Lecture 1: Old Quantum Theory
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Lecture 1: Old Quantum Theory September 3, 2019 1 Early Atom Models 1. Thompson Model ∆p F ∆t ZZ e2=4π" R2 · (2R=v) ZZ e2 θ ∼ ∼ ∼ 1 0 ∼ 1 p p mv RE 4π"0 e2 e02 = = 1:44fm Mev 4π"0 5 Z −5 - For R ∼ 1:0Å = 10 fm, α particle Z = 2, θ ∼ E × 10 (very small) - The effect of electrons can be neglected (because the mass of electron is very small) 2. Rutherford Model (a) Coulomb scattering formula 1 a θ ZZ e02 Coulomb Scattering Formula: b = cot with a = 1 scattering factor 2 2 E dv ZZ e02 m = 1 ^r dt r2 dφ mr2 = L = mvb dt The trajectory is then ZZ e02 dv = 1 ^rdφ mvb Integrate both sides Z f θ dv = vf − vi = 2v sin i 2 Z f Z π−θ θ θ θ ^rdφ = (i cos φ + j sin φ) dφ = i sin θ+j (1 + cos θ) = 2 cos i sin + j cos i 0 2 2 2 θ θ Note the direction of vf − vi is (π − θ) =2, i.e, i sin 2 + j cos 2 , the R f same with i ^rdφ. Thus we have θ θ ZZ e02 2v sin = 2 cos · 1 2 2 mvb ZZ e02 θ a θ b = 1 cot = cot mv2 2 2 2 with Coulomb scattering factor ZZ e02 a = 1 : E Clearly, θ ! 0 for b ! 1, and θ = π for b = 0. 2 (b) Rutherform Scattering Formula i. Question: Given N incident particles, how many scattering in the solid angle dΩ = 2π sin θdθ per unit area, per target particle? ii. This gives the differential cross section (DCS) 1 dN 0 σ (θ) = c Nnd dΩ where n is the number density and d the sample depth (ndA gives the number of target particles with A the area) iii. Particles in b ! b + db, scattering between angle θ ! θ + dθ. Assume the target particles scatter the incident particle inde- pendently. dN 0 2πbjdbj a2 cot (θ=2) = · (ndA) = 2π dθ · nd N A 8 sin2 (θ=2) Thus 1 dN 0 a2 cot (θ=2) 1 a2 σc (θ) = = 2π dθ = Nnd dΩ 8 sin2 (θ=2) 2π sin θdθ 16 sin4 (θ=2) iv. Written explicitly 02 2 ZZ1e 1 σc (θ) = · 4E sin4 (θ=2) v. This DCS can be verified experimentally. It has the dimension of area (Note e02 = 1:44 fm Mev). If this formula holds, then dN 0 sin4 (θ=2), dN 0 · E2, and dN 0=Z2 should be constants given other paramaters fixed. These were verified in 1913. (c) Rutherford Planet model fails for the following reasons. A charged particle like the electron circulating in orbit would be expected to radiate light, with the same frequency as the orbital motion. The frequencies of these orbital motions could be anything. Worse, as the electron lost energy to radiation it would spiral down into the atomic nucleus. Atoms cannot remain stable. 2 Planck Quantum Concept 1. Rayleigh-Jeans formula 3 (a) Question: The energy density E (ν; T ) in the frequency range ν ! ν + dν at temperature T ? (b) According to classical physics, the radiation field is the sum of Fourier normal modes proportional to exp (iq · r) with q = 2πn=L where n = (n1; n2; n3) and L is the length of a cubic box. n1;2;3 are intergers. 3 (c) # of modes within dq = dqxdqydqz is dn = (L=2π) dq. # of modes V 2 V 2 with q ! q + dq is then 8π3 4πq dq = 2π2 q dq. Since q = 2π/λ, ν = c/λ, thus q = 2πν=c. (d) The number of normal modes in the range ν ! ν + dν is then V V 2πν 2 2π 8πν2V N (ν) dν = 2 × q2dq = 2 × dν = dν 2π2 2π2 c c c3 where the first factor 2 stands for two polorization directions. (e) At temperture T , according to classical statistical mechanics, the average energy of the normal modes (viewed as harmonic oscillators) is given by R 1 0 exp (−βE) EdE −1 E = R 1 = β = kBT 0 exp (−βE) dE (f) The energy density is then N (v) dν · E 8πν2 E (ν; T ) dν = = · k T dν L3 c3 B which is Rayleigh-Jeans (RJ) formula (g) The R-J formula fails in two aspects: Deviates badly for large ν, called ultraviolet catastrophe. Also the integration R E (ν; T ) dν di- verges if it holds for any ν. 4 2. Planck Formula (a) 1900, proceedings of German Physical Society, Planck noted the data can be fitted very well by (a guess work) 8πv2 hν E (ν; T ) dν = 3 · dν c exp (hν=kBT ) − 1 with fitting constants −23 −34 −27 kB ' 1:4 × 10 J/K and h ' 6:6 × 10 J·s = 6:63 × 10 erg s (b) Planck’s quantization assumption: the radiation was the same as if it were in equilibrium with a large number of charged oscillators with different frequencies, the energy of any oscillator of frequency ν being an integer multiple of hν. Later, Lorenz gave a derivation in 1910, simply assuming discrete energy: 1 1 ! P exp (−βnhν) · (nhν) @ X E = 0 = − ln exp (−βnhν) P1 exp (−βnhν) @β 0 0 @ 1 hν exp (−βhν) hν = − ln = = @β 1 − exp (−βhν) 1 − exp (−βhν) exp (βhν) − 1 (c) Hence 8πν2 8πv2 hν E (ν; T ) dν = 3 · Edν = 3 · dν c c exp (hν=kBT ) − 1 Z 1 5 4 4 8π kB E (ν; T ) dv = αBT with αB = 3 3 0 15h c (d) Application of Planck’s work: Calculate NA = R=kB, the size of atom, the electron charge ... 3 Einstein Photon Assumption 1. Photo-Electronic Effect Some basic facts: 5 (a) About 10−9s, electronic current becomes saturated immediately (b) Threshold frequency, below which no photo-electron comes out Classically, the energy of lights are proportional to the intensity, which cannot understand the above facts. Einstein assumes light quanta: E = hν and the kinetic energy of photo-electron is given by K = hν − W 2. Compton Scattering (1922-23) Accordingly, the momentum of the photon is given by p = E=c = h/λ m2c4 = E2 − p2c2 = 0 For X-ray scattering by electron, assuming the light scattering backwards with frequency ν0, then the electron forwards with momentum hν=c + hν0=c, and energy conservation gives s hν + hν0 2 hν + m c2 = hν0 + m2c4 + c2 e e c Thus 2 0 2 2 0 0 mec ν 2h (ν − ν ) mec = 4h νν ! ν = 2 2hν + mec Written in the wave-length 2h λ0 = c/ν0 = λ + = λ + 2 × 2:425 × 10−10 cm mec −10 h=mec = 2:425 × 10 cm is known as Comptom wavelength of electron. If scattering at an angle θ, the factor 2 before h=mec is replaced by 1 − cos θ. Later after verification of this relation, chemist Lewis gave the name photon. 4 Bohr Model 1. Bohr was a visitor to Rutherford’s group in 1913. He proposed in the first place that the energies of atoms are quantized, in the sense that atoms exist in only a discrete set of states. • Discrete stable orbital: v2 Ze02 m = r r2 6 • Angular momentum quantization: (~ is a new parameter) n mvr = n (n = 1; 2; 3 ::: ) ! r = ~ ~ mv • Substitute into the first Eq., we get Ze02 vn = n~ 2 2 2 n~ n ~ 2 a1 ~ rn = = 02 = n with a1 = 02 mvn mZe Z me • Energy 02 2 04 1 2 Ze 1 Z mee En = mvn − = − 2 2 2 rn 2 n ~ • Transition frequency E − E 1 1 Z2m e04 ν = m n = − e h n2 m2 2~2h For transition n ! n + 1 and n ! 1, 2 Z2m e02 Z2m e04 ν ' · e = e n3 2~2h n3~2h • Correspondence principle: If an electron in an atom is moving on an orbit with period T, classically the electromagnetic radiation will repeat itself every orbital period. If the coupling to the electromag- netic field is weak, so that the orbit doesn’t decay very much in one cycle, the radiation will be emitted in a pattern which repeats every period, so that the Fourier transform will have frequencies which are only multiples of 1/T. This is the classical radiation law: the fre- quencies emitted are integer multiples of 1/T. For n ! 1, ν should be the frequency of the orbital 02 2 04 vn Ze =n~ Z mee f = = 2 2 02 = 3 2 2πrn 2πn ~ =meZe n ~ (2π~) Letting f = ν, we thus obtain as expected ~ = h=2π • Results Ze02 Z e02 Z e02 1 vn = = c = αc with α = ' n~ n ~c n ~c 137 n2 2 n2 2 r = ~ = a with a = ~ = 0:529 × 10−8cm n mZe02 Z 1 1 me02 2 04 2 1 Z mee 1 Z En = − = − × 27:2 eV 2 n2 ~2 2 n2 7 2. Atomic spectrum of Hydrogen 2 04 1 ν 1 1 Z mee 1 1 ν~ = = = − = − RH λ c n2 m2 2~2hc n2 m2 2 04 1 Z mee −E1 RH = = hc 2~2 hc For Hydrogen, E1 = −13:6 eV, 13:6 × 1:6 × 10−19(J) R = ' 109677:58 cm−1 H 6:63 × 10−34(J s) × 3 × 1010(cm/s) Actually, for hydrogen atom, me should be replaced by reduced mass memH 1837 µH = ' me + mH 1838 If we consider Deuterium, then the reduced mass is memD 3674 µD = ' me + mD 3675 Thus we have slight isotrope effect for Hydrogen spectrum.