Electrons. Unit H Chapter 6 - PDF Free Download

Electrons Unit H Chapter 6 1

Electrons were discovered by 1. Dalton 2. Lavoisier 3. Proust 4. Mendeleev 6. Rutherford 7. Bohr 8. Schrodinger 9. Dirac 5. Thomson 2

Electrons were discovered by 1. Dalton 2. Lavoisier 3. Proust 4. Mendeleev 6. Rutherford 7. Bohr 8. Schrodinger 9. Dirac 5. Thomson 3

What experiment, law, or tool did Thomson invoke or implement? 1. Atomic Theory 2. Law of Conservation of Mass 3. Law of Constant Composition 4. Law of Multiple Proportions 5. Periodic Law 6. Cathode Ray Tubes 7. Gold Foil Experiment 8. Atomic Emission Spectra 9. Wave Equation 4

What experiment or tool did Thomson use? 1. Atomic Theory 2. Law of Conservation of Mass 3. Law of Constant Composition 4. Law of Multiple Proportions 5. Periodic Law 6. Cathode Ray Tubes 7. Gold Foil Experiment 8. Atomic Emission Spectra 9. Wave Equation 5

6 J.J. Thomson (1856-1940) Electricity (which we now know is nothing more than moving electrons) was applied to cathode ray tubes. The beam of electrons itself can not be seen. The green beam is caused by the electrons effect on the phosphor screen or the small amount of gas in the cathode ray tube.

7 Discovery of the Electron (and Proton) The cathode ray tube caused electrons to travel from an electrode (the cathode) at one end to an electrode (the anode) at the other end. The beam of electrons can be bent by a magnetic force - repelled by the negative end of a magnet. The beam behaved the same regardless of which gas was in tube or which metal was at the cathode. Thomson named these negatively charged particles, electrons. Scientists knew that matter was neutral, they quickly discovered the proton as well. electrodes magnet

Write the entire electron configuration for 84Po Turn it into a condensed electron configuration. 8

Write the entire electron configuration for 84Po Turn it into a condensed electron configuration. 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2 4f 14 5d 10 6p 4 9

Write the entire electron configuration for 84Po Turn it into a condensed electron configuration. 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2 4f 14 5d 10 6p 4 [Xe] 6s 2 4f 14 5d 10 6p 4 10

Write the condensed electron configuration for 33As Sketch an orbital notation for this. 11

Write the condensed electron configuration for 33As Sketch an orbital notation for this. [Ar] 4s 2 3d 10 4p 3 12

Write the condensed electron configuration for 33As Sketch an orbital notation for this. [Ar] 4s 2 3d 10 4p 3 13

Write electron configuration for the single highest energy subset of orbital(s) that describes 32Ge in its ground state. 14

Write electron configuration for the single highest energy subset of orbital(s) that describes 32Ge in its ground state. 4p 2 15

Write condensed electron configuration for describes 35Br in its ground state. 16

Write condensed electron configuration for describes 35Br in its ground state. Br: [Ar] 4s 2 3d 10 4p 5 Br : [Ar] 4s 2 3d 10 4p 6 17

Write entire electron configuration for 12Mg 2+ in its ground state. 18

Write entire electron configuration for 12Mg 2+ in its ground state. Mg: 1s 2 2s 2 2p 6 3s 2 Mg 2+ : 1s 2 2s 2 2p 6 19

Write condensed electron configuration that describes 26Fe 3+ in its ground state. 20

Write condensed electron configuration for describes 26Fe 3+ in its ground state. Fe: [Ar] 4s 2 3d 6 Fe 3+ : [Ar] 3d 5 NOT [Ar] 4s 2 3d 3 21

Disscuss with your mate why the electron configuration for 26Fe 2+ in its ground state is [Ar] 4s 1 3d 5 not [Ar] 3d 6 nor [Ar] 4s 2 3d 4 22

Name the element that is described by for an atom electron configuration of the highest energy orbital of the element in its ground state: 4d 4 23

Name the element that is described by the electron configuration of the highest energy orbital of the element in its ground state: 4d 4 42Mo 24

Write a generic valence electron configuration that could describe any halogen. 25

Write a generic valence electron configuration that could describe any halogen. ns 2 np 5 in most other contexts, n means mole, but in reference to electrons, n means principal quantum number = energy level 26

Name three (common) naturally occurring particles (atoms or ions) that are isoelectronic with Cl 27

Name three (common) naturally occurring particles (atoms or ions) that are isoelectronic with Cl same number of electrons, 18 Ar, S 2, K +, Ca 2+, V 5+ 28

What is it that makes this an impossible ground state electron configuration? 1s 2 2s 3 2p 6 2d 10 3s 2 3p 3 4s 2 4d 10 4p 6 6s 1 29

What is it that makes this an impossible ground state electron configuration? 1s 2 2s 3 2p 6 2d 10 3s 2 3p 3 4s 2 4d 10 4p 6 6s 1 1s 2 2s 2 2p 6 2d 10 3s 2 3p 6 4s 2 3d 10 4p 6 5s 1 30

What are the valence electrons of selenium? 1. 4s 2 2. 4p 4 3. 4d 4 4. 4s 2 4p 4 5. 4s 2 3d 10 4p 4 6. 4s 2 4d 10 4p 4 1 H 2 He 3 Li 4 Be 5 B 6 C 7 N 8 O 9 F 10 Ne 11 Na 12 Mg 13 Al 14 Si 15 P 16 S 17 Cl 18 Ar 19 K 20 Ca 21 Sc 22 Ti 23 V 24 Cr 25 Mn 26 Fe 27 Co 28 Ni 29 Cu 30 Zn 31 Ga 32 Ge 33 As 34 Se 35 Br 36 Kr 37 Rb 38 Sr 39 Y 40 Zr 41 Nb 42 Mo 43 Tc 44 Ru 45 Rh 46 Pd 47 Ag 48 Cd 49 In 50 Sn 51 Sb 52 Te 53 I 54 Xe 55 Cs 56 Ba 71 Lu 72 Hf 73 Ta 73 W 75 Re 76 Os 77 Ir 78 Pt 79 Au 80 Hg 81 Tl 82 Pb 83 Bi 84 Po 85 At 86 Rn 31

What are the valence electrons of selenium? 1. 4s 2 2. 4p 4 3. 4d 4 4. 4s 2 4p 4 5. 4s 2 3d 10 4p 4 6. 4s 2 4d 10 4p 4 Some textbooks consider the d electrons to be valence electrons, but most do not. AP will refer only to s and p as valence electrons. 1 H 2 He 3 Li 4 Be 5 B 6 C 7 N 8 O 9 F 10 Ne 11 Na 12 Mg 13 Al 14 Si 15 P 16 S 17 Cl 18 Ar 19 K 20 Ca 21 Sc 22 Ti 23 V 24 Cr 25 Mn 26 Fe 27 Co 28 Ni 29 Cu 30 Zn 31 Ga 32 Ge 33 As 34 Se 35 Br 36 Kr 37 Rb 38 Sr 39 Y 40 Zr 41 Nb 42 Mo 43 Tc 44 Ru 45 Rh 46 Pd 47 Ag 48 Cd 49 In 50 Sn 51 Sb 52 Te 53 I 54 Xe 55 Cs 56 Ba 71 Lu 72 Hf 73 Ta 73 W 75 Re 76 Os 77 Ir 78 Pt 79 Au 80 Hg 81 Tl 82 Pb 83 Bi 84 Po 85 At 86 Rn 32

How many unpaired electrons does selenium have? 33 1 H 2 He 3 Li 4 Be 5 B 6 C 7 N 8 O 9 F 10 Ne 11 Na 12 Mg 13 Al 14 Si 15 P 16 S 17 Cl 18 Ar 19 K 20 Ca 21 Sc 22 Ti 23 V 24 Cr 25 Mn 26 Fe 27 Co 28 Ni 29 Cu 30 Zn 31 Ga 32 Ge 33 As 34 Se 35 Br 36 Kr 37 Rb 38 Sr 39 Y 40 Zr 41 Nb 42 Mo 43 Tc 44 Ru 45 Rh 46 Pd 47 Ag 48 Cd 49 In 50 Sn 51 Sb 52 Te 53 I 54 Xe 55 Cs 56 Ba 71 Lu 72 Hf 73 Ta 73 W 75 Re 76 Os 77 Ir 78 Pt 79 Au 80 Hg 81 Tl 82 Pb 83 Bi 84 Po 85 At 86 Rn

How many unpaired electrons does selenium have? 1. 2. 4s 2 4p 4 orbital notation: orbital notation showing spin: 1 H 2 He 3 Li 4 Be 5 B 6 C 7 N 8 O 9 F 10 Ne 11 Na 12 Mg 13 Al 14 Si 15 P 16 S 17 Cl 18 Ar 19 K 20 Ca 21 Sc 22 Ti 23 V 24 Cr 25 Mn 26 Fe 27 Co 28 Ni 29 Cu 30 Zn 31 Ga 32 Ge 33 As 34 Se 35 Br 36 Kr 37 Rb 38 Sr 39 Y 40 Zr 41 Nb 42 Mo 43 Tc 44 Ru 45 Rh 46 Pd 47 Ag 48 Cd 49 In 50 Sn 51 Sb 52 Te 53 I 54 Xe 55 Cs 56 Ba 71 Lu 72 Hf 73 Ta 73 W 75 Re 76 Os 77 Ir 78 Pt 79 Au 80 Hg 81 Tl 82 Pb 83 Bi 84 Po 85 At 86 Rn 34

How many unpaired electrons does chromium have? Input your answer 35 1 H 2 He 3 Li 4 Be 5 B 6 C 7 N 8 O 9 F 10 Ne 11 Na 12 Mg 13 Al 14 Si 15 P 16 S 17 Cl 18 Ar 19 K 20 Ca 21 Sc 22 Ti 23 V 24 Cr 25 Mn 26 Fe 27 Co 28 Ni 29 Cu 30 Zn 31 Ga 32 Ge 33 As 34 Se 35 Br 36 Kr 37 Rb 38 Sr 39 Y 40 Zr 41 Nb 42 Mo 43 Tc 44 Ru 45 Rh 46 Pd 47 Ag 48 Cd 49 In 50 Sn 51 Sb 52 Te 53 I 54 Xe 55 Cs 56 Ba 71 Lu 72 Hf 73 Ta 73 W 75 Re 76 Os 77 Ir 78 Pt 79 Au 80 Hg 81 Tl 82 Pb 83 Bi 84 Po 85 At 86 Rn

How many unpaired electrons does chromium have? you might think 4, 4s 2 3d 4 actually has 6, 4s 1 3d 5 because 4s and 3d electrons are close in energy, the cost of moving to the slightly higher energy d orbital is paid back in the minimization of repulsion. 1 H 3 Li 11 Na 19 K 37 Rb 55 Cs 4 Be 12 Mg 20 Ca 38 Sr 56 Ba 21 Sc 39 Y 71 Lu 22 Ti 40 Zr 72 Hf 23 V 41 Nb 73 Ta 24 Cr 42 Mo 73 W 25 Mn 43 Tc 75 Re 26 Fe 44 Ru 76 Os 27 Co 45 Rh 77 Ir 28 Ni 46 Pd 78 Pt 29 Cu 47 Ag 79 Au 30 Zn 48 Cd 80 Hg 5 B 13 Al 31 Ga 49 In 81 Tl 6 C 14 Si 32 Ge 50 Sn 82 Pb 7 N 15 P 33 As 51 Sb 83 Bi 8 O 16 S 34 Se 52 Te 84 Po 9 F 17 Cl 35 Br 53 I 85 At 2 He 10 Ne 18 Ar 36 Kr 54 Xe 86 Rn 36

Which element is represented by the orbital diagram shown below? 1. Carbon 2. Nitrogen 3. Oxygen 4. Neon 1s 2s 2p 37

Which element is represented by the orbital diagram shown below? 1. Carbon 2. Nitrogen 3. Oxygen 4. Neon 1s 2s 2p 38

Considering only the n = 1 through the n = 5 states in the hydrogen atom, which transition listed will emit the most energy? 1. n = 5 to n = 4 2. n = 5 to n = 2 3. n = 3 to n = 1 4. n = 3 to n = 2 5. n = 4 to n = 2 39

Considering only the n = 1 to n = 5 states in the hydrogen atom, which transition listed will emit the most energy? 1. n = 5 to n = 4 2. n = 5 to n = 2 3. n = 3 to n = 1 4. n = 3 to n = 2 5. n = 4 to n = 2 40

L, ml = liter(s), milliliter(s) g = gram(s) NCED PLACEMENT CHEMISTRY EQUATIONS AND CONSTANTS the following symbols have the definitions=specified unless otherwise noted. nm nanometer(s) L = liter(s), milliliter(s) millimeters of mercury atm mm Hg= = atmosphere(s) Calculations = gram(s) = nanometer(s) = atmosphere(s) in this unit? J, kj V mol = joule(s), kilojoule(s) = volt(s) = mole(s) ATOMIC STRUCTURE TURE E=h c= E = energy = frequency = wavelength Planck s constant, h = 6.626 10 34 J s Speed of light, c = 2.998 108 m s 1 Avogadro s number = 6.022 1023 mol 1 Electron charge, e = 1.602 10 19 coulomb 41

The energy of radiant light is sometimes capable of breaking molecular bonds. The longest wavelength of light with enough energy capable of breaking the bond in a single oxygen molecule is 242 nm. 1. Calculate the frequency, in s 1 ( cycles /sec) of this light. 2. Calculate the energy, in J, of a photon of the light. 3. Calculate the minimum energy, in kj mol 1, of the oxygen bond. 42

atm = The energy of radiant light is sometimes capable of breaking molecular The longest wavelength of light with ADVANCED PLACEMENT CHEMISTRY EQUATIONS ANDenough CONSTANTS Throughout the test the following symbols have the definitions specified unless otherwise noted. ATOMIC STRUCTUR energy capable of L,breaking the bond a single ml = liter(s), milliliter(s) mm Hg = in millimeters of mercury g = gram(s) J, kj = joule(s), kilojoule(s) = nanometer(s) V = volt(s) oxygen molecule isnm 242 nm. =h atm = atmosphere(s) mol = E mole(s) bonds. 1. Calculate the frequency, in ATOMIC STRUCTURE of this light. ce == h c ν= λ c= 1 s E = energy = frequency = wavelength Planck s constant, h = 6.626 10 34 J s Speed of light, c = 2.998 108 m s 1 Avogadro s number = 6.022 1023 mol 1 Electron charge, e = 1.602 10 19 coulomb EQUILIBRIUM c d Kc = [C] a[d]b, where a A + b B cc+dd [A] [B] Equilibrium Constants Kc Kp Ka Kb Kw EQUILIBRIUM (molar concentrations) 3.0 10 m / sec 10 nm 15 c [D]d 1 [C] =, where ν= = 1.24 K c 10 sec a b [A] [B] 242nm 1m c d 8p = K (PC )c (PD )d a (PA ) (PB ) 9 b Ka = [H ][A ] [HA] Kb = [OH ][HB ] [B] Kw = [H+][OH ] = 1.0 = Ka Kb 10 14 at 25 C (gas pressures) (weak acid) (weak base) (water) Kp = (PC ) (PD ) (P ) a (P ) b 44

atm = The energy of radiant light is sometimes capable of breaking molecular The longest wavelength of light with ADVANCED PLACEMENT CHEMISTRY EQUATIONS ANDenough CONSTANTS Throughout the test the following symbols have the definitions specified unless otherwise noted. ATOMIC STRUCTUR energy capable of L,breaking the bond a single ml = liter(s), milliliter(s) mm Hg = in millimeters of mercury g = gram(s) J, kj = joule(s), kilojoule(s) = nanometer(s) V = volt(s) oxygen molecule isnm 242 nm. =h atm = atmosphere(s) mol = E mole(s) bonds. c= 1 s 1. Calculate the frequency, in ATOMIC STRUCTURE of this light. ce == h 15 ν = 1.24 10 sec 2. Calculate the energy, in J, of a photon of the light. 1 E = energy = frequency = wavelength Planck s constant, h = 6.626 10 34 J s Speed of light, c = 2.998 108 m s 1 Avogadro s number = 6.022 1023 mol 1 Electron charge, e = 1.602 10 19 coulomb EQUILIBRIUM E = hν c d Kc = [C] a[d]b, where a A + b B [A] [B] Kp = 34 cc+dd Equilibrium Constants (PC )c (PD )d (PA )a (PB )b 1 Kc Kp Ka Kb Kw EQUILIBRIUM (molar concentrations) (gas pressures) d (weak acid) [C]c [D] 19 (weak base) c (water) [A]a [B]b K = 10 E = (6.63 10 K =J sec)(1.24 10 sec ) = 8.22 a 15 [H ][A ] [HA] Kb = [OH ][HB ] [B] Kw = [H+][OH ] = 1.0 = Ka Kb 10 14 at 25 C Kp = J, where (PC )c (PD )d a (P ) (P ) b 46

The energy of radiant light is sometimes capable of breaking molecular bonds. The longest wavelength of light with enough energy capable of breaking the bond in a single oxygen molecule is 242 nm. 1. Calculate the frequency, in s 1 of this light. ν = 1.24 10 15 sec 1 2. Calculate the energy, in J, of a photon of the light. E = 8.22 10 19 J 3. Calculate the minimum energy, in kj mol 1, of the oxygen bond. 47

The energy of radiant light is sometimes capable of breaking molecular bonds. The longest wavelength of light with enough energy capable of breaking the bond in a single oxygen molecule is 242 nm. 1. Calculate the frequency, in s 1 of this light. 2. Calculate the energy, in J, of a photon of the light. E = 8.22 10 19 J ATOMIC STRUCTURE E hv c " v h " = p m! m! 18 2.178 10 En joule 2 n 23 1 Avogadro s number 6.022 10 mol 19 3. Calculate the minimum energy, in kj mol 1, of the oxygen bond. 8.22 10 19 J 1kJ 1000J 6.022 1023 bonds mol = 495kJmol 1 48

Considering only the n = 1 to n = 5 states in the hydrogen atom, which transition listed will emit light of the longest wavelength? 1. n = 5 to n = 4 2. n = 5 to n = 2 3. n = 3 to n = 1 4. n = 3 to n = 2 5. n = 4 to n = 2 6. n = 1 to n = 5 7. n = 4 to n = 5 8. n = 6 to n = 7 50

Considering only the n = 1 to n = 5 states in the hydrogen atom, which transition listed will emit the longest wavelength? 1. n = 5 to n = 4 2. n = 5 to n = 2 3. n = 3 to n = 1 4. n = 3 to n = 2 Longest wavelength = lowest energy E = hν 5. n = 4 to n = 2 c = λν E = h c λ 51

Which statement about the top EMR wave compared to the bottom EMR wave is true? Select as many as apply. 1. Top is higher energy. 2. Top is longer λ (wavelength) 3. Top is higher ν (frequency) 4. Top is the same amplitude. 5. Top could be red light while the lower could be blue. 6. Top moves faster than bottom. 7. The top could be IR while the bottom could be microwaves. 55

Which statement about the top wave compared to the bottom wave is true? 1. Top is higher energy. 2. Top is longer λ (wavelength) 3. Top is higher ν (frequency) 4. Top is same amplitude. 5. Top could red light while the lower could be blue. 6. Top moves faster than bottom. 7. The top could be IR while the bottom could be microwaves. 56

That s it for now. The material on the following slides is no longer part of the AP curriculum, however it is on the SAT subject test. We will go over this in late May for those of you taking the June SAT subject. 57

Electron Configuration & Quantum Numbers Schrodinger s Wave equation has many solutions. Each solution is represented by a set of 4 quantum numbers. No two electrons can have the same exact set of 4 numbers 58

The Four Quantum Numbers Principal quantum number, n Integer values: 1, 2, 3, etc Indicates the energy level Second quantum number, l (Azimuthal number) n 1 or lower to 0 values allowed Indicates the orbital type 0 = s, 1 = p, 2 = d, 3 = f Third quantum number, ml (Magnetic number) l = +, 0,- values allowed Indicates the particular orbital within the same subgroup Orbitals with the same energy are said to be degenerate orbitals Fourth quantum number, spin, ms +½, -½values allowed 59

Degenerate orbitals means same energy orbitals, i.e. a particular subset. So the px, py, pz orbitals are degenerate orbitals 60

Principles of Electron Configuration Pauli Exclusion Principle Each electron has a unique set of quantum numbers. Aufbau Principle (Aufbau: in German means building up) Placing electrons in the lowest energy (most stable) orbital that is not already filled. Hund s Rule Electrons will spread out in a subset of orbitals. (An atom will try to maximize its number of unpaired electrons.) 61

For the principal (first) quantum number n=3, what are the allowed numerical values for the second quantum number l =? Answer on the white board What orbitals (subshells) do these l values describe? 62

For the principal (first) quantum number n=3, what are the allowed numerical values for the second quantum number l =? 0, 1, 2 s, p, d 63

Which of the following is not an allowed set of the first three quantum numbers? 1. n = 3, l = 1, ml = 1 2. n = 3, l = 3, ml = -2 3. n = 4, l = 0, ml = 0 4. n = 3, l = 1, ml = -1 5. n = 2, l = 1, ml = -1 64

Which of the following is not an allowed set of the first three quantum numbers? 1. n = 3, l = 1, ml = 1 2. n = 3, l = 3, ml = -2 3. n = 4, l = 0, ml = 0 4. n = 3, l = 1, ml = -1 5. n = 2, l = 1, ml = -1 65

How many total electrons will fit in this subset of orbitals AND what is the l quantum number of the orbitals represented below. 1. 5 electrons, l = 2 2. 5 electrons, l = 3 3. 10 electrons, l = 3 4. 10 electrons, l = 2 5. 18 electrons, l = 4 66

How many total electrons will fit in this subset of orbitals AND what is the L quantum number of the orbitals represented below. 1. 5 electrons, l = 2 2. 5 electrons, l = 3 3. 10 electrons, l = 3 4. 10 electrons, l = 2 5. 18 electrons, l = 4 67

shape of s orbitals 68

shape of p orbitals 69

shape of d orbitals 70

shape of f orbitals 71

What is the maximum number of orbitals described by the quantum numbers: n = 3, l = 2 Input your answer 72

What is the maximum number of orbitals described by the quantum numbers: n = 3, l = 2 5 orbitals these quantum numbers refer to the subset, 3d 73

In an iodine atom, what are the total number of electrons described by the quantum number: l = 1 Input your answer 74

In an iodine atom, what are the total number of electrons described by the quantum number: l = 1 2p 6, 3p 6, 4p 6, 5p 5 = 23 75

What is the maximum number of electrons that can be described by the quantum number: n = 4 Input your answer 76

What is the maximum number of electrons described by the quantum number: n = 4 32 4th energy level can contain: s = 2 electrons p = 6 electrons d = 10 electrons f = 14 electrons no g orbitals 77

The maximum number of electrons in an atom that can exist in the 4f subshell and have ml = 1 is Input your answer 78

The maximum number of electrons in an atom that can exist in the 4f subshell and have ml = 1 is Just 2e- Quantum numbers would be 4, 3, 1, +½ OR 4, 3, 1, ½ 79

Atomic emission lines are associated with particular electron transitions. A certain line of atomic hydrogen is associated with the electronic transition in the Hydrogen atom from the fourth to the second energy levels. 1. Which of the bright lines is associated with this transition? ATOMIC STRUCTURE E hv c " v h " = p m! m! 18 2.178 10 En joule 2 n Calculate the energy of the electron at both energy levels, then calculate the difference in those two energy values 80

Atomic emission lines are associated with particular electron transitions. A certain line of atomic hydrogen is associated with the electronic transition in the Hydrogen atom from the fourth to the second energy levels. 1. Which of the bright lines is associated with this transition? ATOMIC STRUCTURE E hv c " v h " = p m! m! 18 2.178 10 En joule 2 n E 4 = 2.178 10 18 J 4 2 = 1.361 10 19 J E 2 = 2.178 10 18 J 2 2 = 5.445 10 19 J E 4 E 2 = ( 1.361 10 19 J) ( 5.445 10 19 J) = 4.084 10 19 J 81

Atomic emission lines are associated with particular electron transitions. A certain line of atomic hydrogen is associated with the electronic transition in the Hydrogen atom from the fourth to the second energy levels. 1. Which of the bright lines is associated with this transition? ATOMIC STRUCTURE E hv c " v h " = p m! m! 18 2.178 10 En joule 2 n E 4 = 2.178 10 18 J 4 2 = 1.361 10 19 J E 2 = 2.178 10 18 J 2 2 = 5.445 10 19 J E = hν ν = c λ E = hc λ 6.626 10 34 J sec 4.084 10 19 J 3 108 m sec E 4 E 2 = ( 1.361 10 19 J) ( 5.445 10 19 J) = 4.084 10 19 J 109 nm 1m = 486nm Thus it must be the green line 82