sol manual atomic physics PDF

Title sol manual atomic physics
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Course atomic physics
Institution Seoul National University of Science and Technology
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foot. atomic physics solution manual....
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Solutions to the Exercies of C. J. Foot’s Atomic Physics Chenchao Zhao Department of Physics, Beijing Normal University, Beijing, China (Dated: June 28, 2011)

1

Early atomic physics Key formulas:  1 1 − n2 n′2 mH RH = R me + mH ~2 = 0.529 × 10−10 m a0 = 2 (Ze /4πǫ 0 )me Ze2 /4πǫ0 1 ∝ Z2 E =− 2a0 n2 α2 ∆E = 2 E n e2 /4πǫ 0 α= = 1/137 ~c f = R∞ ((Z − σK )2 − (Z − σL )2 ) c ~ω 3 1 ρ(ω) = 2 3 π c exp ~ω/kT − 1   ~ω N2 N1 exp − = g1 g2 kT e2 x20 ω 4 P = 12πǫ0 c3 eB ΩL = 2me 6πǫ0 me c3 τ= e2 ω 2 1 =R λ



1

(1.1) (1.2) (1.3) (1.4) (1.5) (1.6) (1.7) (1.8) (1.9) (1.10) (1.11) (1.12)

1.1

Isotope shift

To find wavelengths of Balmer-α transitions, we set n = 2 and n′ = 3 as 36 in Eq (??) Then λ = 5R , hence 36 (1/RH − 1/RD ) 5   1 36me 1 = − mD 5R∞ mH 18me ≈ = 0.18 nm 5mH R∞

λH − λD =

where mD ≈ 2mH .

1.2

The energy levels of one-electron atoms

Since (1.4), and let m, n be the quantum numbers of He+ and H, neglecting the isotope shifts, the energy levels in agreement are those with 1/n2 = 4/m2 , namely m = 2n. Those wavelengths should have the ratio λH RH e = λH e RH mH e (mH + me ) = mH (mH e + me ) 4mH (mH + me ) ≈ mH (4mH + me ) ≈ 1.00041 which is in accord with the data set, 1.00041.

1.3

Relativistic effects

With n = 4, Eq (1.5) gives λ 4 E = 2 = 75076 = α ∆E ∆λ for the fact that λ = ch/E and dλ = −ch dE/E 2

= −(ch/E) (dE/E ) = −λ (dE/E )

This corresponds to a grating of ∼ 105 grooves.

2

(1.13)

1.4

X-rays

Eq (1.7) reduces to f = 3cR∞ Z 2 /4 when Z ≫ σ. Therefore,

1.5

√ f ∝Z

X-rays

Since E = hf = 3/4(hcR∞ )Z 2 = 13.6 eV × 0.75Z 2 , then it predicts absorptions at around 6.4 keV and 6.9 keV.

1.6

X-ray experiments

See http://www.physics.ox.ac.uk/ history.asp?page=Exhibit10

1.7

Fine structure in X-ray transitions

Energy of the electron in the L-shell should be E = (82/2)2 × 13.6 = 22.9 keV and Eq (1.5) gives ∆E =

α2 E = E/75076 = 0.3 eV n2

But Kα transition means an energy of Eα = 10.2(Z − 1)2 eV = 66.9 keV then

1.8

∆E = 4.5 × 10−4 % Eα

Radiative life time

Eq (1.10) provides the power of dipole radiation which is rate of change of energy, for the circular motion, the power doubles since a circular motion can be decomposed into two linear oscillations, τ = E/P = ~ω

12πǫ0 c3 ∝ 1/(er)2 ω 3 2e2 r 2 ω 4

For photons of wavelength 650 nm, ω = 2.89 × 1015 rad/s; let r = a0 , the life time will be 2.7 × 10−7 s.

3

Table 1: Frequency shifts B [T] 3 × 10−5 1 ΩL [MHz] 2.6 8.8 × 104 14 −8 ΩL /(10 Hz) 2.6 × 10 8.8 × 10−4

1.9

Black-body radiation

Setting N2 = 0.1N, N1 = 0.9N, g1 = 1, g2 = 3, Eq (1.9) gives exp(~ω/kT ) = 27 provided the wavelength λ = 600 nm, ω = 2πc/λ = 3.13 × 1015 rad/s. Therefore, T = 7.23 × 103 K and Eq (1.8) gives the density ρ = 4.70 × 10−16 J s/m3

1.10

Zeeman effect

The Larmor frequency is given by Eq (1.11), and the Earth magnetic field is about 3 × 10−5 T, then the frequency shifts are listed in Table 1.

1.11

Relative intensities in the Zeeman effect

One circular motion can be decomposed into to two orthogonal sinusoidal motions. Let I denote the intensity of three eigen-oscillations of the electron. Then, we have • Along the magnetic field, only circularly polarized lights can be observed and the intensities are Iσ+ = Iσ− = I ; • Perpendicular to the magnetic field, motion along z direction and projected horizontal motions can all be observed, and • 2Iσ+ = 2Iσ− = Iπ = I since the projected horizontal motions are only half of the circular motions. Therefore, (a) Total intensity perpendicular to the field is 2I ; (b) Ratio of intensities received along to perpendicular to the field is 2I/2I = 1.

4

1.12

Bohr theory and the correspondence principle

Energy of hydrogen atoms takes the form 1 E = K + V = −K = − me v 2 2 But me Then E =−

v2 e2 /4πǫ 0 = r2 r

1 e2 /4πǫ 0 , 2 r

dE =

e2 /4πǫ0 dr 2r 2

Hence

e2 /4πǫ0 ∆r 2r 2 ~ The angular frequency is also given by ω = ∆E/~ =

ω2 =

v 2 e2 /4πǫ 0 = r2 me r 3

Equating the two expressions of ω, we have s r~2 √ ∆r = 2 = 2 a0 r 2 me e /4πǫ 0 which is equivalent to

∆r = 2(a0 r)1/2 . ∆n Approximate the equation above by the corresponding differential equa√ tion, namely r ′ = 2 a0 r, and the solution turns out to be r = a0 n2 .

1.13

(1.14)

Rydberg atoms

Eq (1.4), then dE e2 /4πǫ0 d (n−2 ) e2 /4πǫ0 = Ry/n3 = =− a0 n3 2a0 dn dn and for n = 50, ∆E = 1.1 × 10−4 eV. The radius of such atoms is around 2500a0 or 132 nm according to Eq (1.14).

5

2

The hydrogen atom

Key formulas Ylm (θ, φ) = (−1)m Z

π

−π Z π

Z−π π

s

(2l + 1) (l − m)! imφ m e Pl (cos θ) 4π (l + m)!

dθ eimθ e−inθ = 2πδmn   cos mθ cos nθ = πδmn dθ sin mθ sin nθ

(2.1) (2.2) (2.3)

dθ sin mθ cos nθ = 0

(2.4)

−π

Z

dΩ =

Z





0

Z

π

sin θdθ =

0

Z





Z

1

d(cos θ)

β (j (j + 1) − l(l + 1) − s(s + 1)) 2 ~2 1 e2 β= 2 2 2me c 4πǫ0 (na0 )3 l(1 + 12 )(l + 1)     1 ∂ 1 ∂2 ∂ 2 L =− + sin θ sin θ ∂θ ∂θ sin2 θ ∂ 2 φ Es-o = β hS · Li =

2.1

(2.5)

−1

0

(2.6) (2.7) (2.8)

Angular-momentum eigenfunctions

From the table of spherical harmonics and Eq (2.2), (2.5), we have • h l1 m| l2 ni ≡ 0 if m 6= n, therefore h 11| 00i = 0 And also h 10| 00i = (constant) ×

Z

1

cos θ d(cos θ) = 0 −1

• For l = 1, 2, we only need to show that h 10| 20i = h 11| 21i = 0. Through inspections, the integrands as functions of cos θ are both odd, hence the integrals vanish.

2.2

Angular-momentum eigenfunctions

• According to Eq (2.1), Yl,l−1 = (−1)l−1

s

2l + 1 i(l−1)φ l−1 e Pl (cos θ ) 4π (2l − 1) 6

• It is convenient to write, ˆ− )† |l − 1, l − 1i h l, l − 1| l − 1, l − 1i = hl, l| ( L = hl, l| Lˆ+ |l − 1, l − 1i =0

2.3

Radial wavefunctions

With n = 2, l = 1 the integral reads,   Z ∞ 1 1 2 = R (r )r 2 dr 3 2,1 r r3 0   2  Z ∞ dr Z 5 2 √ r 2 e−r/a0 = 2a0 r 3 0 = 1/(24a30 ) Invoking the rhs formula, namely 

1 r3



1 = 1 l(l + 2 )(l + 1)



Z na0

2

(2.9)

yields the same result 1/(24a30 ).

2.4

Hydrogen

The probability is given by Z Z rb 2 2 r dr |ψ(r )| = 0

Z

rb

(r/a0 )2 d (r/a0 ) −2r/a0 e π

0 r ǫ:= ab

y2 dy −2y e π 0 ≈ ǫ3 e−2ǫ /π ∼ (rb /a0 )3 =

0

The electronic charge density of this region is ρe ≈ e|ψ(rb /2)|2 =

2.5

e(1 − rb /a0 ) a03π

Hydrogen: isotope shift, fine structure and Lamb shift

The mass ratio of electron to proton is ∼ 5 × 10−4 and isotope shift is of the same order, namely, if λ = 600 nm, order of isotope shift will be δ ν¯isotope ∼ (5 × 10−4 ) · (5 × 105 GHz) = 250 GHz 7

Relativistic effect is of the order of α2 ≈ 5 × 10−5 , then the wave number difference reads δ ν¯fs ∼ (5 × 10−5 ) · (5 × 105 GHz) = 25 GHz Lamb shift is 1/10 of the fine structure shift, that is δ ν¯Lamb ∼ 2.5 GHz A Fabry-Perot ´etalon of finesse F = 100, width d = 1 cm is supposed to resolve 1 δ ν¯ = = 0.5 × 3 × 108 ∼ 0.1 GHz (2.10) n(2d)F but Doppler effect attenuates the resolution to ∼ 0.7 GHz ∼ 1 GHz. Therefore, isotope shift, fine structure can be resolved but Lamb shift approaches the limit of the apparatus and hence cannot be accurately observed.

2.6

Transitions

From Eq (1.12), we have Ultraviolet

100 nm

Infrared 1000 nm

2.7

0.45 ns 450 ns

Selection rules

Following similar arguments as in Problem 2.1 and notice there is an additional cos θ in the integrand.

2.8

Spin-orbit interaction

Calculations based on Eq (2.6) give Ej =

β l 2

β Ej ′ = − (l + 1) 2

and the mean of the two is ¯ = (2j + 1)Ej + (2j ′ + 1)Ej ′ = β[(l + 1)l − l(l + 1)] = 0 E

2.9

Selection rule for the magnetic quantum number

The integral is readily obtained in text of section 2.2.1, and the result follows the same arguments of Problem 2.1.

8

2.10

Transitions

(a) The wavefunction takes the form Ψ(t, r, θ) = Aψ1 eiE1 t/~ + Bψ2 eiE2 t/~ with A ≫ B , ψ1 = R1,0 Y00, ψ2 = R2,1 Y1,0 and |Ψ|2 ≈ A2 |ψ1 |2 + 2|AB||ψ1∗ ψ2 | cos(ω12 t) The second term can be written as f (r)r cos θ cos(ω12 t) = f (r )r · zˆ cos(ω12 t) = f (r) z cos(ω12t) The sketch of the orbital is as follows (Figure 1)

time

Figure 1: Contour of electron density, or the orbital of wavefunction Ψ(t, r, θ) = Aψ1 eiE1 t/~ + Bψ2 eiE2 t/~ with A ≫ B, ψ1 = R1,0 Y00, ψ2 = R2,1 Y1,0 during one period of oscillation. ˆ |Ψi = ±1 |Ψi, namely the state exhibits parity, then (b) If Π ˆ |Ψi = − hr i hri = hΨ| Πˆ† rΠ hence hri = 0. But the Hamiltonian of hydrogen atom commutes with ˆ then the eigenstates are of specific parities. The only possibility for a Π, none-vanishing hri is to require ψ1 and ψ2 to possess opposite parities. (c) Now set a0 = 1, the radial integral yields  5 Z ∞ dr 4 − 3 r 1 2 2 √ √ r e = 6 3 6 0 The angular integral is exactly 1. The total electric dipole moment is  5 2 1 ea0 cos ωt zˆ −eD = − √ 6 3 where a0 is put back through dimension analysis. 9

(d) The density distribution should be more or less similar to Figure 1 for a constant φ but it becomes apparent if one writes c = ωt − φ or φ = ωt − c that a flock of charge is circulating about the z-axis. (e) The case of (a) is akin to the vertical motion of the electron while (d) corresponds to right-handed circular motion. The motions are characterized by the ml but the role of ground state 1s is crucial as pointed out in (b). The 1s state is an exponential decay which binds the electron to the region around nucleus and hence it is the quantum analogue of the classical restoring force −ω 2 r.

2.11

Angular eigenfunctions: Yll

(a) Raising operator is given as L+ = eiφ (∂θ + i cot θ∂φ )

(2.11)

Then we write (∂θ + i cot θ∂φ ) (Θeimφ ) = 0 it is equivalent to

Θ′ cos θ =m sin θ Θ

(b) The solution is Θ = sinm θ. Applying L2 (see (2.8)) yields L2 (Θeimφ ) = m(m + 1)Θeimφ

2.12

Parity and selection rules

If l1 − l2 is even, then Iang = (−1)l1 −l2 +1Iang = −Iang which implies the integral vanishes. To have non-vanishing angular integrals, l1 − l2 must be odd. But the parity of spherical harmonics is ΠYlm = (−1)l Ylm

(2.12)

Therefore, the initial and final states must exhibit different parities.

2.13

Selection rules in hydrogen

The wavelengths corresponds to energy 0.306, 0.663, 1.890 eV and they are transitions from 5 to 4, 4 to 3 and 3 to 2 respectively. 10

3 3.1

Helium Estimate the binding energy of helium

(a) The total Hamiltonian is the sum of two individual and one interaction Hamiltonians, ˆ =H ˆ1 + H ˆ 2 + Hˆint H

(3.1)

2

2

ˆ 2 − Z e˜ ˆi = − ~ ∇ H 2m i ri 2 e˜ ˆ int = H |r 1 − r 2 |

(3.2) (3.3)

where e˜2 = e2 /4πǫ0 . (b) The energy E(r) =

Z e˜2 ~2 − 2mr 2 r

assumes minimum at rm =

~2 Zm˜ e2

and it is E(rm ) = −

Z 2 m˜ e4 2~2

(c) The repulsive energy, namely electron-electron interaction, is Eint =

Zm˜ e4 e˜2 e˜2 = ≈ ~2 rm r12

The ionization energy of one electron, according to the estimated energies, is Eion = 0 But experiment gives Eion = 24 eV, then the average distance between the two electrons should be greater than rm . (d) For Si12+ , Eion = (142 − 14 × 2) × 13.6 = 2285 eV, excluding the repulsion we have E = 2666 eV. Comparing with results of helium, repulsion gets irrelevant for larger Z .

3.2

Direct and exchange integrals for an arbitrary system

(a) The direct and exchange integrals are respectively, Z e˜2 J = d 3 x1 d 3 x2 |uα (r1 )|2 |uβ (r2 )|2 r12 Z 2 e ˜ uα (r1 )uβ (r2 ) K = d 3 x1 d 3 x2 u∗α(r1 )uβ∗(r2 ) r12 11

(3.4) (3.5)

Yet, |Ai = (2)−1/2 (|αβ i − |βαi), 1 (hαβ | − hβα|)Hˆ ′ (|αβ i − |βαi) 2 hαβH ′ αβ i − hαβH ′ βαi + α ↔ β = 2

hA| Hˆ ′ |Ai =

Note that

(3.6) (3.7)



   αβH ′ αβ = βαH ′ βα    ∗ αβH ′ βα = βαH ′ αβ

Therefore, it holds for real-valued u(r) that     hA|Hˆ′ |Ai = αβH ′ αβ − αβH ′ βα = J − K (b) The symmetric wavefunction is construct as

|Si = (2)−1/2 (|αβi + |βαi) The inner product reads h A| Si =

(1 − 1 + h αβ | βαi − c.c) = Im h αβ | βαi 2

Again, for real u(r), h A| Si are orthogonal. ˆ ′ is invariant under the interchange of particle labels, let Σ (c) Since H denote such an operation, then ˆ ′ |Si hA| Hˆ ′ |Si = hA| Σ†Hˆ ′ Σ |Si = − hA| H ˆ ′ |Si = 0. Hence hA| H

3.3

Exchange integrals for a delta-function interaction

(a) The Hamiltonian is simply the kinetic energy, H =−

~2 2 ∂ 2m x

For the fact that u0′′ = −(π/l)2 u0

u1′′ = −(2π/l)2 u1 ,

then the energies are E0 =

~2 π 2 2ml2

E1 = 12

2~2 π 2 ml2

(b) The direct integral is  2 Z 2 πx2 πx1 ) )aδx1 −x2 sin2 ( dx1 dx2 sin2 ( l l l  2 Z l 2 sin4 (πx/l)dx =a l 0 Z π 3a 4 sin4 x′ dx′ = =a πl 0 2l

J=

The exchange integral is the same thing, K = J. Then the energy shift will be 3a/l. There is only one state, the symmetric one. (c) All the eigenstates takes the form, r 2 nπx un (x) = sin l l and trivially Jmn = Kmn since the delta potential identifies the two coordinates. Hence, the antisymmetric state gives no rise to energy shift. Actually, the antisymmetric part does not exist at all when interaction is considered, because delta interaction rules out the possibility of ‘no-touch’, otherwise there is no interaction, and hence no energy shift! (d) See Figure 2.

Figure 2: The horizontal axes are the coordinates x1 and x2 , and the vertical axis marks the values of function ψ = u1 (x1 )u2 (x2 ).

(e) Possible total spins are 0 and 1. The symmetric spatial functions corresponds to 0-spin, and the antisymmetric spatial functions should have spin-1 or spin-0 but they do not exist.

13

(f ) The wavefunctions are independent of particle masses, therefore all the mathematics are invariant once the states are given. The energy levels are still E+ = E1 + E2 + 2J E− = E1 + E2 but the difference is that the levels are not related to the exchange symmetries of the particles.

3.4

A helium-like system with non-identical particles

The exoticon-exoticon system (identical fermions) is indifferent with the electron-electron system except for richer spin configurations for the two symmetries. Yet no restriction upon symmetries of particle exchange is placed on exoticon-electron system, therefore, the spatial orbitals can be freely occupied by the two fermions, and for the fact that spin is not included in the Hamiltonian, hence the energies levels are left unaltered.

3.5

Integrals in helium

Set a0 = 2Z, the integral in the curly brackets is −r2 e−2r2 /2 − e−2r2 /4 + 1/4 and J=

3.6

e˜2 5 Z = 34 eV 2a0 4

Calculation of integrals for 1s2p configuration 6

R 10 R20 R 21

5

4

Rnl

3

2

1 0

-10

2

4

r / a0

6

8

10

Figure 3: The plot of R10 , R20 and R21 with Z = 2. The integral J1s2p = −

e˜2 × 0.00208 = 0.0283 eV 2a0 14

3.7

Expansion of 1/r12

Expansion in terms of spherical harmonics is the following  ∞  k 1 X r1 k 4π X ∗ 1 = Y (θ1 , φ1 )Ykq (θ2 , φ2 ) r12 r2 2k + 1 q=−k kq r2 k=0

(3.8)

(a) Setting k = 0, 1, we have 1 1 {4π/4π ≈ r2 r12  r1 4π 3 + (cos θ1 cos θ2 + sin θ1 sin θ2 cos(φ1 + φ2 )) r2 3 4π   1 r1 = cos θ 1+ 12 r2 r2 (b) Mathematically, the expectation values of 1/r12 are always sandwiched by ‘bra-ket’ where the phase factors cancels; this is seen in the expression for K1s nl . Physically, the quantum number m is responsible for magnetic interactions but Coulomb repulsions have no interests in that. (c) Due to the orthogonality relations of spherical harmonics, the terms in 1/r12 are eliminated excluding the one with quantum numbers lm. For l = 1, this corresponds to the second term in the expansion as shown in part (a). (d) This follows the arguments in part (c), the given l samples out the order k in Eq (3.8).

4

The alkalis

4.1

Configuration of the electrons in francium Fr = [Rn]7s1 Rn = [Xe]5d10 5f 14 6s2 6p6 6d10 Xe = 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 4f 14 5s2 5p6

4.2

Finding the series limit for sodium

Energy levels of sodium obey the rule E =−

Ry 13.6 eV =− (n − δ )2 (n∗ )2 15

(4.1)

Then ∆E ∝ E 3/2

(∆E )2 ∝ E 3

The topmost level must have ∆E = 0, and that energy corresponds ionization energy which was found to be 5.1 eV. The effective principal number p n∗ = 13.6/5.1 = 1.63.

4.3

Quantum defects of sodium

From formula Eq (4.1), the quantum defects are 3s 1.37

4s 1.34

5s 1.33

6s 1.35

the average quantum defect is 1.35 ± .02. Assume that the quantum defect for 8s is still 1.35, the binding energy will be ENa 8s = −13.6/(8 − 1.35)2 = 0.31 eV while for hydrogen EH 8s = 0.21 eV.

4.4

Quantum defect

Quantum defect of Rb 5s is 3.19 with which that of 7s is approximated. Therefore the energy difference reads   1 1 − = 3.22 eV ∆E = 13.6 (5 − 3.19)2 (7 − 3.19)2 Then the wavelength of the two identical photons are λ=

4.5

hc = 771 nm ∆E/2

Application of quantum defects to helium and heliumlike ions

Throught a direct calculation, the wavelength from 1s3d to 1s2p is obtained as 625 nm, compared with 656 nm of hydrogen Balmer-α. Quantum defects can be evaluated by Eq (4.1), they are listed in Table 2. It is readily seen that δs > δp > δd . Table 2: Quantum defects of helium 1s2s 1s2p 1s3s 1s3p 1s3d 0.2356 0.0275 0.2265 0.0287 0.0026 To estimate binding energy of 1s4l states, assume the quantum defects for l = 1, 2, 3, 4 as 0.23, 0.28, 0.0026, 0.00. Then the binding energies read 0.96, 0.98, 0.85, 0.85 eV 16

The second ionization energy of Li+ is IE2 = −(E ′ − E1s4f ) = 72.24 +

Z 2 Ry = 75...


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