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CHEM 1320 Study Guide - Hmu Language, Diamagnetism, Fluorine

15 Pages
Fall 2013

Course Code
CHEM 1320

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Chapter 7: Quantum Theory and the Electronic Structure of
Wavelength ( ): peak-to-peak distance on successive wavesλ
Amplitude: vertical distance from midline to peak
Frequency ( ): # of waves pass through a point in 1 sec. ν
Speed ( ) of the wave: X υ λ ν
Higher the frequency, shorter the wavelength
Lower the frequency, longer the wavelength
Inverse relationship
Maxwell (1873): visible light is electromagnetic (EM) waves.
EM radiation: emission & transmission of energy in the form of
electromagnetic waves.
Speed of light (c)= 3.00 X 10^8m/s
ALL EM radiation: λX ν=c
Violent=shortest wavelength, longest frequency
Red=longest wavelength, shortest frequency
Units of frequency:
Sec^-1 S^-1 Cycles per second (cps) Hz (Hertz)
Symbol: (ν) (Greek nu)
Unit: (λ)
Different λ=different kind of light
Measured in nm (one billionth of a meter)
250nm=250 X 10^-9m
Frequency Wavelength
Can convert λ and frequency (ν):
In meters  λ=c/ν c=3.00 X 10^8m/s
A radio station is 103.5 FM. What wavelength does it broadcast?
103.5FM=103.5MHz=103.5 million s^-1
λ=3.00 X 10^8 m/s
---------------------- = 2.8 m (~10ft)
What is the frequency of light if its wavelength is 200 m?
=3.00 X 10^8 m/s
---------------------- = 1,500,000s^-1 1.5 X 10^6s^-1
Visible=most intense
Wavelength=most abundant
UV=most damaging
Mystery #1
“Heated Solids Problem”
-Heated solids emit EM radiation over a wide range of
White-hot object: hotter than red-hot
Different wavelengths  different amounts of energy
Energy (light) is emitted or absorbed in discrete units (quantum).
E=h X ν
Planck’s constant (h)
h=6.63 X 10^-34 Js
What is the energy of UV light with a frequency of
1.00 X 10^15 s^-1?
E=hν=(6.63 X 10^-34 Js)(1 X 10^15 s^-1)
E=6.63 X 10^-18 J
E=hν=hc/λ Since ν=c/λ
Compare to a radio wave of λ 3 X 10^9 m
E= hc/λ =(6.63 X 10^-34 Js)(3.00 X 10^8 m/s)
3 X 10^9 m
E=6.63 X 10^-35J
What happens when light hits matter?
IR-molecules vibrate, matter heats up

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Chapter 7: Quantum Theory and the Electronic Structure of  Atoms 10/11  Wavelength (λ): peak­to­peak distance on successive waves Amplitude: vertical distance from midline to peak Frequency (ν): # of waves pass through a point in 1 sec.  (Hz=1cycle/s) Speed (υ) of the wave: λ X ν Higher the frequency, shorter the wavelength Lower the frequency, longer the wavelength Inverse relationship Maxwell (1873): visible light is electromagnetic (EM) waves. EM radiation: emission & transmission of energy in the form of  electromagnetic waves. Speed of light (c)= 3.00 X 10^8m/s ALL EM radiation: λ X ν=c Violent=shortest wavelength, longest frequency Red=longest wavelength, shortest frequency Frequency: Units of frequency: Sec^­1   S^­1  Cycles per second (cps)   Hz (Hertz) Symbol: (ν) (Greek nu) Wavelength:  Unit: (λ) Different λ=different kind of light Measured in nm (one billionth of a meter) 250nm=250 X 10^­9m Frequency  Wavelength  Can convert λ and frequency (ν): In meters  ▯ λ=c/ν                   c=3.00 X 10^8m/s A radio station is 103.5 FM. What wavelength does it broadcast? 103.5FM=103.5MHz=103.5 million s^­1 λ=3.00 X 10^8 m/s    ­­­­­­­­­­­­­­­­­­­­­­ = 2.8 m    (~10ft)    103,500,000s^­1 What is the frequency of light if its wavelength is 200 m? ν=c/λ =3.00 X 10^8 m/s   ­­­­­­­­­­­­­­­­­­­­­­ = 1,500,000s^­1        1.5 X 10^6s^­1           200m Visible=most intense Wavelength=most abundant UV=most damaging Mystery #1 “Heated Solids Problem” ­Heated solids emit EM radiation over a wide range of  wavelengths. White­hot object: hotter than red­hot Different wavelengths  ▯different amounts of energy Energy (light) is emitted or absorbed in discrete units (quantum). E=h X ν Planck’s constant (h) h=6.63 X 10^­34 Js What is the energy of UV light with a frequency of 1.00 X 10^15 s^­1? E=hν=(6.63 X 10^­34 Js)(1 X 10^15 s^­1) E=6.63 X 10^­18 J E=hν=hc/λ        Since ν=c/λ Compare to a radio wave of λ 3 X 10^9 m E= hc/λ       =(6.63 X 10^­34 Js)(3.00 X 10^8 m/s)                           ­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­ 3 X 10^9 m E=6.63 X 10^­35J What happens when light hits matter? IR­molecules vibrate, matter heats up Visible­excite electrons, chemical reactions can occur  (photosynthesis) UV­knock electrons off  ▯ions High frequency­ UV­break covalent bonds (!) (disrupt biological processes,      damage tissues, cancer) Radiation and matter Einstein: photon=“bundle of light energy” New way of thinking about light: ­Light is a particle ­Has no mass ­Moves in waves Mystery #2 “Photoelectric Effect” Light has both: 1.wave nature 2.particle nature Photon: “particle” of light Hν=KE + W KE=hν­W W: the work function Depends how strongly the metal holds electrons  Minimum energy to induce the photoelectric eff. Why do we care about energy? Reactions require energy If light has enough E, the reaction occurs If not, the reactions will not occur, no matter how much light you  put in! You won’t get a tan from standing in front of a radio ­no matter how long It requires a photon of 4.41 X 10^­9J to emit electrons from  sodium. What is the minimum frequency of light required? E=hν ν=E/h 6.65 X 10^14s^­1 What is the minimum wavelength of light required? E=hc/λ λ=3.00 X 10^8 s    ­­­­­­­­­­­­­­­­­­­­­­ = 4.51 X 10^­9 m       1 X 10^4 nm   = 451nm    6.65 X 10^14 s^­1 *m▯nm= X 10^­9 What is the maximum KE emitted electrons if 439 nm light were  used? KE=hν+w         ν=c/λ  = 6.83 X 10^14 (use numbers from above)       =(6.63 X 10^­34Js)(6.83 X 10^14Hz) + 4.41 X 10^­2 J)       =4.0 X 10^­9.,J Max number of electrons freed if 1.00 μJ of energy are used? 1.00μJ          1e­                           1J             x ­­­­­­­­­­­­­­­­­­­­ x ­­­­­­­­­­­­­­  = 2.27 X 10^12 electrons              4.41 X 10^­19  J     1 X 10^6 J 10/14  If molybdenum is irradiated with 120. nm light and the photons  ejected have a KE of 2.38 X 10^­18J, what is molybdenum’s work  function? KE=hν+w W=KE­hν     =2.38 X 10^­18J – (6.63 X 10^­34Js)(3.00 X 10^8 m/s) ­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­                            120 X 10^­9 m W= 7.20 X 10^­19J Bohr’s Postulates: ­Only orbits of certain radii are allowed ­e­ in allowed orbits have specific energies >Allowed energy states >Won’t crash into the nucleus ­e­ can only absorb/emit energy if they wind up in an allowed  energy state >Energy is a photon \ of E=hν Rules: 1. e­ can have specific (quantized) energy values 2. When e­ moves to a lower energy level, light is emitted >E =­R H (1/n^2) >n(principal quantum number) = 1,2,3,…. >R H(Rydberg constant for H)=2.18 X 10^­18 J E phot=ΔE=E f­Ei E =­R H (1/n^2f) E =­R H (1/n^2i) ΔE=­R H(1/n^2 ­1/n^2 ) What is the energy of an electron in H if n=2? E =­R H(1/n^2) E =(­2.18 X 10^­18J)(1/2^2) E =­5.45 X 10^­19 J What is the energy of an electron in H if n=6? E =­R H(1/n^2) E =(­2.18 X 10^­18J)(1/6^2) E =­6.05 X 10^­20 J What is the energy change if an election in H goes from n=5 to  n=3? ΔE=­R H(1/n^2 ­1/n^2 ) ΔE=(­2.18 X 10^­18J)(1/3^2­1/5^2) ΔE= ­1.55 X 10^­19 J Higher energy level to lower energy level What is the energy change if an electron in H goes from n=1 to  n=4? ΔE=­R H(1/n^2 ­1/n^2 ) ΔE=(­2.18 X 10^­18J)(1/4^2­1/1^2) ΔE= 2.04 X 10^­18 J Lower energy level to higher energy level What is the wavelength of light emitted if an electron in H goes  from n=6 to n=2? ΔE=­R H(1/n^2 ­1/n^2 ) ΔE=(­2.18 X 10^­18J)(1/2^2­1/6^2)=hc/λ ΔE= ­4.84 X 10^­19J = λ=hc/E         E=­4.84 X 10^­19J λ=(6.63 X 10^­34Js)(3.00 X 10^8m/s)     ­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­ = 4.11 X 10^­7m/411nm                       4.84 X 10 ^­19 J What is the frequency of light emitted if an electron in H goes  from n=4 to n=2? ΔE=­R H (1/n^2f­1/n^2 ) ΔE=(­2.18 X 10^­18J)(1/2^2­1/4^2)=hν ΔE= ­4.09 X 10^­19J = hν ν=4.09 X 10^­19J       ­­­­­­­­­­­­­­­­­­­­­ = 6.17 X 10^14 s    6.63 X 10^­34Js Why is the e­ quantized? De Broglie (1924) reasoned that e­ is both particle and wave. 2πr=nλ λ=h/mu 2πr=orbit circumference  u= velocity of e­ m= mass of e­ What is the wavelength of an 85­
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