11 Quantum theory: introduction and principles

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1 Part 2: Structure

3 Quantum theory: introduction and principles Solutions to exercises E.b E.2b E.3b E.4b E.5b E.6b Discussion questions A successful theory of black-body radiation must be able to explain the energy density distribution of the radiation as a function of wavelength, in particular, the observed drop to zero as λ. Classical theory predicts the opposite. However, if we assume, as did Planck, that the energy of the oscillators that constitute electromagnetic radiation are quantized according to the relation E nhν n/λ, we see that at short wavelengths the energy of the oscillators is very large. This energy is too large for the walls to supply it, so the short-wavelength oscillators remain unexcited. The effect of quantization is to reduce the contribution to the total energy emitted by the black-body from the high-energy short-wavelength oscillators, for they cannot be sufficiently excited with the energy available. In quantum mechanics all dynamical properties of a physical system have associated with them a corresponding operator. The system itself is described by a wavefunction. The observable properties of the system can be obtained in one of two ways from the wavefunction depending upon whether or not the wavefunction is an eigenfunction of the operator. When the function representing the state of the system is an eigenfunction of the operator, we solve the eigenvalue equation eqn.3 ω in order to obtain the observable values, ω, of the dynamical properties. When the function is not an eigenfunction of, we can only find the average or expectation value of dynamical properties by performing the integration shown in eqn.39 dτ. No answer. Numerical exercises The power is equal to the excitance M times the emitting area P MA σt 4 2rl Wm 2 K 4 33 K m 5. 2 m W Comment. This could be a 25 W incandescent light bulb. Wien s displacement law is Tλ max c 2 /5 so λ max c 2 5T.44 2 mk.5 6 m.5 µm 525 K The de Broglie relation is λ h p h mv so v h mλ Js kg 3. 2 m v.3 5 ms

4 72 INSTRUCTOR S MANUAL E.7b The de Broglie relation is λ h p h mv so v h mλ Js 9. 3 kg.45 9 m v.6 6 ms E.8b The momentum of a photon is p h λ Js 35 9 m kg m s The momentum of a particle is p mv so v p m kg m s kg mol / mol v.565 m s E.9b The energy of the photon is equal to the ionization energy plus the kinetic energy of the ejected electron E photon E ionize + E electron so λ E ionize + 2 mv2 and λ E ionize Js ms mv J kg ms m 38.4nm E.b The uncertainty principle is p x 2 h so the minimum uncertainty in position is x h 2 p h 2m v m Js kg ms E.b E hν λ ; Eper mole N AE N A λ Js ms Jm N A mol Jm.96 J m mol Thus, E Jm ; Eper mole λ We can therefore draw up the following table.96 J m mol λ

5 QUANTUM THEORY: INTRODUCTION AND PRINCIPLES 73 λ E/J E/kJ mol a 2 nm b 5 pm c. cm E.2b Assuming that the 4 He atom is free and stationary, if a photon is absorbed, the atom acquires its momentum p, achieving a speed v such that p mv. v p m m kg kg a b c p h λ p Js 2 9 m kg m s v p m kg m s kg.499 m s p Js 5 2 m kg m s v p m kg m s kg 665 m s p Js. 2 m kg m s v p m kg m s kg ms E.3b Each emitted photon increases the momentum of the rocket by h/λ. The final momentum of the Nh rocket will be Nh/λ, where N is the number of photons emitted, so the final speed will be. λm rocket The rate of photon emission is the power rate of energy emission divided by the energy per photon /λ, so N tpλ tpλ h and v tp λm rocket cm rocket v.yr 365 day yr 24 h day 36 s h.5 3 W ms.kg 58 m s E.4b Rate of photon emission is rate of energy emission power divided by energy per photon /λ a b rate rate Pλ. W 7 9 m Js ms s E.5b Wien s displacement law is.w 7 9 m Js ms s Tλ max c 2 /5 so T c 2 5λ max.44 2 mk 56 9 m 8 K

6 74 INSTRUCTOR S MANUAL E.6b Conservation of energy requires E photon + E K hν /λ so E K /λ /2 and E K 2 m ev 2 2EK so v m e a E K Js ms ev.6 9 JeV m But this expression is negative, which is unphysical. There is no kinetic energy or velocity because the photon does not have enough energy to dislodge the electron. b E K Js ms 95 9 m J /2 J and v ms kg 2.9 ev.6 9 JeV E.7b E hν h/τ, so a b c E Js/2.5 5 s J 6 kj mol E Js/2.2 5 s 3. 9 J 8 kj mol E Js/. 3 s J 4. kj mol E.8b The de Broglie wavelength is λ h p The momentum is related to the kinetic energy by E K p2 2m so p 2mE K /2 The kinetic energy of an electron accelerated through V is ev.6 9 J, so λ h 2mE K /2

7 QUANTUM THEORY: INTRODUCTION AND PRINCIPLES 75 a λ b λ c λ Js kg ev.6 9 JeV /2.23 m Js kg. 3 ev.6 9 JeV /2 3.9 m Js kg 3 ev.6 9 JeV / m E.9b The minimum uncertainty in position is pm. Therefore, since x p 2 h p h 2 x Js 2 2 m kg m s v p m kg m s ms kg E.2b Conservation of energy requires E photon E binding + 2 m ev 2 hν /λ so E binding /λ 2 m ev 2 and E binding Js ms 2 2 m kg ms J Comment. This calculation uses the non-relativistic kinetic energy, which is only about 3 per cent less than the accurate relativistic value of.52 5 J. In this exercise, however, E binding is a small difference of two larger numbers, so a small error in the kinetic energy results in a larger error in E binding : the accurate value is E binding.26 6 J. Solutions to problems Solutions to numerical problems P.3 θ E hν k, [θ Js s E] JK K In terms of θ E the Einstein equation [.9] for the heat capacity of solids is 2 θe C V 3R T e θ E/2T e θ E/T 2, classical value 3R It reverts to the classical value when T θ E or when hν as demonstrated in the text kt Section.. The criterion for classical behaviour is therefore that T θ E. θ E hν k JHz ν JK ν/hzk

8 76 INSTRUCTOR S MANUAL a For ν Hz, θ E K 223 K b For ν Hz, θ E K 343 K Hence a b C V 3R C V 3R 223 K 298 K 343 K 298 K 2 e 223 / e 223/ e 343/ e 343/ Comment. For many metals the classical value is approached at room temperature; consequently, the failure of classical theory became apparent only after methods for achieving temperatures well below 25 C were developed in the latter part of the nineteenth century. P.5 The hydrogen atom wavefunctions are obtained from the solution of the Schrödinger equation in Chapter 3. Here we need only the wavefunction which is provided. It is the square of the wavefunction that is related to the probability Section.4. ψ 2 a 3 e 2r/a, δτ 4 3 r3, r.pm If we assume that the volume δτ is so small that ψ does not vary within it, the probability is given by ψ 2 δτ 4r3 3a 3 e 2r/a e 2r/a 53 a r : ψ 2 δτ b r a : ψ 2 δτ 4. 3 e Question. If there is a nonzero probability that the electron can be found at r how does it avoid destruction at the nucleus? Hint. See Chapter 3 for part of the solution to this difficult question. P.7 According to the uncertainty principle, p q 2 h, where q and p are root-mean-square deviations: q x 2 x 2 /2 and p p 2 p 2 /2. To verify whether the relationship holds for the particle in a state whose wavefunction is 2a/ /4 e ax2, We need the quantum-mechanical averages x, x 2, p, and p 2. x x 2 dτ 2a /4 e ax2 x 2a /4 e ax2 dx,

9 QUANTUM THEORY: INTRODUCTION AND PRINCIPLES 77 2a /2 x xe 2ax2 dx ; x 2 x 2 2a 2a /4 e ax2 x 2 2a /4 e ax2 dx /2 /2 22a 3/2 4a ; so q 2a /2. h p d dx and p 2 i dx We need to evaluate the derivatives: d 2a /4 dx 2axe ax2 2a /2 x 2 e 2ax2 dx, h 2 d2 dx 2 dx. and d 2 dx 2 2a /4 [ 2ax 2 e ax2 + 2ae ax2 ] 2a /4 4a 2 x 2 2ae ax2. So p 2 h i p 2 2a p 2 2a h 2 p 2 2a h 2 /4 e ax2 h i 2a /2 xe 2ax2 dx ; 2a /4 2axe ax2 dx 2a /4 e ax2 h 2 2a /4 4a 2 x 2 2ae ax2 dx, 2a /2 2a /2 2a /2 2ax 2 e 2ax2 dx, /2 22a 3/2 2a /2 a h 2 ; and p a /2 h. Finally, q p 2a /2 a/2 h /2 h, which is the minimum product consistent with the uncertainty principle.

10 78 INSTRUCTOR S MANUAL Solutions to theoretical problems P.9 We look for the value of λ at which ρ is a maximum, using as appropriate the short-wavelength high-frequency approximation ρ 8 λ 5 dρ dλ 5 λ ρ + e /λkt [.5] e /λkt λ 2 kt e /λkt ρ at λ λ max Then, 5 + λkt e/λkt e /λkt Hence, 5 5e /λkt + λkt e/λkt If [short wavelengths, high frequencies], this expression simplifies. We neglect the initial 5, λkt cancel the two exponents, and obtain 5λkT for λ λ max and λkt or λ max T 2.88 mm K, in accord with observation. 5k Comment. Most experimental studies of black-body radiation have been done over a wavelength range of a factor of to of the wavelength of visible light and over a temperature range of 3 K to K. Question. Does the short-wavelength approximation apply over all of these ranges? Would it apply to the cosmic background radiation of the universe at 2.7 K where λ max.2cm? P. ρ 8 λ 5 e /λkt [.5] As λ increases, decreases, and at very long wavelength /λkt. Hence we can expand λkt the exponential in a power series. Let x /λkt, then e x + x + 2! x2 + 3! x3 + [ ] ρ 8 λ 5 + x + 2! x2 + 3! x3 + lim ρ 8 [ ] λ λ x λ 5 /λkt 8kT λ 4 This is the Rayleigh Jeans law [.3].

11 QUANTUM THEORY: INTRODUCTION AND PRINCIPLES 79 P.2 ρ 8 λ 5 e /λkt [.5] ρ λ 4 λ 6 e /λkt 8 λ 5 e /λkt 5 [ ρ λ 5λkT 8 λ 5 [ 5 λ + λ 2 kt ] e /λkt e /λkt λ 2 kt e /λkt 2 ] e /λkt e /λkt ρ λ when λ λ max and 5λ max kt [ ] e /λ maxkt 5λ max kt e /λ maxkt Let x λ max kt ; then 5 e x or x 5 x e x The solution of this equation is x Then h 4.965λ maxkt c However M σt k 4 5c 2 h 3 T 4 2 Substituting into 2 yields 2 5 k 4 M 5c ckt 835.9λ 3 max k 835.9λ3 max M 2 5 ct c 4.965λ max kt 3 T m W ms 2 K. m 2 k JK 3

12 8 INSTRUCTOR S MANUAL Substituting 3 into h m JK 2 K ms h Js Comment. These calculated values are very close to the currently accepted values for these constants. P.4 In each case form Nψ; integrate Nψ Nψ dτ set the integral equal to and solve for N. a ψ N 2 ra e r/2a ψ 2 N 2 2 r 2 e r/a a ψ 2 dτ N 2 4r 2 4r3 a N 2 hence N 32a 3 + r4 a 2 e r/a dr 4 2a 3 4 6a4 a + 24a5 a 2 /2 where we have used x n e ax dx n! [Problem.3] an+ 2 sin θ dθ dφ a 3 N 2 ; b ψ Nr sin θ cos φ e r/2a 2 ψ 2 dτ N 2 r 4 e r/a dr sin 2 θ sin θ dθ cos 2 φ dφ N 2 4!a 5 cos 2 θd cos θ N 2 4!a /2 32a 5 3 N 2 ; hence N 32a 5 where we have used cos n θ sin θ dθ cos n θ d cos θ x n dx and the relations at the end of the solution to Problem.8. [See Student s solutions manual.]

13 QUANTUM THEORY: INTRODUCTION AND PRINCIPLES 8 P.6 Operate on each function with i; if the function is regenerated multiplied by a constant, it is an eigenfunction of i and the constant is the eigenvalue. a b f x 3 kx ix 3 kx x 3 + kx f Therefore, f is an eigenfunction with eigenvalue, f cos kx i cos kx cos kx cos kx f Therefore, f is an eigenfunction with eigenvalue, + c f x 2 + 3x ix 2 + 3x x 2 3x constant f Therefore, f is not an eigenfunction of i. P.9 The kinetic energy operator, ˆT, is obtained from the operator analogue of the classical equation E K p2 2m that is, Then ˆT ˆp2 2m ˆp x h i T N 2 d dx [.32]; hence ˆp2 d2 x h2 dx 2 and ˆT h2 2m ψ ˆp x 2 ψ ˆp 2 2m ψ dτ ψ dτ 2m ψ ψ dτ d 2 dx 2 [ ] N 2 ψ ψ dτ h 2 2m h 2 2m ψ d 2 dx 2 e ikx cos χ + e ikx sin χdτ ψ ψ dτ ψ k 2 e ikx cos χ + e ikx sin χdτ ψ ψ dτ h2 k 2 ψ ψ dτ 2m ψ ψ dτ h2 k 2 2m P.2 p x h d i dx [.32] p x N 2 ψ ˆp x ψ dx; N 2 ψ ψ dτ ψ ˆp x ψ dx ψ ψ dx h i ψ dψ dx dx ψ ψ dx a ψ e ikx, Hence, dψ dx ikψ p x h i ik ψ ψ dx ψ ψ dx k h

14 82 INSTRUCTOR S MANUAL b dψ ψ cos kx, ksin kx dx ψ dψ dx k cos kx sin kx dx dx Therefore, p x dψ c ψ e αx2, dx 2αxe αx2 ψ dψ dx 2α xe 2αx2 dx dx [by symmetry, since x is an odd function] P.23 No solution. Therefore, p x Solution to applications P.27 a Consider any infinitesimal volume element dx dy dz within the hemisphere Figure. that has a radius equal to the distance traveled by light in the time dt c dt. The objective is to find the total radiation flux perpendicular to the hemisphere face at its center. Imagine an infinitesimal area A at that point. Let r be the distance from dx dy dz to A and imagine the infinitesimal area A perpendicular to r. E is the total isotropic energy density in dx dy dz. E dx dy dz is the energy emitted in dt. A /4r 2 is the fraction of this radiation that passes through A. The radiation flux that originates from dx dy dz and passes through A in dt is given by: J A A 4r 2 E dx dy dz E dx dy dz 4r 2 dt A dt The contribution of J A to the radiation flux through A, J A, is given by the expression J A A cos θ/ A J A cos θ. The integration of this expression over the whole hemisphere gives an c dt A θ A c dt J A θ J A dx dy dz Figure.

15 QUANTUM THEORY: INTRODUCTION AND PRINCIPLES 83 expression for J A. Spherical coordinates facilitate to integration: dx dy dz r 2 sin θ dθ dφ dr r 2 dcos θdφ dr where θ 2 and θ /2. { } E dx dy dz J A cosθ 4r 2 dt hemisphere hemisphere { cosθ E 4 r 2 dt } { r 2 dcos θ dφ dr} cos/2 2 cdt E cosθ dcos θ dφ dr 4 dt cos E w dw 2 c dt 4 dt J A ce 2 {Subscript A has been a bookkeeping device. It may be dropped.} 4 2 J ce 4 or dj c 4 de 8 dλ de λ 5 e /λrt [eqn.5] By eqn 6. ν /λ. Taking differentials to be positive, d ν dλ/λ 2 or dλ λ 2 d ν d ν/ ν 2. The substitution of ν for λ gives: de 8 ν3 e ν/kt d ν Thus, dj f νd ν where f ν 22 ν 3 e ν/kt The value of the Stefan Boltzmann constant σ is defined by the low n dj ν σt 4. n is called the total exitance. Let x ν/kt or ν ktx/, substitute the above equation for dj ν into the Stefan Boltzmann low, and integrate. 2 2 ν 3 d ν n e ν/kt 2k4 T 4 x 3 dx h 3 c 2 e x o 2k 4 T k 4 h 3 c h 3 c 2 T 4 Thus, σ 25 k 4 5h 3 c Wm 2 K 4 The function f ν gives radiation density in units that are compatible with those often used in discussions of infrared radiation which lies between about 33 cm and 2 8 cm Fig..2.

16 84 INSTRUCTOR S MANUAL 6 Blackbody radiation density at K. 4.5 Fν / Jm Wavenumber, ν/cm Figure.2 By graphing f νat the observed average temperature of the Earth s surface K we easily see that the Earth s black-body emissions are in the infrared with a maximum at about 6 cm. b Let R represent the radius of the Earth. Assuming an average balance between the Earth s absorption of solar radiation and Earth s emission of black-body radiation into space gives: Solar energy absorbed black-body energy lost R 2 albedosolar energy flux 4R 2 σ T 4 Solving for T gives: [ ] albedosolar energy flux /4 T 4σ [ ] W cm 2 / Wcm 2 K K This is an estimate of what the Earth s temperature would be in the absence of the greenhouse effect.

Physical Chemistry II Exam 2 Solutions

Chemistry 362 Spring 208 Dr Jean M Standard March 9, 208 Name KEY Physical Chemistry II Exam 2 Solutions ) (4 points) The harmonic vibrational frequency (in wavenumbers) of LiH is 4057 cm Based upon this