZrVIronNiW

Atoms of the same element that differ in charge.(They have the same # of p+, but different # of e-)ION Ex: How many electrons does Sodium have?The atomic number equals the number of protons and the number of electrons. 11 electronsNa1122.99Positive Ions(cations)Negative Ions(anions) Na+ (lost 1 e-) Ca2+ (lost 2 e-) Al3+ (lost 3 e-) Pb4+ (lost 4 e-) H+ (lost 1 e-)

Cl- (gain 1 e-) O2- (gain 2 e-) P3- (gain 3 e-) S2- (gain 2 e-) OH- (gain 1 e-)

If an atom GAINS electrons, its overall charge becomes more negative.If it LOSES electrons, its charge becomes more positiveLearning Check CountingState the number of protons, neutrons, and electrons in each of these ions. 39 K+ 16O -241Ca +2 198 20#p+ ___________________ #no ___________________#e- ___________________

Atoms of the same element that differ in mass.(They have the same # of p+, but different # of N)ISOTOPE Isotopes are CHEMICALLY the SAME as atoms, but DIFFER PHYSICALLY because they have different masses. A few examples of isotopes

Learning Check Counting Naturally occurring carbon consists of three isotopes, 12C, 13C, and 14C. State the number of protons, neutrons, and electrons in each of these carbon atoms. 12C 13C14C 6 6 6#p+ _______ _______ _______ #no _______ _______ _______ #e- _______ _______ _______

Answers12C 13C14C 6 6 6#p+ 6 6 6 #no 6 7 8 #e- 6 6 6

One Last Learning CheckWrite the nuclear symbol form for the following atoms or ions:

A. 8 p+, 8 n, 8 e-___________

B.17p+, 20n, 17e-___________C. 47p+, 60 n, 46 e-___________ 19So, why do the elements on the PT have masses with decimals??? The periodic table has the average atomic mass of all of the isotopes of a substance

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

10B11BAverage atomic mass is calculated by: Mass each isotope and multiply by amount Add calculations for each isotope Divide sum by the total amount