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Transcript of Electronic Structure of Atoms Chapter 6. Light is a Wave Electromagnetic wave Wavelength ( ), m...
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Electronic Structure of Atoms
Chapter 6
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Light is a Wave
• Electromagnetic wave• Wavelength (), m• Frequency (), Hz or s-1
• Travels at c (3.00 X 108 m/s) in a vacuum
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• Wavelength and frequency are inversely proportional
• c =
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•How many complete waves are shown above?•What is the wavelength of light shown above?
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Electromagnetic Spectrum
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Blu-Ray = 405 nanometers (blue light)DVD = 650 nanometers (red light)
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• Visible light = 4 X 10-7 m to 7X 10-7 m (400 to 700 nm)
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Waves: Ex 1
Calculate the wavelength of a 60 Hz EM wave
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Waves: Ex 2
Calculate the wavelength of a 93.3 MHz FM radio station
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Waves: Ex 3
Calculate the frequency of 500 nm blue light.
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Planck’s Quantum Hypothesis: Light is a Particle
• Photon –light particle• Photons emitted in “packets” (whole numbers)• Light is both a wave and a particle
E = h(for one photon)h = 6.63 X 10-34 J s (Planck’s
constant)
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Photons: Ex 1
Calculate the energy of a photon of wavelength 600 nm.
600 nm 1 X 10-9m = 6 X10-7 m1 nm
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Photons: Ex 2
Calculate the energy of a photon of wavelength 450 nm (blue light). Also, calculate the energy per mole.
Ans: (4.42 X 10-19 J , 266 kJ/mol)
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Photons: Ex 3
Calculate the energy of laser light with a frequency of 4.69 X 1014 s-1 . Also, calculate the energy per mole.
Ans: (3.11 X 10-19 J, 187 kJ/mol)
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Photons: Ex 3a
If the laser emits 1.3 X 10-2 J per pulse, how many photons (quanta) are released during this pulse?
Ans: 4.18 X 1016 photons
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Photoelectric Effect (Einstein)
• When light shines on a metal, electrons are emitted
• Can detect a current from the electrons• Used in light meter, scanners, digital cameras
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Three Key Points1.Below a certain frequency, no electrons are
emitted2.Greater intensity light produces more electrons3.Greater Frequency light produces no more
electrons, but they come off with greater speed
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Low Frequency High FrequencyNot enough energy to eject electron
Can eject electronEnergy of photon is greater than W (Work function)
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2. More intensity– More photons– More electrons ejected with same KE
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3. Greater Frequency
– No more electrons ejected
– Electrons ejected with greater speed (KE)
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Photon/Matter Interactions
1. Electron excitation (photon disappears)2. Ionization/photoelectric effect (photon
disappears)3. Scattering by nucleus or electron4. Pair production (photon disappears)
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Electron Excitation
•Photon is absorbed (disappears)
•Electron jumps to an excited state
Ionization/Photoelectric Effect
•Photon is absorbed (disappears)
•Electron is propelled out of the atom
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Scattering•Photon collides with a nucleus or electron•Photon loses some energy•Speed does not change, but the wavelength increases
Pair Production•Photon closely approaches a nucleus•Photon disappears•An electron and positron are created.
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Principle of Complimentarity
• Any experiment can only observe light’s wave or particle properties, not both
• Different “faces” that light shows
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Line Spectra• Discharge tube
– Low density gas (acts like isolated atoms)– high voltage
• Light emitted only at certain (discrete) wavelengths
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Hydrogen
Helium
Solar absorption spectrum
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• Electrons orbit in ground state (without radiating energy)
• Jumps to excited state by absorbing a photon
• Returns to ground state by emitted a photon
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h = Ee - Eg
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4. Ways to make something glow Bohr
Model
Photon Absorption Collision-Glow in the dark -Heat
-Electricity-Chemical Reaction
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Photon Absorption Collision
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Bohr’s Equation
• Works only for H and other 1-electron atoms (He+, Li2+, Be3+, etc…)
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En = -RH
n2 En = Energy of an orbital
RH = Rydberg constant (2.18 X 10-18 J)
n = Principal quantum number of orbital
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Bohr’s Equation: Ex 1
Calculate the energy of the first three orbitals of hydrogen
En = - 2.18 X 10-18 J
n2 E1 = - 2.18 X 10-18 J
12 E1 = - 2.18 X 10-18 J
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E2 = - 2.18 X 10-18 J
22 E2 = - 5.45 X 10-19 J
E3 = - 2.18 X 10-18 J
32 E3 = - 2.42 X 10-19 J
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Bohr’s Equation: Ex 2
What wavelength of light is emitted if a hydrogen electron drops from the n=2 to the n=1 orbit?
E1 = - 2.18 X 10-18 J
E2 = - 5.45 X 10-19 J
E = (- 2.18 X 10-18 J - - 5.45 X 10-19 J)E = -1.64 X 10-18 JE = h = 2.48 X 1015 s-1
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c = c/= (3 X 108m/s)(2.48 X 1015 s-1 )= 122 nm
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E = hc = E = hc/ = hcE = (6.626 X 10-34 J s)(3.0 X 108 m/s)
(1.63 X 10-18J) = 1.22 X 10-7 m = 122 nm (UV)
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Bohr’s Equation: Ex 3
Calculate the wavelength of light emitted if a hydrogen electron drops from the n=6 to the n=2 orbit?
(ANS: 410 nm (violet))
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Bohr’s Equation: Ex 4
Calculate the wavelength of light that must be absorbed to exite a hydrogen electron from the n=1 to the n=3 orbit?
(ANS: 103 nm (UV))
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Bohr Model: Other Atoms
En = (Z2)(- 2.18 X 10-18 J)
n2 Z = Atomic number of the element (H=1, He=2, etc)
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The Batcave Intruder Alert Laser uses a beam that has a frequency of 3.53 X 1014 Hz.
a)Calculate the wavelength in nanometers. (850)
b)Calculate the energy per photon. (2.34X10-19 J)c)Calculate the energy per mole of photons.
(1.41 X 105 J/mol)d)Calculate the number of photons in a 50.0 mJ
pulse of this laser. (2.13 X 1017 photons)e)Calculate the moles of photons in that pulse.
(3.54 X 10-7 moles)f) Identify the range of the electromagnetic
spectrum of this laser.
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Wave Nature of Matter
• Louis DeBroglie• All matter has wave and particle properties• “matter waves”
= h mv
momentum• Everything has a wavelength
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Diffraction pattern of electrons scattered off aluminum foil
•Wavelike properties only matter for small objects
•Electrons, protons, light, etc…
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DeBroglie Wavelength: Ex 1
Calculate the wavelength of a baseball of mass 0.20 kg moving at 40.25 m/s (90 mph)
= h mv
= (6.626 X 10-34 J s) (0.20 kg)(40.25 m/s)
= 8.2 X 10-35 m
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DeBroglie Wavelength: Ex 2
Calculate the wavelength of an electron (9.109 X 10-31 kg) moving at 2.2 X 106 m/s
= h mv
= (6.626 X 10-34 J s) (9.11 X 10-31 kg)(2.2 X 106 m/s)
= 3.3 X 10-10 m or 0.33 nm
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DeBroglie Wavelength: Ex 3
What velocity must a neutron (1.67 X 10-27 kg) move to have a wavelength of 500 pm (1 pm = 1X10-12 m)?
500 pm1X10-12 m = 5.0 X 10-10 m1 pm
= h mv
v = h = (6.626 X 10-34 J s) = 794 m/s m(1.67 X 10-27 kg)(5.0 X 10-10 m/s)
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1. Electron is both wave and particle2. Probabilistic view – Where does the electron
“hang out” rather than where the electron “is
Quantum Mechanical Model
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Electron as a particle• Heisenberg Uncertainty Principle – can never know
both the position and velocity of an electron at the same time
• Resultsa. Electron moves randomly (not like a planet)b. Electron cloud – 90% probabilityc. Important for transistors/chips
Quantum Mechanical Model
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nucleus
Random electron cloud
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Electron as Wave• Schrodinger Wave
Equation (1926) – treats electron solely as a wave
Quantum Mechanical Model
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Result OneExplains the forbidden zone (waves do not match)
Quantum Mechanical Model
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Forbidden Zone
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Result TwoOrbital are not circular
a. Orbital – region of space where there is a significant chance of finding an electron
b. Does not move like a planet
Quantum Mechanical Model
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Three Major DiscoveriesLight is both a wave and a particle
Electron Orbitals are Quantized
The electron (and all matter) is both a wave and a particle
PlanckEinstein
Bohr De Broglie(Later Schrodinger/Heisenberg)
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1. First (principal) QN (n)– how far the electron is from the nucleus (larger the number, farther away) – Level or shell
n = 2
n = 1
Quantum Numbers
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2. Second (azimuthal) QN (l) – the shape of the orbital
Value of l 0 1 2 3Letter used s p d f
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Quantum Mechanical Model
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3. Third (magnetic) QN (ml)– the suborbital
Orbital # suborbitals Total e-
s 0 2p 3 (px,py,pz) 6d 5 10f 7 14
Link to Periodic table
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• In a magnetic field, you can “see” the three p suborbitals in a line spectrum
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4. Fourth (spin) QN (ms)– spin of the electron
Pauli Exclusion Principle – No two electrons in an atom can have the same four quantum numbers
– Two electrons in the same suborbital (ex: py) must have opposite spins
– Can have values of +1/2 or -1/2
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Shows the two electrons, spin +½ and spin -½
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n Possible Values of l
Possible Values of ml Subshell name
1 0 0 1s
2 01
0-1, 0, +1
2s2p
3 012
0-1, 0, +1
-2, -1, 0, +1, +2
3s3p3d
4 0123
0-1, 0, +1
-2, -1, 0, +1, +2-3, -2, -1, 0, +1, +2, +3
4s4p4d4f
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Quantum Numbers: Ex 1
What is the designation for a subshell n=5 and l =1?
How many suborbitals are in this subshell?
What are the values of ml for each of the orbitals?
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Quantum Numbers: Ex 1
What is the designation for a subshell n=5 and l =1?5p
How many orbitals are in this subshell?3
What are the values of ml for each of the orbitals?-1, 0, +1
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Quantum Numbers: Ex 2
Which of the following sets of quantum numbers are not allowed in the hydrogen atom?
A. n=2, l=0, ml =0, ms=1/2
B. n=1, l=0, ml=0, ms= -1/2
C. n=3, l=1, ml= 2, ms=1/2
D. n=4, l=2, ml= -2, ms=1/2
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Quantum Numbers: Ex 2
Which of the following sets of quantum numbers are not allowed in the hydrogen atom?
A. n=2, l=0, ml =0, ms=1/2
B. n=1, l=0, ml=0, ms= -1/2
C. n=3, l=1, ml= 2, ms=1/2
D. n=4, l=2, ml= -2, ms=1/2
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Quantum Numbers: Ex 3
Which of the following sets of quantum numbers are not allowed in the carbon atom?
A. n=2, l=0, ml =0, ms=1/2
B. n=1, l=1, ml=0, ms= -1/2
C. n=2, l=1, ml= -1, ms=1/2
D. n=4, l=2, ml= -3, ms=1/2
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Quantum Numbers: Ex 3
Which of the following sets of quantum numbers are not allowed in the carbon atom?
A. n=2, l=0, ml =0, ms=1/2
B. n=1, l=1, ml=0, ms= -1/2
C. n=2, l=1, ml= -1, ms=1/2
D. n=4, l=2, ml= -3, ms=1/2
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1. Electron Configuration – shorthand notation to tell you the locations of all the electrons in an atom or ion
2. Notation
2p3
Orbit Shape # e-
Electron Configurations
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Electron Configuration
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Electron Configurations
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Why is “d” one less?
• Energy is slightly above the next s orbital
• Complete the orbital filling diagram for manganese
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1. Easy ExamplesH He O Fe S
2. Write e- configuration
Li N Sr P SeV F Ar Mg Kr
Electron Configurations
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3. Which element is represented by the following electron configurations?
1s22s22p63s23p64s23d5
1s22s22p63s23p64s23d104p65s24d7
1s22s22p63s23p64s1
1s22s22p63s23p3
1s22s22p63s1
1s22s22p63s23p2
1s22s22p63s23p64s23d104p6
Electron Configurations
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1. Rule – Use the noble gas in the previous row2. Examples
Ne and PRuKr
You try:
Br Ar S Ca I Xe
Noble Gas (Condensed) Shortcut
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Electron Configuration Exceptions
• Mostly with transition metal elements• There is a special stability to filled and half-filled
orbitals
Element Actual configuration Instead ofCr [Ar]4s13d5 [Ar]4s23d4
Mo [Kr]5s14d5 [Kr]5s24d4
Cu [Ar]4s13d10 [Ar]4s23d9
Ag [Kr]5s14d10 [Kr]5s24d9
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p e e- configuration
SrSr+
Sr2+
Al2+
Al3+
Ions
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p e e- configuration
SS1-
S2-
Br1-
BaBa2+
B3+
Ions
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Transition Metal IonsLose their “s” electrons first.Fe [Ar]4s23d6
Fe1+
Fe2+
Fe3+
CuCu2+
RuRu3+
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Transition Metal Ions
Zn2+
Co1+
Co2+
V3+ Sc2+
Sc3+
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4. a) Increase b) Decreasec) line spectrum(forbidden zone)
14. a) Gamma < (d)yellow < (e)red < (b)93.1 radio < (c)680 AM
16. a) 3.00 X 1017 Hzb) 3.94 X 10-3 mc) (a) is X-Ray and (b) is microwaved) 7.64 X 10-6 m
18. UV has a higher frequency and shorter than IR. UV produces more energy
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22.a) 5.09 X 1014Hz b) 20.3 kJc) 3.37 X 10-19 J d) Na+
24. AM: 6.69 X 10-28 J FM: 6.51 X 10-26 J26.1.56 X 10-18 J/photon, 127 nm28. a) microwave b) 6.4 X 10-11 J/hr34.a) Absorbed b) Emitted c) Absorbed36.a) 9.7 X10-8 m, emitted b) 434 nm, emitted
c) 1.06 X 10-6m, absorbed38. a) n=1 larger E b) 121 nm, 103 nm, 97.2
nm40. a)2.626X10-6m (IR) b) 6 4 transition
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42. 3.97 X 10-10 m (3.97 A)44. 7.75 X 10-11 m (0.775 A)50. a) n = 3 (l=2,1,0) (9 ml values)
b) n = 5 (l = 4,3,2,1,0) (25 ml values)
52. a) 2,1,1 2,1,0 2,1,-1b) 5,2,2 5,2,1 5,2,0 5,2,-1 5,2,-2
54. 2p not allowed1s 4dnot allowed 5f3dnot allowed
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Solution to 41c
6.941 g X 1kg =0.006941kg/mol1mol 1000g
0.006941 kg X 1mol = 1.15 X 10-26kg/atom
1mol 6.022X1023 atoms
= h = 2.3 X 10-13 m(1.15 X 10-26kg)(2.5 X 105 m/s)
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66C P Ne
Config [He]2s2sp2 [Ne]3s23p3 [He]2s22p6
Core 2 10 2
Valence 4 5 8
Unpaired 2 3 0
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68. a) [Ar]4s23d104p1 (1 unpaired)b) [Ar]4s2 (0 unpaired)c) [Ar]4s23d3 (3 unpaired)d) [Kr]5s24d105p5 (1 unpaired)e) [Kr]5s24d1 (1 unpaired)f) [Xe]6s14f145d9 (2 unpaired)g) [Xe]6s24f145d1 (1 unpaired)
70. a) [Ar]3d10 b) [Xe]4f145d8
c) [Ar]3d3 d) [Ne]3s23p6
72.a) 7A b) 4B c) 3Ad) Sm and Pm (f-block)
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74.a) [He]2s22p3 b) [Ar]4s23d104p4
c) [Kr]5s24d7
76.a) Ba Ca K Na b) Au (shortest) Na(longest)c) 455 nm, Ba
78 a) 9.37X1014 s-1 b) 374 kJ/molc) UV-B Shorter , higher energy d)UV-B
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100. O3 O2 + O
H = 105.2 kJ/molH = 1.052 X 105 J/mol
1.052 X 105 J 1 mol = 1.75 X 10-19 J1 mol 6.022X1023 molec.
E = h = 2.63 X 1014 Hz= c/ = 1140 nm
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Extra Problem
A certain biomolecule requires 598 kJ/mol to break one of its bonds. What wavelength of light (nm)would a single photon need to break this bond?
What region of the EM spectrum is this?
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A photon of wavelength 550 nm will break a bond in a synthetic dye. Calculate the energy per mole of that bond (kJ/mol)
(218 kJ/mol)
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a) Calculate the wavelength of light emitted from a 72 transition in a hydrogen atom. (398 nm)
b) Calculate the speed an electron would need to have the wavelength. (1830 m/s)
c) Calculate the wavelength of a proton moving at that same speed.
(2.17 X 10-10 m or 0.217 nm)d) Why is the proton’s wavelength so much
smaller? Which has more energy, e- or p+?
me = 9.109 X 10-31 kg
mp = 1.673 X 10-27 kg
![Page 122: Electronic Structure of Atoms Chapter 6. Light is a Wave Electromagnetic wave Wavelength ( ), m Frequency ( ), Hz or s -1 Travels at c (3.00 X 10 8 m/s)](https://reader031.fdocuments.in/reader031/viewer/2022013004/56649c9a5503460f94957b2f/html5/thumbnails/122.jpg)
a) Calculate the wavelength of a proton that has been accelerated to 2 X 106 m/s. (1.98X10-13 m)
b) Calculate the frequency. (1.51 X 1021 Hz)c) Calculate the energy. (1.00 X 10-12
J/photon)
mp = 1.673 X 10-27 kg