Physics 43 HW #9 Chapter 40 Key

Transcription

1 Pysics 43 HW #9 Captr 4 Ky Captr 4 1 Aftr many ours of dilignt rsarc, you obtain t following data on t potolctric ffct for a crtain matrial: Wavlngt of Ligt (nm) Stopping Potntial (V) a) Plot t stopping potntial vs t frquncy of ligt b) Wat is t cutoff frquncy for t potolctric ffct in tis matrial? c) Wat is t cutoff wavlngt? d) Wat is t work function in V for t mtal ) Wat is t valu of Planck s constant implid by tis data? Wat is t prcnt rror from t accptd valu? f) Try to idntify t mtal Wavlngt of Ligt (nm) Frquncy of Ligt (1^15 Hz) Stopping Potntial (V) Cut Off frquncy: 56/415= wavlngt Work Function (f) (6661^-34)(51^14) 3313E-19J 765v = slop* = 6491 Js Off by %

2 A 7-MV poton scattrs off a fr lctron suc tat t scattring angl of t poton is twic t scattring angl of t lctron Dtrmin (a) t scattring angl for t lctron and (b) t final spd of t lctron (a) Tanks to Compton w av four quations in t unknowns φ, v, and λ : = + γ mc λ λ mc (nrgy consrvation)[1] λ = cosφ+ γ mv cosφ λ (momntum in dirction) [] = sin φ γ mv sin φ λ (momntum in y dirction) [3] λ λ = ( 1 cosφ) (Compton quation) [4] mc Using sin φ = sin φcos φ in Equation [3] givs γ mv = cos φ λ Substituting tis into Equation [] and using cosφ = cos φ 1 yilds ( cos φ 1) cos φ ( 4cos φ 1) = + = λ λ λ λ λ = 4λ cos φ λ [5] Substituting t last rsult into t Compton quation givs ( ) ( ) 4 λ cos φ λ 1 cos φ 1 1 cos φ = = mc mc, or Wit t substitution λ =, tis rducs to E cos For φ mc E mc + E 1+ E = = wr + + mc 7 M V = = 137, tis givs φ 511 M V 1 1+ = cos = 33 + (b) From Equation [5], ( ) λ = λ 4cos φ 1 = λ 4 1 = λ + + Tn, Equation [1] bcoms + = + γ mc λ λ + 3 mc or E E = γ mc + 3 mc + Tus, γ = Trfor, v c γ, and wit = 137 w gt γ = 1614 = 1 = = 785 or v = 785c

3 3 T total powr pr unit ara radiatd by a black body at a tmpratur T is t ara undr t I(λ, T)-vrsus-λ curv, as sown in Figur 43 Sow tat tis powr pr unit ara is I 4 ( λ, T ) dλ = σt wr I(λ, T) is givn by Planck s radiation law and σ is a constant indpndnt of T Tis rsult is known as Stfan s law To carry out t intgration, you sould mak t cang of variabl = /λkt and us t fact tat 3 4 d π = 1 15 T tacr s solution manual is an cllnt guid Plas mak tis vry nat and plain all t stps (a) Starting wit Planck s law, I ( λ, T) = 5 λ B λ π k T 1 t total powr radiatd pr unit ara I ( λ, T) dλ = dλ 5 λ kbt λ Cang variabls by ltting = λkbt dλ and d = kt B λ Not tat as λ varis from, varis from Tn π 1 (b) B B I ( λ, T ) dλ = d = 3 3 π kt π kt π 15 ( 1) Trfor, (, ) From part (a), 5 4 π kb 4 4 I λ T dλ = T σt 3 = π ( J K ) 8 ( ) ( ) 5 4 kb 3 3 π σ = = J s 3 1 m s 8 4 σ = W m K

4 4 Aftr larning about d Brogli s ypotsis tat particls of momntum p av wav caractristics wit wavlngt λ = /p, an 8-kg studnt as grown concrnd about bing diffractd wn passing troug a 75-cm-wid doorway Assum tat significant diffraction occurs wn t widt of t diffraction aprtur is lss tat 1 tims t wavlngt of t wav bing diffractd (a) Dtrmin t maimum spd at wic t studnt can pass troug t doorway in ordr to b significantly diffractd (b) Wit tat spd, ow long will it tak t studnt to pass troug t doorway if it is in a wall 15 cm tick? Compar your rsult to t currntly accptd ag of t Univrs, wic is s (c) Sould tis studnt worry about bing diffractd? (a) T wavlngt of t studnt is λ = p = mv If w is t widt of t diffracting aprtur, tn w nd w 1λ 1 = mv so tat J s v 1 = 1 = 11 1 m s mw ( 8 kg)( 75 m) (b) Using d t = w gt: v 15 m 33 t = s 11 1 m s (c) No T minimum tim to pass troug t door is ovr Univrs 15 1 tims t ag of t 5 T sing ability or rsolution of radiation is dtrmind by its wavlngt If an atom is approimatly 1 1 m in diamtr, ow fast must an lctron travl to av a wavlngt smallr tan t siz of an atom? T dbrogli wavlngt is givn by: λ = St tis qual to an angstrom and solv for v: mv ( J s) 1 ( )( 1 m) v = = = mλ 6 Wat is t dbrogli wavlngt of an lctron jctd by 4 nm potons from a matrial wit a binding nrgy of 71V? Answr: Combin bot potolctric ffct quation wit dbrogli: p KE = f BE = p = m( / λ BE) m λ = / p = / m( / λ BE) = λ = 457nm (9111 kg)((6631 Js(31 m / s) / (41 m) 71 V (161 J / V )) Js

5 7 Litium, bryllium, and mrcury av work functions of 3 V, 39 V, and 45 V, rspctivly Ligt wit a wavlngt of 4 nm is incidnt on ac of ts mtals Dtrmin (a) wic mtals ibit t potolctric ffct and (b) t maimum kintic nrgy for t potolctrons in ac cas (c) Find t trsold frquncis and wavlngts for ac lmnt (a) λc = Li: φ B: Hg: λ c λ c λ c 8 ( J s) ( 3 1 m s) 19 ( 3 V ) ( 16 1 J V ) = = 54 nm 8 ( J s) ( 3 1 m s) 19 ( 39 V ) ( 16 1 J V ) = = 318 nm 8 ( J s) ( 3 1 m s) 19 ( 45 V ) ( 16 1 J V ) = = 76 nm λ < λ c for poto currnt Tus, only litium will ibit t potolctric ffct Knigt Captr 39 (4 in som books) Wav Functions and Uncrtainty 1 You toss a coin 6 tims Wat is t probability of gtting at last 5 ads? Tr ar ^6 diffrnt combinations for 6 tosss Tr ar 7 diffrnt ways to av at last 5 ats So t probability is 7/64 = 11% In t doubl-slit primnt, lctrons ar fird at an opaqu barrir aving two small slits, and a scrn byond t barrir rcords wr t lctrons it Aftr many lctrons ar fird, t standard intrfrnc pattrn is sown on t scrn If t lctrons ar fird wit a igr vlocity, t spacing btwn t dark frings on t scrn will A) dcras B) incras C) rmain t sam EXPLAIN YOUR ANSWER λ = / p so t gratr t momntum, t sortr t wavlngt and t fring spacing will DECREASE 3 In an primnt potons pass troug an apparatus How many potons fall on a 1 mm-wid strip wr t probability dnsity is m -1? N = N ψ d = 811 potons( / m)(1 m) = 16 1 potons total 4 You want to confin an lctron in a bo in suc a way tat you know for crtain tat t lctron's spd is no mor tan 67 m/s Wat is t smallst bo in wic you can confin t lctron? 6331 Js Us t HUNC: p = = 85nm 31 4π 4πmv 4 π(9111 kg)(67 m / s)

6 5 T probability dnsity for an lctron tat as passd troug an primntal apparatus is givn in t figur If 41 lctrons ar usd, wat is t pctd numbr tat will land in a 1 nm-wid strip at = nm? N = N ψ d = 41 lctrons(5 / nm)(1 nm) = 5lctrons total