Lecture 1: Old Theory

September 3, 2019

1 Early Atom Models

1. Thompson Model

∆p F ∆t ZZ e2/4πε R2 · (2R/v) ZZ  e2  θ ∼ ∼ ∼ 1 0 ∼ 1 p p mv RE 4πε0 e2 e02 = = 1.44fm Mev 4πε0 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 ˆr dt r2 dφ mr2 = L = mvb dt The trajectory is then

ZZ e02 dv = 1 ˆrdφ mvb Integrate both sides

Z f θ  dv = vf − vi = 2v sin i 2

Z f Z π−θ θ   θ θ  ˆrdφ = (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 ˆrdφ. 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 → ∞, 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 2πb|db| 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 . The frequencies of these orbital 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 , the average energy of the normal modes (viewed as harmonic oscillators) is given by R ∞ 0 exp (−βE) EdE −1 E = R ∞ = 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: ∞ ∞ ! P exp (−βnhν) · (nhν) ∂ X E = 0 = − ln exp (−βnhν) P∞ 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 ∞ 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 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.

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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 • 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 → ∞, 2 Z2m e02 Z2m e04 ν ' · e = e n3 2~2h n3~2h • : 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 → ∞, 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 , 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. This is the evidence of the existence of Deuterium. For instance, for transition n = 0 2 → n = 3 (the Hα line), λ for H is 656.279 nm, we can obtain that for D is 656.10 nm (µD > µH , thus ν˜D > ν˜H and λD < λH ). 3. Failures of Bohr model: How do the transitions take place? Fails for other simple atoms even like He.

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