ATOMIC STRUCTURE CHAPTER 7 All Bold Numbered Problems

Chapter 7 Outline Events leading to Quantum Mechanics Newton Planck Einstein Bohr de Broglie Schrödinger Heisenberg Using Quantum Numbers

ATOMIC STRUCTURE From the ERA of Newtonian Physics to Quantum Physics

ELECTROMAGNETIC RADIATION

Electromagnetic Radiation Most subatomic particles behave as PARTICLES and obey the physics of waves. Define properties of waves Figure 7.1 and7.2. Wavelength,  Node Amplitude

Figures 7.1 Electromagnetic Frequency

Electromagnetic Radiation wavelength Visible light Wavelength () Ultraviolet radiation Amplitude Node There are no LIMITS to ... there are an  .

Electromagnetic Radiation wavelength Visible light Ultaviolet radiation Amplitude Node Node in a standing wave

Electromagnetic Radiation Waves have a frequency Use the Greek letter “nu”, , for frequency, and units are “cycles per sec” All radiation:  •  = c where c = velocity of light = 3.00x108 m/sec Long wavelength ----> small frequency Short wavelength ----> high frequency

Electromagnetic Radiation Long wavelength -----> small frequency Short wavelength -----> high frequency increasing wavelength increasing frequency See Figure 7.3

Figure 7.3 Long wavelength -----> small frequency Short wavelength -----> high frequency

Electromagnetic Radiation Red light has  = 700. nm. Calculate the frequency. 700 nm • 1 x 10 -9 m 1 nm = 7.00 x 10 -7 Freq = 3.00 x 10 8 m/s 7.00 x 10 -7 m 4.29 x 10 14 sec -1 Examples

Standing Waves 1st vibration 2nd vibration 1st vibration = ½ 3rd vibration = 3(½) 2nd vibration See Figure 7.4

Newtonian Physics Breakdown -Quantization of Energy- It was believed that like wave theory, energy was also continuous. Max Planck (1858-1947) Solved the “ultraviolet catastrophe”

Figure 7.5 Intensity should Increase with Decreasing  . As you add more energy, atoms should vibrate with a higher energy, in a continuous fashion. Objects can gain or lose energy by absorbing or emitting radiant energy in QUANTA.

Quantization of Energy Energy of a vibrating system (electro-magnetic radiation) is proportional to frequency. Ep = h •  h = Planck’s constant = 6.6262 x 10-34 J•s We now MUST abandon the idea that Energy acts as a continuous wave!

From Planck on to Einstein

Photoelectric Effect A. Einstein (1879-1955) Experiment demonstrates the particle nature of light. (Figure 7.6) Classical theory said that E of ejected electron should increase with increase in light intensity—not observed! No e- observed until light of a certain minimum E is used & Number of e- ejected depends on light intensity.

Photoelectric Effect Experimental observations says that light consists of particles called PHOTONS having discrete energy. It takes a high energy particle to bump into an atom to knock it’s electron out, hence the use of a ½ mv2 term. It would take some minimum energy i.e. critical energy to knock that electron away from it’s atom.

Energy of Radiation PROBLEM: Calculate the energy of 1.00 mole of photons of red light.  = 700. nm ( c = l n )  = 4.29 x 1014 sec-1 Ep = h• = (6.63 x 10-34 J•s)(4.29 x 1014 sec-1) = 2.85 x 10-19 J/photon Notice Einstein's use of Planck's formula.

Energy of Radiation Energy of 1.00 mol of photons of red light. Ep = h• = (6.63 x 10-34 J•s)(4.29 x 1014 sec-1) = 2.85 x 10-19 J per photon E per mol = (2.85 x 10-19 J/ph)(6.02 x 1023 ph/mol) = 171.6 kJ/mol This is in the range of energies that can break bonds.

Photoelectric Effect A minimum frequency is required to cause any current flow. Above that frequency, the current is related to the intensity of the light used. The ejected electrons (since we are talking about collisions between photons and electrons) also have more kinetic energy when higher frequencies are used. EK = 1/2 meve2 = Einput - Eminimum Einstein finds: Ep = h• = 1/2 meve2, evidence that photons have both wave/particle properties

Photoelectric Effect Light is used to eject an electron from a metal. Calculate the velocity of the ejected electron if the photon used to eject the electron has a wavelength of 2.35 x 10 -7 m and the minimum frequency required to eject an electron is 8.45 x 10 14 s-1. Step by step!!

The Final Crack in Classical, Newtonian Physics MONUMENTAL Edifice Planck---Energy is NOT Continuous like waves Einstein---Energy comes in packets or is Quantized and energy also has some wave and particle behavior Bohr---Applies Quantized idea to atomic particles….the H1 Atom to explain…..

Atomic Line Spectra and Niels Bohr (1885-1962) Bohr’s greatest contribution to science was BUILDING a SIMPLE MODEL of the ATOM. It was based on an understanding of the LINE SPECTRA of excited atoms and it’s relationship to quantized energy.

Line Spectra of Excited Atoms Excited atoms emit light of only certain wavelengths (Planck). The wavelengths of emitted light depend on the element.

Figure 7.7

Figure 7.8

Figure 7.9

Line Spectra of Excited Atoms High E Short  High  Low E Long  Low  Visible lines in H atom spectrum are called the BALMER series.

Shells or Levels!! Why??

Figure 7.12

Excited Atoms Emit Light

Atomic Spectra and Bohr One view of atomic structure in early 20th century was that an electron (e-) traveled about the nucleus in an orbit. Electron + Orbit 1. Any orbit (like a wave-see slide 3) should be possible and so should any energy. 2. But a charged particle would always be accelerating from the nucleus (vector velocity is always changing) and since it is moving in an electric field would continuously emit energy. End result should be destruction since the energy mentioned in the previous step is finite!

Atomic Spectra and Bohr Bohr said classical (Newtonian) view is wrong!!!. Need a new theory — now called QUANTUM or WAVE MECHANICS. e- can only exist in certain discrete orbits — called stationary states. e- is restricted to QUANTIZED energy states. Energy of state, En = - C/n2 where n = quantum no. = 1, 2, 3, 4, .... this describes the potential energy of an electron

Atomic Spectra and Bohr Energy of quantized state, En = - C/n2 Only orbits where n = integral numbers are permitted. Radius of allowed orbitals, Rn, Rn= n2 R0 with Ro = 0.0529 nm Note the same equations come from modern wave mechanics approach. Results can be used to explain atomic spectra.

Atomic Spectra and Bohr If e-’s are in quantized energy states, then DE of states can have only certain values. This explain sharp line spectra. n = 1 n = 2 E = -C ( 1/2 2 ) E = -C ( 1/1

Atomic Spectra and Bohr = 1 2 E - C ( / ) ENERGY Calculate DE for e- “falling” from high energy level (n = 2) to low energy level (n = 1). DE = Efinal - Einitial = - C [ (1/1)2 - (1/2)2 ] DE = - (3/4) C Note that the process is exothermic!

Atomic Spectra and Bohr = 1 2 E - C ( / ) ENERGY DE = - (3/4)C C has been found from experiment and is proportional to RH, the Rydberg constant. RHhc = C = 1312 kJ/mole. n of emitted light = (3/4)C = 2.47 x 1015 sec-1 and l = c/n = 121.6 nm This is exactly in agreement with experiment!

Line Spectra of Excited Atoms DE = Efinal - Einitial = - RHhc [ (1/nfinal2) - (1/ninitial2) ] A photon of light with frequency 8.02 x 1013 s-1 is emitted from a hydrogen atom when it de-excites from the n = 8 level to the n = ? level. Calculate the final quantum number state of the electron.

Atomic Line Spectra and Niels Bohr Bohr’s theory was a great accomplishment. Rec’d Nobel Prize, 1922 Problems with theory — theory only successful for H and only 1e- systems He+, Li2+. introduced quantum idea artificially. However, Bohr’s model does not explain many e- systems….So, we go on to QUANTUM or WAVE MECHANICS Niels Bohr (1885-1962)

Quantum or Wave Mechanics de Broglie (1924) proposed that all moving objects have wave properties. For light: E = mc2 E = h = hc /  Therefore, mc = h /  and for particles (mass)(velocity) = h /  , the wave-nature of matter. L. de Broglie (1892-1987) l = h mv

Quantum or Wave Mechanics Baseball (115 g) at 100 mph  = 1.3 x 10-32 cm e- with velocity = 1.9 x 108 cm/sec  = 0.388 nm Experimental proof of wave properties of electrons

Quantum or Wave Mechanics Schrödinger applied idea of e- behaving as a wave to the problem of electrons in atoms. He developed the WAVE EQUATION. E. Schrödinger 1887-1961

Quantum or Wave Mechanics Solution of the wave equation give a set of mathematical expressions called WAVE FUNCTIONS, . Each describes an allowed energy state of an e-. Quantization is introduced naturally. E. Schrodinger 1887-1961

WAVE FUNCTIONS,  is a function of distance and two angles. Each  corresponds to an ORBITAL — the region of space within which an electron is found.  does NOT describe the exact location of the electron. 2 is proportional to the probability of finding an e- at a given point.

Uncertainty Principle Problem of defining nature of electrons in atoms solved by W. Heisenberg. Cannot simultaneously define the position and momentum (= m•v) of an electron. We define e- energy exactly but accept limitation that we do not know exact position. W. Heisenberg 1901-1976

QUANTUM NUMBERS n --> shell l --> subshell Each orbital is a function of 3 quantum numbers: n, l, and ml Electrons are arranged in shells(levels) and subshells(sublevels). n --> shell l --> subshell ml --> designates an orbital within a subshell

Symbol Values Description QUANTUM NUMBERS Symbol Values Description n (major) 1, 2, 3, .. Orbital size and energy where E = - RHhc(1/n2) l (angular) 0, 1, 2, .. n-1 Orbital shape or type (subshell) ml (magnetic) - l..0..+ l Orbital orientation # of orbitals in subshell = 2 l + 1

All 4 Quantum Numbers Principle quantum number (n) Azimuthal quantum number (l) Magnetic quantum number (m) Spin quantum number (s)

Atomic Orbitals the result of Quantum Mechanics Calculations

Shells and Subshells When n = 1, then l = 0 and ml = 0 . Therefore, if n = 1, there is 1 type of subshell and that subshell has a single orbital. (ml has a single value ---> 1 orbital) This subshell is labeled s Each shell has 1 orbital labeled s, and it is SPHERICAL in shape.

1s Orbital

2s Orbital

3s Orbital

Atomic Orbitals the result of Quantum Mechanics Calculations

p Orbitals When n = 2, then l = 0 and 1 Typical p orbital When n = 2, then l = 0 and 1 Therefore, in the n = 2 shell there are 2 types of orbitals — 2 subshells For l = 0 ml = 0 this is an s subshell For l = 1 ml = -1, 0, +1 this is a p subshell with 3 orbitals planar node When l = 1, there is a PLANAR NODE thru the nucleus. See Figure 7.16

p Orbitals A p orbital The three p orbitals lie 90o apart in space

2px Orbital

2py Orbital

2pz Orbital

3px Orbital

3py Orbital

3pz Orbital

d Orbitals l = 0, 1, 2 When n = 3, what are the values of l? and so there are 3 subshells in the shell. For l = 0, ml = 0 ---> s subshell with a single orbital For l = 1, ml = -1, 0, +1 ---> p subshell with 3 orbitals For l = 2, ml = -2, -1, 0, +1, +2 ---> d subshell with 5 orbitals

d Orbitals typical d orbital planar node See Figure 7.16 s orbitals have no planar node (l = 0) and so are spherical. p orbitals have l = 1, and have 1 planar node, and so are “dumbbell” shaped. This means d orbitals, ( l = 2) have 2 planar nodes

3dxy Orbital

3dxz Orbital

3dyz Orbital

3dz2 Orbital

3dx2- y2 Orbital

f Orbitals For l = 3, ml = -3, -2, -1, 0, +1, +2, +3 When n = 4, l = 0, 1, 2, 3 so there are 4 subshells in the shell. For l = 0, ml = 0 ---> s subshell with single orbital For l = 1, ml = -1, 0, +1 ---> p subshell with 3 orbitals For l = 2, ml = -2, -1, 0, +1, +2 ---> d subshell with 5 orbitals For l = 3, ml = -3, -2, -1, 0, +1, +2, +3 ---> f subshell with 7 orbitals

Orbitals and Quantum Numbers n l ml 1 0 0 1s 2 0 0 2s 2 1 1 2 1 0 2 1 -1 2p

Orbitals and Quantum Numbers n l ml 3 0 0 3s 3 1 1 3 1 0 3 1 -1 3p 3d 3 2 2 3 2 1 3 2 0 3 2 -1 3 2 -2

Sample Problems No Yes One 9 5d 3d 6f 1. Is it possible to have a d orbital in level 1? 2. Is it possible to have a 6s subshell? 3. How many orbitals are in a 7s sublevel? 4. How many orbitals are possible if n = 3? 5. What type of orbital has the quantum numbers a) n = 5, l = 2, ml = 1 b) n = 3, l = 2, ml =-1 c) n = 6, l = 3, ml = -3 No Yes One 9 5d 3d 6f

Practice Problems 1. Calculate the wavelength of a photon having an energy of 2.58 x 10-18 J. 2. In the hydrogen atom, which transition, 3 --> 2 or 2 --> 1, has the longer wavelength? 3. Calculate the wavelength of an object (mass = 545 lbs) with a speed of 45 miles/hour. 4. Give all possible sets of quantum numbers for 4p, 3d, and 5s. 5. How many orbitals are in the a. the third level? b. l = 3 sublevel?

Practice Problems Answers 1. 7.71 x 10-8 2. 3 --> 2 3. 1.3 x 10-37 m 5. a) 9 b) 7 4. 4p n l ml 4 1 1 4 1 0 4 1 -1 Problem 4 continued on next slide.

Practice Problems Answers 3d n l ml 3 2 2 3 2 1 3 2 0 3 2 -1 3 2 -2 5s 5 0 0

 •  = c Sample Problem 1 x 10-7m . n = 3.00 x 108 m/s 1. Calculate the frequency of light having a wavelength of 1 x 10-7m.  •  = c 1 x 10-7m . n = 3.00 x 108 m/s n = 3 x 1015 /s

 •  = c . 1.5 x 108 /s = 3.00 x 108 m/s Sample Problem  = 2.0 m 2. Calculate the wavelength of light having a frequency of 1.5 x 108 hz.  •  = c . 1.5 x 108 /s = 3.00 x 108 m/s  = 2.0 m

 •  = c Sample Problem 1 x 10-6m . n = 3.00 x 108 m/s 3. Calculate the frequency of light having a wavelength of 1 x 103nm.  •  = c 1 x 10-6m . n = 3.00 x 108 m/s n = 3 x 1014 /s

Practice Problem Ep = h• Ep = 6.63 x 10-34 Js • 3 x 1015/s 1. Calculate the energy of a photon having a frequency of 3 x 1015/s. Ep = h• Ep = 6.63 x 10-34 Js • 3 x 1015/s = 2 x 10-18 J

Practice Problem Ep = h•  = 5.0 x 1014 /s 2. Calculate the frequency of light having an energy of 2.0 x 105 J/mole. Ep = h• 2.0 X 105 J mole mole 6.02 x 1023 photon 3.3 x 10-19 J = 6.63 x 10-34 Js •   = 5.0 x 1014 /s

 •  = c Practice Problem Ep = h• 3. Calculate the energy of a photon with a wavelength of 575 nm.  •  = c 5.75 x 10-7 m •  = 3.00 x 108 m/s  = 5.22 x 1014/s Ep = h• Ep = 6.63 x 10-34 Js • 5.22 x 1014/s = 3.46 x 10-19 J

 •  = c Calculate the energy of the photon: Ep = h• Photon wavelength = 2.35 x 10 -7 m  •  = c 2.35 x 10-7 m •  = 3.00 x 108 m/s  = 1.28 x 1015/s Ep = h• Ep = 6.63 x 10-34 Js • 1.28 x 1015/s = 8.49 x 10-19 J

Ep = h• Calculate the min. energy to eject an electron: Min. n = 8.45 x 10 14 s-1. Ep = h• Ep = 6.63 x 10-34 Js • 8.45 x 1014/s = 5.60 x 10-19 J

Calculate the extra energy of the electron: 8.49 x 10-19 J - 5.60 x 10-19 J = 2.89 x 10-19 J Calculate the velocity of the electron: E = 1/2 m v2 2.89 x 10-19 J = (1/2) 9.11 x 10-31 kg v2 = 7.96 x 105 m/s

Calculate the energy of the photon: Ep = 6.63 x 10-34 Js • 8.02 x 1013/s 5.32 X 10-20 J 6.02 x 1023 photon photon mole = 3.20 x 104 J = 32.0 kJ

Calculate the level number : Ep = h• DE = -C [(1/n)2 - (1/n)2] -32. kJ = -1312 kJ [(1/n)2 - (1/8)2] n = 5