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)

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 108 m/s) in a vacuum

• Wavelength and frequency are inversely proportional

• c =

•How many complete waves are shown above?•What is the wavelength of light shown above?

Electromagnetic Spectrum

Blu-Ray = 405 nanometers (blue light)DVD = 650 nanometers (red light)

• Visible light = 4 X 10-7 m to 7X 10-7 m (400 to 700 nm)

Waves: Ex 1

Calculate the wavelength of a 60 Hz EM wave

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.

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)

Photons: Ex 1

Calculate the energy of a photon of wavelength 600 nm.

600 nm 1 X 10-9m = 6 X10-7 m1 nm

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)

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)

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

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

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

Low Frequency High FrequencyNot enough energy to eject electron

Can eject electronEnergy of photon is greater than W (Work function)

2. More intensity– More photons– More electrons ejected with same KE

3. Greater Frequency

– No more electrons ejected

– Electrons ejected with greater speed (KE)

Photon/Matter Interactions

1. Electron excitation (photon disappears)2. Ionization/photoelectric effect (photon

disappears)3. Scattering by nucleus or electron4. Pair production (photon disappears)

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

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.

Principle of Complimentarity

• Any experiment can only observe light’s wave or particle properties, not both

• Different “faces” that light shows

Line Spectra• Discharge tube

– Low density gas (acts like isolated atoms)– high voltage

• Light emitted only at certain (discrete) wavelengths

Hydrogen

Helium

Solar absorption spectrum

• Electrons orbit in ground state (without radiating energy)

• Jumps to excited state by absorbing a photon

• Returns to ground state by emitted a photon

h = Ee - Eg

4. Ways to make something glow Bohr

Model

Photon Absorption Collision-Glow in the dark -Heat

-Electricity-Chemical Reaction

Photon Absorption Collision

Bohr’s Equation

• Works only for H and other 1-electron atoms (He+, Li2+, Be3+, etc…)

En = -RH

n2 En = Energy of an orbital

RH = Rydberg constant (2.18 X 10-18 J)

n = Principal quantum number of orbital

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

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


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

c = c/= (3 X 108m/s)(2.48 X 1015 s-1 )= 122 nm

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)


Calculate the wavelength of light emitted if a hydrogen electron drops from the n=6 to the n=2 orbit?

(ANS: 410 nm (violet))


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))

Bohr Model: Other Atoms

En = (Z2)(- 2.18 X 10-18 J)

n2 Z = Atomic number of the element (H=1, He=2, etc)

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.

Wave Nature of Matter

• Louis DeBroglie• All matter has wave and particle properties• “matter waves”

= h mv

momentum• Everything has a wavelength

Diffraction pattern of electrons scattered off aluminum foil

•Wavelike properties only matter for small objects

•Electrons, protons, light, etc…

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


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


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)

1. Electron is both wave and particle2. Probabilistic view – Where does the electron

“hang out” rather than where the electron “is

Quantum Mechanical Model

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


nucleus

Random electron cloud

Electron as Wave• Schrodinger Wave

Equation (1926) – treats electron solely as a wave


Result OneExplains the forbidden zone (waves do not match)


Forbidden Zone

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


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)

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

2. Second (azimuthal) QN (l) – the shape of the orbital

Value of l 0 1 2 3Letter used s p d f


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

• In a magnetic field, you can “see” the three p suborbitals in a line spectrum

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

Shows the two electrons, spin +½ and spin -½

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

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?


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


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


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


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


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

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

Electron Configuration


Why is “d” one less?

• Energy is slightly above the next s orbital

• Complete the orbital filling diagram for manganese

1. Easy ExamplesH He O Fe S

2. Write e- configuration

Li N Sr P SeV F Ar Mg Kr


3. Which element is represented by the following electron configurations?

1s22s22p63s23p64s23d5

1s22s22p63s23p64s23d104p65s24d7

1s22s22p63s23p64s1

1s22s22p63s23p3

1s22s22p63s1

1s22s22p63s23p2

1s22s22p63s23p64s23d104p6


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

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

p e e- configuration

SrSr+

Sr2+

Al2+

Al3+

Ions

p e e- configuration

SS1-

S2-

Br1-

BaBa2+

B3+

Ions

Transition Metal IonsLose their “s” electrons first.Fe [Ar]4s23d6

Fe1+

Fe2+

Fe3+

CuCu2+

RuRu3+

Transition Metal Ions

Zn2+

Co1+

Co2+

V3+ Sc2+

Sc3+

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

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

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

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)

66C P Ne

Config [He]2s2sp2 [Ne]3s23p3 [He]2s22p6

Core 2 10 2

Valence 4 5 8

Unpaired 2 3 0

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)

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

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

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?

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)

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

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

Documents

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)