Physics 102: Lecture 25
Atom ic S pect roscopy & Q uant um At oms
Physics 102: Lecture 25, Slide 1 From last lecture – Bohr model
Angular momentum is quantized
Ln = nh/2π n = 1, 2, 3 ... Energy is quantized mk24 e Z 213.6⋅ Z 2 Ehn =−22 ≈− 2 eV ()where h ≡ / 2π 2h nn Radius is quantized
2 22+Ze ⎛⎞hn1 n rn ==⎜⎟ 2 ()0.0529 nm ⎝⎠2π mke Z Z
2 2 2 Velocity too! En= -13.6 Z /n = ½ mvn Physics 102: Lecture 25, Slide 2 Transitions + Energy Conservation
• Each orbit has a specific energy: energ : 2 2 En= -13.6 Z /n E2 • Photon absorbed when electron jfljumps from low energy t thiho high energy. Photon emitted when eltlectron j umps f rom high hih energy t o low energy orbit: E1
E2 – E1 = h f = h c / λ
Physics 102: Lecture 25, Slide 3 Demo: Line Spectra
In addition to the continuous blackbody spectrum, elements emit a discrete set of wavelengths which show up as lines in a diffraction grating. H n=3 2 2 656 nm En= -13.6 Z /n n=2
This is how neon signs & Na lamps work!
Spectra give us information on n=1 atitomic stttructure
Physics 102: Lecture 25, Slide 4 Checkpoint 1.1
Electron A falls from energy level n=2 to energy level n=1 (ground st tt)ate), causi ng a ph htoton t o b e emitt ittded.
Electron B falls from energy level nn3=3 to energy level nn1=1 (ground state), causing a photon to be emitted.
n=3 Which photon has more energy? n=2 1) Photon A A B 2) Photon B
n=1 Physics 102: Lecture 25, Slide 5 Spectral Line Wavelengths
Calculate the wavelength of photon emitted when an electron in the hydroge n ato m dr ops fr om th e n=2 state to t he g rou nd state (n=1). n=3 Z 2 E = − . eV n=2 n 13 6 2 E2= -3.4 eV n
hf = E2 − E1 = −3.4eV − (−13.6eV) =10.2eV E = -13.6 eV 1 n=1
hc hc 1240 E = λ = = ≈124nm photon λ 10.2eV 10.2
Physics 102: Lecture 25, Slide 6 ACT: Spectral Line Wavelengths
Compare the wavelength of a photon produced from a transition from n=3 to n=2 w it h t hat o f a p hoto n p roduced fro m a t ra ns it io n n=2 to n=1.
(Α) λ32 < λ21 n=3 n=2 (Β) λ32 = λ21
(C)λ32 > λ21
E32 < E21 so λ32 > λ21 n=1
Physics 102: Lecture 25, Slide 7 ACT/Checkpoint 1.2
The electrons in a large group of hydrogen atoms are exc ited to t he n=3 leve l. How ma ny spect ra l lines w ill be produced? A. 1 n=3 B. 2 n=2 C. 3 D. 4 E. 5 n=1
Physics 102: Lecture 25, Slide 8 The Bohr Model is incorrect!
To be consistent with the Heisenberg Uncertainty Principle, which of these properties cannot be quantized (have the exact value known)? Electron Radius Would know location Electron Energy Electron Velocity Would know momentum Electron Angular Momentum BihBhdlBut, in the Bohr model:
2 ⎛⎞hn1 22 nQuantized radii rn ==⎜⎟ 2 (0. 0529 nm) and velocities for 2π mke Z Z ⎝⎠ electron orbitals Physics 102: Lecture 25, Slide 9 Checkpoint 2
+Ze
Bohr Model Quantum Atom
So what keeps the electron from “sticking” to the nucleus? Centripetal Acceleration Pauli Exclusion Principle Heisenberg Uncertainty Principle
Physics 102: Lecture 25, Slide 10 Quantum Mechanics Theory used to predict probability distributions
QM
Physics 102: Lecture 25, Slide 11 Quantum Mechanical Atom • Predicts available energy states agreeing with Bohr. • Don’t have definite electron position, only a probability bilit f uncti on. • Each orbital can have 0 angular momentum! • Each electron state labeled by 4 numbers: n = pppqrincipal quantum number ( 1, 2, 3, …) l = angular momentum (0, 1, 2, … n-1)
ml = compp(onent of l (-l < ml < l)
ms = spin (-½ , +½) Physics 102: Lecture 25, Slide 12 Quantum Mechanics (vs. Bohr)
Electrons are described by a probability function, not a definite radius!
Physics 102: Lecture 25, Slide 13 Quantum Numbers Each electron in an atom is labeled by 4 #’s n = Principal Q uantum Number (,,,(1, 2, 3, … ) • Determines the Bohr energy
l = Orbital Quantum Number (0 , 1 , 2 , … n -1) • Determines angular momentum h L =+ll(1) • l < n always true! 2π
ml = Magnetic Quantum Number (-l , … 0, … l ) • z-component of h l Lm= z l • | ml | <= l always true! 2π
ms = SiSpin Quan tum Num ber (-½+½)½ , +½) • “Up Spin” or “Down Spin” Physics 102: Lecture 25, Slide 14 ACT: Quantum numbers
For which state of hydrogen is the orbital angular momentum required to be zero?
1. n=1 The allowed values of l are 2. n=2 0, 1, 2, …, n-1. When n=1, l must be zero. 3. n=3
Physics 102: Lecture 25, Slide 15 Spppectroscopic Nomenclature “Shells” “Subshells” n=1 is “K shell” l =0 is “s state” n=2 is “L shell” l =1 is “p state” n=3 is “M shell” l =2 is “d state” n=4 is “N shell” l =3 is “f state” n=5 is “O shell” l =4 is “g state”
1 electron in ground state of Hydrogen: n=1, l =0 is denoted as: 1s1
n=1 l =0 1 electron
Physics 102: Lecture 25, Slide 16 Electron orbitals IhildiifiifIn correct quantum mechanical description of atoms, positions of electrons not quantized, orbitals represent probabilities
Carbon orbitals imaged in 2009 using electron microscopy!
Physics 102: Lecture 25, Slide 17 Quantum Numbers
How many unique electron states exist with nn2=2?
l = 0 : 2s2
ml = 0 : ms = ½ , -½ 2 states l = 1 : 2p6
ml = +1: ms = ½ , -½ 2 states ml = 0: ms = ½ , -½ 2 states ml = -1: ms = ½ , -½ 2 sta tes
There are a total of 8 states with n=2
Physics 102: Lecture 25, Slide 18 ACT: Quantum Numbers How many unique electron states exist with n=5
and ml = +3? A) 0 B) 4 C) 8 D) 16 E) 50
l = 0 : ml = 0 Only l = 3 and l = 4 l = 1 : ml = -1, 0, +1 have ml = +3 l = 2 : ml = -2, -1,,, 0, +1, +2
l = 3 : ml = -3, -2, -1, 0, +1, +2, +3
ms = ½ , -½ 2 states
l = 4 : ml = -4, -3, -2, -1, 0, +1, +2, +3, +4
ms = ½ , -½ 2 states
There are a total of 4 states with n=5, ml = +3 Physics 102: Lecture 25, Slide 19 Pauli Exclusion Principle In an atom with many electrons only one electron is allowed in each qq(,uantum state (n, l, ml, ms). This explains the periodic table!
Physics 102: Lecture 25, Slide 20 # Electron Configurations electrons Atom Configuration 1 H 1s1 2 2 He 1s 1s shell filled (n=1 shell filled - noble gas)
3 Li 1s22s1
4 Be 1s22s2 2s shell filled 5 B 1s22s22p1 etc (n=2 shell filled - 10 Ne 1s22s22p6 2p shell filled noble gas)
s shells hold up to 2 electrons p shells hold up to 6 electrons
Physics 102: Lecture 25, Slide 21 The Periodic Table s (l =0) Also s p (l =1)
d (l =2) , 3, ... 22 n = 1,
f (l =3) What determines the sequence? Pauli exclusion & energies Physics 102: Lecture 25, Slide 22 Summary • Each electron state labeled by 4 numbers: n = priilincipal quant um numb b(123er (1, 2, 3, …) l = angular momentum (0, 1, 2, … n-1)
ml = componentfl(t of l (-l < ml < l)
ms = spin (-½ , +½) • Pauli Exclusion Principle explains periodic table • Shells fill in order of lowest energy.
Physics 102: Lecture 25, Slide 23