Catalyst 11/2/10 No New Objective Please have out chart to be checked 1. Which of the following...

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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.

Transcript of Catalyst 11/2/10 No New Objective Please have out chart to be checked 1. Which of the following...

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Slide 2 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 13 ? 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. Slide 3 Agenda Subatomic Particle Review (15-30) Ion and Isotope Notes/Practice (30) Homework: 1, 2, 5 & 7: Pick 10 additional elements to add to chart 4 & 5 atoms, ions, and isotopes worksheet Slide 4 Name 1. 2. 3.. 20. Symbolatomic mass (A) atomic number (Z) protons (=Z) neutrons (=A-Z) electrons (=Z if neutral ) Homework: Slide 5 Symbol Atomic Mass Atomic Number Slide 6 H Hydrogen 1 1 Protons: 1 Neutrons: 0 Electrons: 1 Slide 7 Na Sodium 23 11 Protons: 11 Neutrons: 12 Electrons: 11 Slide 8 Rhenium Re 186 75 Protons: 75 Neutrons: 111 Electrons: 75 Slide 9 Try these:Name/ Atomic # Mass P N E Symbol Niobium In Arsenic Lead Au Zr V Iron Ni W Slide 10 Atoms of the same element that differ in charge. (They have the same # of p +, but different # of e - ) ION Slide 11 Ex: How many electrons does Sodium have? The atomic number equals the number of protons and the number of electrons. 11 electrons Na 11 22.99 Slide 12 Positive Ions (cations) Negative Ions (anions) Na + (lost 1 e - ) Ca 2+ (lost 2 e - ) Al 3+ (lost 3 e - ) Pb 4+ (lost 4 e - ) H + (lost 1 e - ) Cl - (gain 1 e - ) O 2- (gain 2 e - ) P 3- (gain 3 e - ) S 2- (gain 2 e - ) OH - (gain 1 e - ) Slide 13 If an atom GAINS electrons, its overall charge becomes more negative. If it LOSES electrons, its charge becomes more positive Slide 14 Learning Check Counting State the number of protons, neutrons, and electrons in each of these ions. 39 K + 16 O -241 Ca +2 198 20 #p + ___________________ #n o ___________________ #e - ___________________ Slide 15 Atoms of the same element that differ in mass. (They have the same # of p +, but different # of N) ISOTOPE Slide 16 Isotopes are CHEMICALLY the SAME as atoms, but DIFFER PHYSICALLY because they have different masses. Slide 17 A few examples of isotopes Slide 18 Learning Check Counting Naturally occurring carbon consists of three isotopes, 12 C, 13 C, and 14 C. State the number of protons, neutrons, and electrons in each of these carbon atoms. 12 C 13 C 14 C 6 6 6 #p + _______ _______ _______ #n o _______ _______ _______ #e - _______ _______ _______ Slide 19 Answers 12 C 13 C 14 C 6 6 6 #p + 6 6 6 #n o 6 7 8 #e - 6 6 6 Slide 20 One Last Learning Check Write 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 - ___________ Slide 21 So, why do the elements on the PT have masses with decimals??? Slide 22 The periodic table has the average atomic mass of all of the isotopes of a substance Slide 23 AVERAGE ATOMIC MASS Because of the existence of isotopes, the mass of a collection of atoms has an average value.Because of the existence of isotopes, the mass of a collection of atoms has an average value. Boron is 20% 10 B and 80% 11 B. That is, 11 B is 80 percent abundant on earth.Boron is 20% 10 B and 80% 11 B. That is, 11 B is 80 percent abundant on earth. For boron atomic weightFor boron atomic weight = 0.20 (10 amu) + 0.80 (11 amu) = 10.8 amu = 0.20 (10 amu) + 0.80 (11 amu) = 10.8 amu 10 B 11 B Slide 24 Average atomic mass is calculated by: 1. Mass each isotope and multiply by amount 2. Add calculations for each isotope 3. Divide sum by the total amount