Question

KURSVLOEBEUR Question 4 [20 marks] (a) What particular aspect of the Rayleigh-Jeans Law, the classical theory of blackbody radiation, caused it to fail to explain the blackbody radiation spectrum? (b) What aspect of the Boltzmann probability distribution used in formulating Planck's Law of blackbody radiation allows it to avoid the problem that caused the classical model to "crash and burn" i.e. not match experiment? (c) In a low energy electron diffraction (LEED) experiment, 62 eV electrons are scattered from the surface of a nickel crystal. (i) Calculate the de Broglie wavelength of the electrons used in this experiment. (ii) If the interatomic spacing at the surface of the nickel crystal is 0.203 nm, find a value for the angle associated with the first-order maximum for the diffracted electrons. (d) In the derivation for our wave representation of a particle that showed that the group speed, Vg , is equal to the classical speed of the particle we had the following expression for the phase speed, vp(k): mc 2,1/2 6 (i) Is this expression consistent with what we expect for the phase speed for photons? Explain briefly (ii) And, what does it tell us about the phase speeds associated with our wave representation of a particle with non-zero mass? Does this contravene any laws of Physics? Explain briefly. For a normalised wavefunction, 4(x), that is an eigenfunction of the momentum operator, [Px] : (i) Show by explicit calculation of Apx = Kpz) – (Px)2 that the observable, Px , is sharp (ii) What can we infer about the localisation of the particle represented by this (e) wavefunction? [2+3+(3+2)+(2+2)+(5+1) = 20 marks]

Answer 1

classmate ales 4 Date Page solution Rayleigh Jeans law is in terms of frequency- Endre speed of light RB Botimann constant o-frequency V-Volume T- Temperature 87 V ² (RBT) da C² in vacuum. Ev Energy of black body radiation ů enceblure between frequencies y andst do and (RBT) is average energy. Energy density with Frequency range and &t da ugdy Erdo 87 22 KBT do 22 C3 unde = 8TKBT 22d2 c3 In terms of wavelength & undal = 87 KBT da 24 As we can see from above Formulation that when we talk about tota energy density when of too u二 uida or a to

classmate Date Page since there is no maximum. An infinite energy density physically unacceptable this fallire were unacceptable. The uniplicato that behavlour the theory at low wawelingerie Chigh ) was collastrope field -jons of tumed Rayleigh Leans haw (no-maximum Moda TK Experimental curve. X 104 Fr Frequency Actual Black body spectrum (Experimental) It has a maximum So this was the aspect of Rayleigh Jean's law which were not an agreement with Blackloody Blackbody radiation spectrum

classmate be Date Page According to boltamann Probability dielsibution, the number envigy en u given as Planck reolered that some He with of oscillalox N(n) No e-En RBT En = hhy - radical change was required to explain the expeumental ebsexed Spectrum. Quantum hypothesis postulate Planck's The material es cullator lies the walls ot cavity) can have enly diecente chungy levels rather than continous classical If a particle ficquendy Its take only values envegy En = nho range of energies as assumed osullabeing with physics can n=0, 1,2,3 - ha planck's constant 6:625X10-34 Js Now coninuing Boltzmann Dieleebulisin € = N(N) GA nzo Š N(n) nzo

a No enhylkost classmate nzo Date Page mho Z Noe-nholket nzo Let x = e-h BBT 9 E E becomes hox 11+2x+3x2+ 4x3 Hata2 + a² T. hox 1-a E- hox (1-x)-2 (-x) E = ho hyllest - e Now undo 87 22 ēda (3 C² explhmen wody 8T 22 hr da exp/h21-1 KBT In tums of a 8the da A da 15 exp/hch آماما Now this relation is in close agreement with Black body Specleum for all values values of A&T

calulaled was elengths can be classmate Date Page scattering angle Braggle how ad sino interatomic spacing ordee of n wavelength diffraction dig After detailed calculations- debrogue wavelength 12.24 Dodge |2024 (1) Note AdB h 2m qv h = 6.634 x 10-34 JS m = 9.1X10-31 kg 9 - 1. 6x 10-19 W = 62 V filem (1) Å = 1.5545 Odg= 12024 562

classmate Date Page cap Bragg's law ad sino = n AdB ha 1 7 2d sino = A do d = 0.203 nm 2 (0:283) nm sino = 1.5545 À sino = 1.5545 X10-10 8 (0.203) X10-9 3.82878 X10-1 0.38288 sin (0.38288) uno 0 = 22.512°

classmate Solution:- Date Page up(K) = cllt mc est TR i) This expression is consistent with what we Howh? expect for phase speed of photone for photons rest mass is equal to zero In the above expression us see m=0 mc=0 rep LK) = C1t101272 tik rep (k) = c[ito?"2 |20p (K) = 6 so phase speed of photons is a to above relation and as It not exceed exceed clspeed of light in vacuum It is consistent and say replk) equal to c according dous we can dans be

classmate Date Page üy For non zero mass mzo ep (k) =C It tk For m to the tum under sa root is eator than 1 greater square in always greater that ol a number Mathematically co This resut is inconsistent with mc also, the than 1 in Squaseroot or equal to 1 greaton Itlmo 7. If mto Ink Squaring both side Tt Tmc) from here we conclude reptk) 7 6 Now, this is not possible inconsistent with what we expect for phase speeds of paelicle weith Volk) must be lees that os equal toc 2 This is 2 non- zure mass laws of physics

e) 41x) normalked was avefuncion eigen fundión of momentuin operator - in au A 4 Les ax L some real value Apx = V <22 27 - {6x7² bx is sharp means uncertainity in measuring px is zero or spx = 0 <p 27 = { bx 7² Px = - it a dx

classmate Date Page (i) According to uncertaunity Principles Dx spa lithia uncetainity position uncertainity (as y (x) is ergi moment and operator) in in momentium function eregin Abxto Dxo So It as uncertainity lends to Infinity means particue can't be located in its measurment
For part e ii particle cannot be located because wavefunction is an eigenfunction of momentum operator. So uncertainity in momentum measurement is zero due to uncertainity principle, the uncertainity in measuring the location of the particle will be a very large value. So particle can't be located.