Problem 1. Part a. Wayne Witzke ProblemSet #2 PHY 361. Comparison of Rayleigh-Jeans and Plank formulas for the black body spectrum.

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1 Wayn Witzk ProblmSt # PHY 36 Problm omparison of Rayligh-Jans and Plank formulas for th black body spctrum. Part a Draw th nod lins for th (n x, n y ) (, ); (, ); (, ); and (, ) standing wav mods on a squar mdium with sids of lngth L. Show that n x λ x L. Show that th dnsity of mods is G(f) 8πf /c 3. It is dfind as th numbr of mods pr volum pr frquncy intrval, i.. G(f) df d 3 n/v. Rmmb r that n x,y,z > 0 and thr ar two indpndnt polarizations of light. Figur : Standing Wav Mods: (n x, n y ) (, ) Figurs through 4 show th standing wav nods for th givn mods on a squar mdium with sids of lngth L. Figur 3: Standing Wav Mods: (n x, n y ) (, ) Figur 3 shows that: n x λ x L λ x L λ x L Figur : Standing Wav Mods: (n x, n y ) (, ) Figurs and 3 show how a standing wav in th mdium is distributd across th mdium. In Figur, only half of th total wavlngth of th wav is in th mdium. In Figur 3, th ntir wav is within th mdium. Figur shows that: n x λ x L () λ x L λ x L (Not, this nxt dmonstration rquird [Tiplr & Llwllyn, pp ] to complt.) To show that G(f) 8πf /c 3, w start by rcognizing that G(f) dn V df, whr N is th numbr of stats of th systm insid of a givn shll of radius n, V is volum and f is th frquncy of th systm. W continu by rcognizing that th mods of th systm, rprsntd in n x, n y, and n z, can only b positiv intgrs. Th stat spac, thn, can b rprsntd by 8th of a sphr. th part of th sphr in th positiv octant. So w gt: N ( ) ( ) 4πn Pag

2 Wayn Witzk ProblmSt # PHY 36 ( ) π fl c L G(f) V c 4πf L 3 V c 3 Figur 4: Standing Wav Mods: (n x, n y ) (, ) πn3 6 And now, w can asily calculat dn in trms of dn: dn π3n 6 dn πn dn Th infinitsimal frquncy, df, can by found in trms of th infinitsimal numbr of mods in any on dirction, dn, by: G(f) is thn: G(f) dn V df L nλ nλ nc f L nc f n fl c df c L dn G(f) Substituting in n fl c : πn dn V c L dn ( πn πn L V c ) ( L V c ) But L 3 V, and w actually hav twic as many stats sinc thr ar two polaritis of light, so w nd up with: Part b G(f) 8πf Th probability of a mod having nrgy ɛ is proportional to ɛ/kt, th Boltzman distribution. Lt β /kt, and intgrat th total probability Z βɛ dɛ to obtain th normalization factor. Z(β) is also 0 calld th partition function. Show that ɛ d ln Z/ kt. Show that this lads to th Rayligh-Jans formula for th spctral intnsity of black body radiation. 0 βɛ dɛ is rlativly straight- Th intgral Z forward to find: Z c 3 ˆ βɛ dɛ 0 [ ] β βɛ 0 ( ) ( β( ) ) β(0) β β 0 + β Z β Z kt Z kt Pag

3 Wayn Witzk ProblmSt # PHY 36 To find ɛ, w us th dfinition u uf(x) dx f(x) dx, whr, in this cas, f(x) βɛ, and w intgrat ovr th domain of ɛ, which is from 0 to, sinc thr ar not ngativ nrgy stats. W v alrady calculatd f(ɛ) dɛ kt. This lavs ɛ βɛ dɛ. Not, 0 0 howvr, that ɛ βɛ d βɛ, so w can writ: ˆ ɛ, thn, bcoms: Also: 0 ˆ ɛ βɛ dɛ d 0 βɛ dɛ d ˆ βɛ dɛ 0 dz β ɛ d ln( β ) kt (kt ) kt kt β β β β kt β It s also possibl to show this by noting that d ln Z dz, so w hav: Z dz ɛ Z dz Z d (ln Z) Now that w hav G(f) and ɛ, w can calculat th Rayligh-Jans quation, u(f) G(f) ɛ : Part c u(f) G(f) ɛ 8πf c 3 kt Assum th nrgy is not continuous, but quantizd to discrt lvls ɛ n for n 0,,,... Rpat th calculation of th normalization factor Z n0. alculat ɛ. Show that this lads to Planck s βɛn formula. Th normalization factor, in this cas, can b found by rcognizing that Z ( ) n0 β n ( ) n0 /kt n. As in part (b), w /kt can not that: So onc again: ɛ n βɛn d βɛn dz d β ( ) ( β ) ( ) ( β ) ɛ dz Z β ( β ) /kt ( β ( β ) β ( β ) ) ( /kt ) Pag 3

4 Wayn Witzk ProblmSt # PHY 36 β β β β /kt Now, to driv Planck s formula, w put ɛ togthr with G(f) to find u(f): Part d u(f) G(f) ɛ 8πf c 3 /kt Show that ε kt in th limit kt. Show that ɛ 0 in th limit kt, thus solving th ultravilot catastroph. Qualitativly, this is bcaus th tmpratur is too low to xcit vn on photon (n). If kt, w can us th Taylor xpansion for x : + x + x! + x3 3! + Ignoring th x and highr trms, w gt: And: /kt + kt ɛ /kt + /kt kt For kt, w hav: ɛ Sinc, by l Hopital s rul, fastr than, ɛ 0. Problm Som of th fundamntal constants in SI units ar: h J s c m/s k N m / 4πɛ 0 k B J/K m kg m p kg m n kg Not that V V is a compound unit of nrgy. alculat th following usful combinations of constants in th units spcifid: hc [V nm], c h π c [MV fm], k [V nm], α k / c [], kt [mv] at room tmpratur [ T K, and m, m p, m ] n MV/c. W will us ths combinations in natural units rpatdly throughout th rst of th smstr. NOTE: Th mass of th proton and nutron in kg as statd in th original problm is incorrct. Ths should b 0 7, not 0 4. hc [V nm]: hc ( J s)( m/s) J m Pag 4

5 Wayn Witzk ProblmSt # PHY 36 W convrt this by dividing througy by and multiplying by 0 9 nm/m: hc ( ) J m ( nm/m ) V nm c h π c [MV fm]: c ( V nm π MV fm k [V nm]: ) ( MV 0 6 V ) ( 0 6 ) fm nm m, m p, m n [ MV/c ] : For ach, w find mc, dividing th mass by th lmntary charg and multiplying by c to find th nrgy. m c ( ) kg ( m/s) m MV/c m p c ( ) kg ( m/s) m p MV/c k ( N ) m ( ) V m W convrt this by multiplying by 0 9 nm/m: k ( V m ) ( 0 9 nm/m ) α k / c []: V nm α k c V nm V nm kt [mv] at room tmpratur T K: m n c ( ) kg ( m/s) m n MV/c Problm 3.4 Part a Th wavlngths of visibl light rang from about 380 nm to about 750 nm. What is th rang of photon nrgis (in V) in visibl light? Each photon has nrgy E hc/λ. hc 40 V nm. So at 380 nm, ach photon has nrgy: kt J/K (93.5 K) ( V ) ( ) 000 mv V 5.67 mv 40 V E 40 V nm 380 nm At 750 nm, ach photon has nrgy: E 40 V nm 750 nm 3.6 V.65 V Pag 5

6 Wayn Witzk ProblmSt # PHY 36 Part b A typical FM radio station s broadcast frquncy is about 00 MHz. What is th nrgy of an FM photon of th frquncy? NOTE: I assum thy mant of that frquncy. Each photon has nrgy E. ach photon has nrgy: So, at 00 MHz, Part b Find th stopping potntial if th wavlngth of th incidnt light is 300 nm. Th stopping potntial is givn by V 0 φ, or V 0 φ. Howvr, w r givn th wavlngth, not th frquncy, so w must us this form of th sam quation: E ( ) J s (00 MHz) V Problm 3.6 Part a Th work function for csium is.9 V, th lowst of any mtal. Find th thrshold frquncy and wavlngth for th photolctric ffct. Th work function is φ t hc λ t. Sinc h and hc ar just constants, f t and λ t can b asily found: V 0 hc λ φ 40 V nm 300 nm.9 V.3 V (Not that th / cancls with th in V to lav volts. Part c Find th stopping potntial if th wavlngth of th incidnt light is 400 nm. Similar to part b: f t φ h (.9 V) ( ) J s Hz V 0 hc λ φ 40 V nm 400 nm.9 V.0 V λ t hc φ 40 V nm.9 V 653 nm Problm 3.3 Undr optimum conditions, th y will prciv a flash if about 60 photons arriv at th corna. How much nrgy is this in jouls if th wavlngth of th light is 550 nm? Pag 6

7 Wayn Witzk ProblmSt # PHY 36 Th nrgy of ach photon E hc λ : E hc λ 40 V nm 550 nm.5 V If 60 photons with this nrgy arriv at th corna, this bcoms (60 photons) (.5 V/photon) 35 V. To convrt this to Jouls, just multiply by : Th corrsponding valus for f(θ ) cos θ ar: f(0) 0 f(45 ) 0.93 f(90 ) f(35 ).707 Plotting λ vrsus f, w gt Figur 5. E (35 V) ( ) J Problm 3.40 ompton s quation (Equation 3-5) indicats that a graph of λ vrsus ( cos θ) should b a straight lin whos slop h/mc allows a dtrmination of h. Givn that th wavlngth of λ in Figur 3-7 is 0.07 nm, comput λ for ach scattring angl in th figur and graph th rsult vrsus ( cos θ). What is th slop of th lin? Using th calibration of th Bragg spctromtr, θ s , and d sin θ mλ, w can dtrmin d 0.07 m nm sin nm. From that, w can calculat λ valus using λ (θ B ) ( ) d m sin θb, taking θ B from th graphs in [Tiplr & Llwllyn, p.37]. W can also calculat cos θ c using th angls spcifid for ach graph. This yilds th following for th function λ (θ B ): λ (6 4 ) 0.07 nm λ (6 47 ) nm λ (6 56 ) nm λ (7 06 ) nm Figur 5: ompton Scattring for λ 0.07 nm Th y-intrcpt of th bst fit lin for that graph is nm, and th slop is nm λ c. Th χ χ nm, and δy nm. Problm 3.47 Whn a bam of monochromatic x-rays is incidnt on a particular Nal crystal, Bragg rflction in th first ordr (i.., with m ) occurs at θ 0. Th valu of d 0.8 nm. What is th minimum voltag at which th x-ray tub can b oprating? Pag 7

8 Wayn Witzk ProblmSt # PHY 36 Bragg s Law stats that d sin θ mλ, and w also 40 know that λ m V nm V m. Sinc m,w can asily solv for V m : V m 40 V nm d sin θ 40 V nm (0.8 nm) sin V Dividing through by to gt volts yilds 6474 V. Problm 3.49 Show that th maximum kintic nrgy, E k, calld th ompton dg, that a rcoiling lctron can carry away from a ompton scattring vnt is givn by: E k + mc / E γ [Wikipdia hlpd with this on slightly.] W know that th rlationship btwn th wavlngths of th light and th angl of scattring is givn by th quation: ( cos θ) E mc + E E E ( cos θ) E mc + E E E ( cos θ) mc + E E E k givs th nrgy transfrd to th lctron. To maximiz E k, w nd to minimiz E. To minimiz E, w want E ( cos θ) as larg as possibl, which will happn whn cos θ, or whn θ 80. This, along with th substitution for E, (and using E γ E ) givs: E γ E k E γ E γ mc + E γ E γ+mc mc E γ mc E k E γ E γmc E ( γ Eγ + mc ) E γ mc E γ + E γ mc E γ mc λ λ h ( cos θ) mc E k E γ W also know that wavlngth is rlatd to nrgy by th quation: E γ hc λ Substituting into th first quation, w gt: hc hc h ( cos θ) E E mc ( cos θ) E E mc Now that w hav this, w can asily substitu E γ back into th quation to gt th othr form from th problm statmnt: E k E k () + mc () + mc + mc Pag 8