Chapter 30: Quantum Physics

Transcription

1 Answrs to Evn-Numbrd Concptual Qustions. I nrgy is quantizd, as suggstd by Planck, t amount o nrgy or vn a singl ig-rquncy poton can b arbitrarily larg. T init nrgy in a blackbody simply can t produc suc ig-rquncy potons, and tror t ininit nrgy implid by t ultraviolt catastrop cannot occur. In classical pysics, any amount o nrgy can b in t orm o ig-rquncy ligt t nrgy dos not av to b supplid in discrt, larg lumps as in Planck s tory. Tror, classical pysics implis tat all rquncis o ligt av t sam amount o nrgy, no mattr ow ig t rquncy. Tis is wat lads to t catastrop. 4. Planck s tory o blackbody radiation implis a on-to-on rlationsip btwn t absolut tmpratur o a blackbody and t rquncy o ligt at t pak o its radiatd nrgy spctrum. Tis rlationsip is givn by Win s displacmnt law (quation 30-1). Tror, by masuring t pak in t radiatd nrgy rom a star, w can tll its tmpratur. In broad trms, a blu star is vry ot, a rd star muc lss so, and a yllowis star lik our Sun is intrmdiat in tmpratur. 6. A monocromatic sourc o ligt mans litrally tat it mits ligt o a singl color. Tis mans tat all t potons mittd by t sourc av t sam rquncy, and nc ty also av t sam nrgy. 8. (a) A poton rom a grn ligt sourc always as lss nrgy tan a poton rom a blu ligt sourc. (b) A poton rom a grn ligt sourc always as mor nrgy tan a poton rom a rd ligt sourc. T rason or ts rsults is tat t nrgy o a poton dpnds linarly on t rquncy o ligt; tat is, E = ƒ. 10. Classically, it sould b possibl to jct lctrons wit ligt o any rquncy all tat is rquird is to incras t intnsity o t bam o ligt suicintly. T act tat tis is not t cas mans tat t classical pictur is incorrct. In addition, t act tat tr is a lowst rquncy tat will jct lctrons implis tat t nrgy o t poton is proportional to its rquncy, in agrmnt wit E = ƒ. 1. Ys. An lctron and a proton av t sam d Brogli wavlngt p i ty av t sam momntum. Solutions to Problms and Concptual Exrciss. Pictur t Problm: T star Btlgus mits a pak rquncy tat is proportional to its surac tmpratur. Stratgy: Solv Win s Displacmnt Law (quation 30-1) or t tmpratur o Btlgus. Solution: Solv quation 30-1 or t tmpratur: pak s K Hz T s K T 3100 K Insigt: Btlgus, lik all rd-giant stars, as a coolr surac tmpratur tan our own Sun. T irst printing o t tird dition txt ad incorrctly listd t pak rquncy as Hz, wic would corrspond to an incorrct surac tmpratur o 560 K, muc closr to t 5800 K surac tmpratur o our Sun. 30 1

2 4. Pictur t Problm: T initial Big Bang can b considrd as a blackbody radiator wos pak rquncy is a unction o its tmpratur. Stratgy: Solv Win s Displacmnt Law (quation 30-1) or t pak rquncy. Us quation 14-1 to calculat t wavlngt, wr t wav spd is t spd o ligt Solution: 1. (a) Calculat t pak rquncy: pak s K T s K.7 K Hz. (b) Calculat t wavlngt: m/s Hz c 1.9 mm Insigt: Tis wavlngt alls in t microwav portion o t lctromagntic spctrum (s Captr 5). 6. Pictur t Problm: A blackbody mits radiation according to Win s Displacmnt Law. Stratgy: Us quation 30-1 to calculat t ratio o t pak rquncis or tmpraturs o 0.0 K and 40.0 K. Rpat or t Klvin tmpraturs tat corrspond to 0.0C and 40.0C. Solution: 1. (a) Calculat t ratio o pak rquncis or 0.0 K and 40.0 K:. (b) Calculat t ratio o pak rquncis or 0.0C and 40.0C: s K T pak, K pak, s K T 0.0 K 10 T s K pak, C pak, s K T C 0 Insigt: Doubling t tmpratur in Klvin actually doubls t avrag kintic nrgy o t molculs. Howvr, in tis cas doubling t tmpratur in Clsius only corrsponds to a sligt (6.8%) incras in t Klvin tmpratur. 30

3 8. Pictur t Problm: An incandscnt ligt bulb oprats at a coolr tmpratur tan a alogn bulb. Tror, t blackbody radiation rom t two ligt bulbs will av dirnt pak rquncis. Stratgy: According to Win s Displacmnt Law, a ottr bulb will mit ligt wit a igr pak rquncy. Bcaus t alogn bulb is ottr, it will mit a igr pak rquncy tan t incandscnt bulb. Us Win s Displacmnt Law (quation 30-1) to calculat bot o t pak rquncis and calculat tir ratio. Solution: 1. (a) Bcaus pak T, t alogn bulb as t igr pak rquncy.. (b) Calculat t ratio o t pak rquncis: al std s K Tal 3400 K s K Tstd 900 K (c) Calculat t pak rquncis o t two bulbs: al std 4. T alogn bulb producs a pak rquncy tat is closr to s K 3400 K.0 10 Hz s K 900 K Hz Hz tan t standard incandscnt bulb. Insigt: Bcaus t alogn bulb producs a pak rquncy closr to t snsitiv rquncis o t uman y, mor o t radiation rom t alogn bulb is visibl, making it appar brigtr or t sam powr output. 10. Pictur t Problm: T imag sows two oxygn atoms oscillating back and ort, similar to a mass on a spring. T imag also sows t vnly spacd nrgy lvls o t oscillation. Stratgy: Us quation to calculat t priod o oscillation or t two atoms. Calculat t rquncy, using quation 13-1, rom t invrs o t priod. Multiply t priod by Planck s constant to calculat t spacing o t nrgy lvls. Solution: 1. (a) St t rquncy qual to t invrs o t priod: m T k 1 1 k N/m T m kg Hz (b) Multiply t rquncy by Planck s constant: E ( J s)( Hz) J 13 Insigt: Oxygn molculs in t atmospr absorb ligt wavs wit a rquncy Hz, and t nrgy rom t ligt is convrtd into t oscillations o t atoms. 1. Pictur t Problm: A sourc o rd ligt as a igr wattag tan a sourc o grn ligt. Stratgy: Us t xprssion or poton nrgy (quation 30-4) to answr t concptual qustion. Solution: 1. (a) T nrgy o potons mittd by t rd sourc is lss tan t nrgy o potons mittd by t grn sourc, rgardlss o t wattag o t sourc. Tis conclusion ollows rom t rlation E = and t act tat grn ligt as a igr rquncy tan rd ligt.. (b) T bst xplanation is II. T rd-sourc potons av lss nrgy tan t grn-sourc potons bcaus ty av a lowr rquncy. T wattag o t sourc dosn t mattr. Statmnts I and III ar ac als. Insigt: A igr wattag sourc in gnral producs mor potons tan dos a lowr wattag sourc, but t nrgy o ac poton is dtrmind by t rquncy, not t powr o t sourc. 14. Pictur t Problm: Ligt o a particular wavlngt dos not jct lctrons rom t surac o a givn mtal. Stratgy: Us t concpts tat dscrib t potolctric ct to answr t qustion. Solution: 1. (a) To jct lctrons rom tis mtal rquirs potons wit gratr nrgy, wic mans tat t rquncy o t ligt must b incrasd. I t rquncy o t ligt is incrasd, its wavlngt must b dcrasd. 30 3

4 . (b) T bst xplanation is II. T nrgy o a poton is proportional to its rquncy; tat is, invrsly proportional to its wavlngt. To incras t nrgy o t potons so ty can jct lctrons, on must dcras tir wavlngt. Statmnt I rronously assrts tat incrasing t wavlngt will incras t nrgy. Insigt: Incrasing t intnsity o t irst ligt sourc will av no ct, bcaus tat would only dirct mor potons to t mtal surac, non o wic av suicint nrgy to jct an lctron Pictur t Problm: UV potons av an nrgy o J. Tis nrgy corrsponds to a spciic rquncy and wavlngt. Stratgy: Solv quation 30-4 or t rquncy o t poton. Insrt t rquncy into quation 14-1 to calculat t wavlngt. 19 E J 14 Solution: 1. Calculat t rquncy: Hz J s m/s 14 c. Calculat t wavlngt rom t rquncy: 310 nm 0.31 m Hz Insigt: As xpctd, tis rquncy and wavlngt li in t UV spctrum (s Captr 5). 18. Pictur t Problm: A lium atom is ionizd wn it absorbs t nrgy rom a 50.4-nm poton. Stratgy: St t nrgy o t poton (rom quations 30-4 and 14-1) qual to t ionization potntial o lium. Solution: Rplac t rquncy in quation 30-4 wit t spd o ligt dividd by t wavlngt: J s m/s c E m J Insigt: Potons wit wavlngts longr tan 50.4 nm cannot compltly ioniz t lium atom bcaus ty av too littl nrgy. Wn potons wit wavlngts tat ar sortr tan 50.4 nm intract wit t lium atom, ty will bot ioniz t atom and giv xtra kintic nrgy to t mittd lctrons. 0. Pictur t Problm: As potons intract wit silvr, som o t nrgy o t poton rs an lctron rom t silvr atom. T rmaindr o t nrgy bcoms kintic nrgy o t lctron. Stratgy: Solv quation 30-7 or t work unction o silvr. Solution: Calculat t work unction: K W max W0 Kmax J s Hz J 19 = J 14 Insigt: In ordr or a poton to jct an lctron rom silvr, it must av a rquncy gratr tan Hz. At tis rquncy, all o t nrgy o t poton gos into t work unction and t jctd lctron as no kintic nrgy.. Pictur t Problm: Eac poton contains a quantizd amount o nrgy, dtrmind by t poton wavlngt. T total nrgy o.5 J is obtaind by t addition o many potons. Stratgy: Calculat t nrgy o a singl poton rom its wavlngt using quations 30-4 and Divid t total nrgy by t nrgy o t singl poton to calculat t numbr o potons. Solution: 1. (a) Us quations 30-4 and 14-1 to writ t nrgy o a singl poton:. Divid t total nrgy by t nrgy o a singl 350-nm poton: c E n E E m.5 J total total potons Epoton c J s m/s 30 4

5 3. (b) Divid t total nrgy by t nrgy o a 750-nm poton: m.5 J n J s m/s potons Insigt: Mor rd potons tan UV potons ar rquird in ordr to produc t sam amount o nrgy. 4. Pictur t Problm: Two radio stations broadcast at t sam powr lvl, but station A broadcasts at a lowr rquncy tan station B. Stratgy: T rat o poton mission is t powr lvl dividd by t nrgy o a singl poton. Not tat t nrgy o ac poton is proportional to its rquncy (quation 30-4). Solution: 1. (a) T rat o mittd potons is dtrmind by t ratio o t mittd powr to t nrgy o a poton. Tis ratio is invrsly proportional to t rquncy o t mittd potons. Tror, t station wit t lowr rquncy as t igst rat o poton mission, and station A mits mor potons pr scond tan station B.. (b) Bcaus t nrgy pr poton is dirctly proportional to t rquncy o t poton, t station tat broadcasts at t igr rquncy mits t igr nrgy potons. So, station B mits igr nrgy potons tan station A. Insigt: In tis xampl, station A mits potons/s 8 tat ac av an nrgy o J potons/s tat ac av an nrgy o Station B mits J. 30 5

6 6. Pictur t Problm: T ligt rom t lasr can b considrd as t mission o many potons o t sam wavlngt. T powr output is t product o t poton mission rat and t nrgy o ac poton. Stratgy: Divid t mittd powr by t nrgy o ac poton (quation 30-4) to calculat t rat o poton mission, and tn ind t rquncy using quation Solution: 1. (a) Calculat t P P P P n rat o poton mission: E c/ c. Insrt t givn wavlngt and powr: 3. Calculat t rquncy rom quation 14-1: m W J s m/s n 8 m c s m Hz potons/s Insigt: T larg numbr o potons in a lasr bam can caus prmannt damag to t rtina bor you can blink, wic is wy you sould nvr look into a lasr! 8. Pictur t Problm: A poton o wavlngt 64 nm will jct an lctron rom a coppr surac. T rsulting lctron, owvr, as no kintic nrgy. Stratgy: St t work unction o coppr qual to t nrgy o t poton, givn by quations 30-4 and Solution: Calculat t work unction o coppr: J s m/s c W m J Insigt: Elctrons jctd rom coppr by potons wit wavlngts lss tan 64 nm will av kintic nrgy. Potons largr tan 64 nm cannot jct lctrons rom coppr. 30. Pictur t Problm: Wn a poton is absorbd by a mtal, its nrgy is split into t nrgy ndd to jct t lctron and t kintic nrgy o t lctron. T nrgy o t poton is invrsly proportional to its wavlngt. Stratgy: Us quations 30-7 and 14-1 to writ an xprssion o t kintic nrgy o t lctron in trms o t wavlngt o t poton. c Solution: 1. (a) Bcaus K, max W0 W0 a sortr wavlngt implis a largr kintic nrgy. So, t bam wit wavlngt B producs potolctrons wit gratr kintic nrgy tan t bam wit wavlngt. A. (b) Calculat t kintic nrgy o t potolctron wn struck by poton A: 3. Calculat t kintic nrgy o t potolctron wn struck by poton B: K K J s m/s 6010 m J/V 1.9 V 0.1 V max,a J s m/s m J/V 1.9 V 1.1 V max,b 9 19 Insigt: T cuto wavlngt tat will jct an lctron rom csium is 654 nm. Bcaus poton A as a wavlngt sligtly smallr tan t cuto, t rsulting potolctron as a small kintic nrgy. 3. Pictur t Problm: Wn wit ligt is incidnt upon t potassium, t potons wit nrgis gratr tan t work unction o potassium will jct lctrons. T gratr t poton nrgy, t gratr t kintic nrgy o t jctd lctron. Stratgy: Bcaus t poton nrgy is proportional to t rquncy, t potons wit t gratst rquncy will jct lctrons wit t maximum kintic nrgy. Insrt t maximum rquncy into quation 30-7 to calculat t maximum kintic nrgy o t potolctrons. Tn us quation 30-6 to calculat t cuto rquncy. All potons wit smallr rquncis will not jct lctrons. 30 6

7 Solution: 1. (a) Us quation 30-7 to ind max :. (b) Calculat t cuto rquncy: 3. Writ t rquncy rang or wic no lctrons ar mittd: K J s Hz Kmax W J/V.4 V 1.03 V W.4 V J/V Hz J s Hz Hz 14 Insigt: Potons tat ar incidnt upon a potassium surac and tat av rquncis gratr tan Hz will mit potolctrons wit kintic nrgis ranging rom zro to 1.03 V. 34. Pictur t Problm: Wn potons o t givn rquncy ar incidnt on t two mtals, som o t nrgy o t poton will jct t lctron rom t surac (work unction) and t rmaindr o t nrgy will bcom kintic nrgy o t lctron. Stratgy: Us quation 30-4 to calculat t nrgy o t poton. Tn insrt tat nrgy togtr wit t work unction o ac mtal into quation 30-5 to calculat t maximum kintic nrgy. Solution: 1. (a) Bcaus K, max W 0 t potolctrons mittd by t mtal wit t smallr work unction will av t gratr kintic nrgy. T lctrons jctd rom t iron surac will av t gratr maximum kintic nrgy bcaus iron as a smallr work unction tan platinum.. (b) Calculat t poton nrgy: J s Hz 3. Calculat t maximum kintic nrgy or lctrons potojctd rom platinum: 4. Calculat t maximum kintic nrgy or potolctrons jctd rom iron: Insigt: T cuto rquncy or platinum is E 7.79 V J/V Kmax, Pt EW V 6.35 V 1.44 V Kmax, F 7.79 V 4.50 V 3.9 V Hz and or iron is Hz. Potons wit rquncis btwn ts two valus will jct potolctrons rom t iron, but not rom t platinum. 36. Pictur t Problm: T larg pupil in an owl s y allows or mor potons pr scond to ntr t y tan t numbr o potons tat would ntr t uman y. In addition, t rtina o t owl is mor snsitiv to wr potons tan t uman y. Stratgy: Multiply t minimum intnsity o ligt by t ara o t owl s pupil to dtrmin t powr absorbd by t owl s y. Divid tis powr by t nrgy pr poton to dtrmin t minimum poton lux tat t owl can dtct. Solution: 1. Calculat t powr absorbd P IA Ir W/m m W in t owl s y at minimum intnsity:. Calculat t nrgy pr poton: E 17 P W J s Hz J 3. Divid t powr by t nrgy: n 61 potons/s 19 E J Insigt: I t uman y wr snsitiv to 61 potons/s, t smallr pupil o t uman y would rquir a minimum 13 ligt intnsity o W/m. 38. Pictur t Problm: T potons in a microwav av a givn momntum and wavlngt. 30 7

8 Stratgy: Us quation to calculat t wavlngt o t microwavs. Solution: 1. (a) Solv quation or t wavlngt: 13 cm p J s kg m/s. (b) T wavlngt is muc largr tan t siz o t ols in t mtal scrn. Insigt: Bcaus t ols ar smallr tan t microwavs, t microwavs rlct o t scrn and rmain insid t microwav ovn. Howvr, visibl ligt, wic as a muc smallr wavlngt, can xit troug t ols and allow you to s t ood cooking. 40. Pictur t Problm: A poton and an lctron av t sam momntum. T lctron s momntum is dtrmind by its spd and mass. T poton s momntum is dtrmind by its wavlngt. Stratgy: Calculat t momntum o t lctron using quation 9-1. T rlativistic ormula (quation 9-5) is not ncssary bcaus t spd is muc lowr tan t spd o ligt. St t lctron and poton momnta qual and us quation to calculat t poton wavlngt p mv kg 100 m/s kg m/s Solution: 1. Find t momntum o t lctron: J s. Calculat t wavlngt o t poton: 610 nm 0.61 m 7 p kg m/s Insigt: Tis wavlngt corrsponds to visibl, grn ligt. 4. Pictur t Problm: A ydrogn atom tat is initially at rst mits a 1-nm poton. T atom rcoils in t dirction opposit to t propagation o t poton. Stratgy: Bor t mission t ydrogn atom was at rst. T poton and t ydrogn atom tn av qual but opposit momnta. Us quation to calculat t momntum o t poton. St it qual to t momntum o t ydrogn atom and us quation 9-1 to calculat t rcoil spd o t ydrogn atom J s 7 Solution 1: Calculat t momntum o t poton: puv kg m/s m. Solv or t rcoil spd: puv ph mv 7 puv kg m/s v 3.5 m/s 7 m kg Insigt: Not tat t spd o t ydrogn atom is invrsly proportional to t wavlngt o t poton. I t atom ad mittd a poton wit wavlngt 61 nm, t rcoil spd would doubl to 6.5 m/s. 44. Pictur t Problm: A rd poton and a blu poton ac carry momntum in proportion to tir rquncis. Stratgy: Calculat t momnta o t two potons using quation Solution: 1. (a) Bcaus, blu rd and sinc p, blu rd c p p. A poton o blu ligt as t gratr momntum J s4.010 Hz 8. (b) Calculat p rd : prd kg m/s 8 c m/s 3. Calculat p blu : p J s Hz blu m/s kg m/s Insigt: Bcaus t rquncy o t blu poton is about doubl t rquncy o t rd poton, t momntum o t blu poton is about doubl t momntum o t rd poton. 46. Pictur t Problm: As a bam o potons is absorbd by a black surac, t momntum o t potons is absorbd by t surac, and a corrsponding orc is xrtd on t surac. Stratgy: Calculat t rat o poton production by dividing t powr o t lasr bam by t nrgy in ac poton (rom quations 30-4 and 14-1). Calculat t cang in momntum by using quation or t poton momntum. 30 8

9 Multiply t poton rat by t momntum cang to obtain t nt orc on t black surac. Solution: 1. (a) Calculat t nrgy in ac poton:. Divid t powr by t nrgy pr poton: 3. (b) Calculat t momntum cang wn ac poton is stoppd: 4. (c) Multiply t momntum cang by t poton rat: J s m/s c E poton m P n E J p p p 0 p W potons/s i i J s m J kg m/s potons/s kg m/s F np N Insigt: Tis orc, vn toug it is small, as applications in usion rsarc, wr it is usd to conin ydrogn pllts. It is also mployd by an optical twzrs, a lasr dvic tat allows a rsarcr to manipulat vry tiny objcts wil looking at tm troug a microscop. 30 9

10 48. Pictur t Problm: In a Compton scattring xprimnt, t scattrd lctron is obsrvd to mov in t sam dirction as t incidnt X-ray poton. Stratgy: Us t principls tat govrn t Compton ct to answr t concptual qustion. Solution: Suppos t initial poton movs in t positiv x dirction. Tis mans tat t initial y componnt o momntum is zro. Atr t collision, w ar told tat t lctron movs in t positiv x dirction it as no y componnt o momntum. Tror, t scattrd poton must also av zro y componnt o momntum, wic mans it will propagat in itr t positiv x dirction or t ngativ x dirction. I it wr to propagat in t positiv x dirction, t scattring angl would b 0 and t Compton ormula (quation 30-15) indicats t poton will transr no momntum to t lctron. W rjct tat solution bcaus it implis t lctron dos not scattr at all, and w conclud tat t scattring angl o t poton is 180. Insigt: Insrting θ = 180 into quation sows tat t wavlngt o t scattrd poton will incras by t mc.46 pm 4.85 pm. amount 50. Pictur t Problm: T wavlngt o a poton incrass atr it scattrs o an lctron and travls at an angl θ rom its initial dirction. Stratgy: Us t Compton scattring quation (quation 30-15) to ind or ac o t scattring angls. A rcurring valu in similar 1 Compton scattring problms is mc.4610 m.46 pm. Solution: 1. Insrt valus or 1cos t constants in quation 30-15: mc. (a) Insrt 30 : 3. (b) Insrt 90 : 4. (c) Insrt 180 :.46 pm1cos.46 pm 1 cos pm.46 pm 1 cos pm.46 pm 1cos pm Insigt: T maximum cang in wavlngt is 4.85 pm bcaus t maximum scattring angl is Pictur t Problm: T smallst cang in wavlngt in a Compton scattring is zro, wn t poton continus to travl straigt orward. T gratst cang in wavlngt is wn t poton is scattrd dirctly backward. A cang in wavlngt o on-ourt o t maximum will occur at an angl θ btwn t minimum and maximum. Stratgy: St t scattring angl θ in quation to 180 to ind t maximum cang in wavlngt. Tn solv quation or t scattring angl, wn t cang in wavlngt is qual to on-quartr o t maximum. A 1 rcurring valu in similar Compton scattring problms is mc.4610 m.46 pm. max mc mc mc st max max 1cos 1cos mc cos 1 60 Solution: 1. Calculat t maximum cang in wavlngt: 1 cos 1 cos180. St t cang in wavlngt qual to a quartr o t maximum and solv or t scattring angl: Insigt: Tis angl is on-tird o t maximum angl. I t angl wr two-tirds o t maximum, or 10, t cang in wavlngt would b 3 4 max

11 54. Pictur t Problm: A poton o wavlngt 0.40 nm scattrs o a r lctron at rst at an angl o 105. Stratgy: Calculat t initial momntum o t poton using quation Tn us quation to calculat t inal wavlngt o t poton. Insrt t inal wavlngt into quation to calculat t inal momntum. A rcurring valu in similar Compton scattring 1 problms is mc.4610 m nm. Solution: 1. (a) Calculat t initial momntum: p J s i 9 i m kg m/s. (b) Solv quation or t inal wavlngt: i 1cos mc i 1cos mc 0.40 nm nm 1 cos nm 3. (c) Calculat t inal momntum: p J s m kg m/s Insigt: T inal momntum o t lctron can b calculatd rom t vctor cang in t momntum o t poton. 4 T inal lctron momntum is kg m/s at an angl o =

12 56. Pictur t Problm: An X-ray scattrs 180 o an lctron or o a lium atom. As a rsult o t scattring, som o t nrgy o t X-ray is transrrd to t particl, and t wavlngt o t X-ray is incrasd. Stratgy: Us quation to calculat t cang in wavlngt, wr θ = 180. For t scattring o o t lium atom, rplac t mass o t lctron wit t mass o lium. 1 Solution: 1. (a) Bcaus and H m m m, t cang in wavlngt o t X-ray is gratr or t lctron.. (b) Calculat t cang in wavlngt or scattring o o t lctron: 3. Calculat t cang in wavlngt or scattring o o t lium atom: J s1cos180 1cos 4.85 pm kg m/s lctron 31 8 mc J s1cos180 1cos m kg m/s H 7 8 mhc Insigt: Bcaus t lium atom as muc mor mass tan t lctron, t lium atom rquirs lss kintic nrgy to carry away t sam momntum, as suc t poton dos not nd to transr as muc nrgy to t lium atom. 58. Pictur t Problm: T imag sows a poton wit an initial wavlngt o 0.55 nm scattring o an lctron. T wavlngt o t scattrd poton is 3.33 pm longr tan t initial wavlngt. As a rsult o t collision, t lctron scattrs at an angl rom t positiv x-axis. Stratgy: Us quation to calculat t magnitud o t initial and inal momntum o t poton. Calculat t scattring angl o t poton using quation Us t scattring angl to calculat t orizontal and vrtical componnts o t momntum o t scattrd X-ray poton. Us consrvation o momntum to calculat t componnts o t lctron s momntum. Finally, us ts componnts to calculat t angl tat t lctron scattrs. A rcurring valu in similar Compton scattring 1 problms is mc.4610 m.46 pm. Solution: 1. Calculat t initial momntum o t poton: J s pi 0.55 nm i kg m/s. Calculat t inal momntum o t poton: p i J s 0.55 nm nm kg m/s 3. Calculat t scattring angl o t poton: 4. Calculat t orizontal and vrtical componnts o t inal poton momntum: 1cos mc 1 mc pm cos 1 cos pm p p x, y, 4 p cos kg m/s cos kg m/s kg m/s 4 p sin kg m/s sin Calculat t orizontal componnt o t lctron p p p i x,,x 4 5 momntum: p,x pi px, kg m/s kg m/s kg m/s 30 1

13 6. Calculat t vrtical componnt o t lctron momntum: 7. Calculat t angl at wic t lctron rcoils: Insigt: T momntum o t rcoiling lctron is p 0 p y,, y y, p, y p kg m/s kg m/s 4 p 1,y kg m/s tan tan 34 4 p,x kg m/s 4 p, x p, y.0910 kg m/s and its kintic nrgy is 14.9 V

14 60. Pictur t Problm: T d Brogli wavlngt o a particl cangs wn itr its momntum or its kintic nrgy is doubld. Stratgy: Us t d Brogli wavlngt quation (quation 30-16) to answr t concptual qustion. Solution: 1. (a) Bcaus t d Brogli wavlngt is p, it ollows tat doubling t momntum alvs t wavlngt. Tror, t wavlngt cangs by a multiplicativ actor o 1.. Doubling t kintic nrgy mans tat t momntum incrass by a actor o. As a rsult, t d Brogli wavlngt cangs by a multiplicativ actor o 1. Insigt: In ac cas t d Brogli wavlngt o t particl dcrass as t spd incrass. 6. Pictur t Problm: T d Brogli wavlngt o a nutron is invrsly proportional to its momntum. Stratgy: Rplac t momntum in t d Brogli wavlngt quation (quation 30-16) wit quation 9-1. Solv t rsulting quation or t nutron vlocity. Solution: 1. Combin quations and 9-1:. Solv or t nutron vlocity: p mv J s v 1.40 km/s m m kg Insigt: Wn t wavlngt o t nutron is qual to t intratomic spacing, t nutron bam will diract and produc a maximum at an angl o 60 rom t normal

15 64. Pictur t Problm: T d Brogli wavlngt o an lctron is rlatd to bot t momntum and t kintic nrgy o an lctron. Stratgy: Writ t kintic nrgy (quation 7-6) o t lctron in trms o t momntum (quation 9-1). Tn us quation to writ t momntum in trms o t d Brogli wavlngt. Solution: 1. Writ t kintic nrgy in trms o t d Brogli wavlngt:. Insrt t wavlngt: K 1 mv p m m m mv J s K kg m J Insigt: In tis problm w usd t classical quations or t kintic nrgy and t momntum. Tis assumption can 6 b cckd solving quation or t spd. T rsulting v m/s 0.016c is small noug tat rlativistic cts can b nglctd. 66. Pictur t Problm: Bcaus t proton is signiicantly mor massiv tan an lctron, it will av a gratr momntum wn t proton and lctron av t sam spd. T d Brogli wavlngt is invrsly proportional to t particl momntum. Stratgy: Calculat t ratio o t d Brogli wavlngts using quation 30-16, wr t momntum is t mass tims vlocity. Solution: 1. (a) Bcaus mvand m, m p or idntical spds, an lctron as a longr d Brogli wavlngt tan a proton.. (b) Calculat t ratio o wavlngts: mv m 7 p kg 31 p mv p m kg Insigt: In tis problm w assumd a classical spd so tat w could us t classical momntum quation. Howvr, bcaus t rlativistic actor dpnds only upon t spd, t ratio o t wavlngts would still b 1840 at rlativistic spds

16 68. Pictur t Problm: Wn a prson walks troug a doorway, w typically do not s t prson diract. Howvr, i t prson s d Brogli wavlngt wr on t ordr o t door widt, t diraction would b obsrvabl. Stratgy: Solv quation or t spd o a prson wos d Brogli wavlngt quals t widt o t door. For part (b), divid on millimtr by t vlocity to dtrmin t tim lapsd. Solution: 1. (a) Solv q or t prson s spd: p mv J s v m 0.76 m 65 kg m/s x m 31. (b) Calculat t tim to travl 1.0 mm: t s 35 v m/s 14 Insigt: Tis tim is about.0 10 tims longr tan t ag o t univrs. T doorway would not xist long noug to obsrv t diraction! In rality, t prson is rally a collction o many particls (atoms), ac o wic is moving rapidly in trmal motion and as a d Brogli wavlngt tat is smallr tan an atom. It would b impossibl to slow down vry atom in t prson s body to a spd as small as tat ound in part (a). 70. Pictur t Problm: According to t Hisnbrg uncrtainty principl, t product o t uncrtainty in position and uncrtainty in momntum must b gratr tan Planck s constant dividd by. Stratgy: T 5.0% uncrtainty in spd mans tat t magnitud o t momntum (mass tims spd) o t lctron and t basball ar also uncrtain by 5.0%. T minimum uncrtainty in position is givn by quation Solution: 1. Solv t uncrtainty principl y or t uncrtainty in position: py. St p mv: y 3. Calculat t uncrtainty or t basball: 4. Calculat t uncrtainty or t lctron: y 0.050mvy J s ybasball kg 41 m/s J s lctron 31 y kg 41 m/s m 57 m Insigt: T minimum uncrtainty in t position o t basball is smallr tan an atom by a actor o Howvr, t minimum uncrtainty in t position o t lctron is masurabl, rougly t ticknss o a uman air. 7. Pictur t Problm: Wn t position o a cart is masurd wit a spciic accuracy, t Hisnbrg uncrtainty principl rquirs tat t momntum must also av a minimum uncrtainty. Stratgy: Writ t momntum in quation in trms o t mass and uncrtainty in vlocity and solv t inquality or t uncrtainty in vlocity. Solution: 1. Writ quation wit py m vy: pyy mvyy. Solv or t uncrtainty in t vlocity: J s vy my 0.6 kg 0.00 m m/s Insigt: At tis spd it would tak t cart ovr a billion yars to mov t diamtr o a proton! T uncrtainty 30 16

17 principl is not signiicant or larg objcts. 74. Pictur t Problm: Wn t nrgy is masurd to witin a givn uncrtainty, t tim cannot b known to witin a gratr uncrtainty tan is stipulatd by t Hisnbrg uncrtainty principl. Stratgy: Solv quation 30-0 or t minimum uncrtainty in t tim. Solution: Calculat t minimum uncrtainty in tim: Et t V J/V J s min 19 Emax s Insigt: As t nrgy is masurd wit gratr prcision, t tim at wic t particl ad tat givn nrgy is known wit lss crtainty. It is impossibl to simultanously masur bot nrgy and tim wit arbitrary crtainty. 76. Pictur t Problm: T litim o t particl limits t tim ovr wic an nrgy masurmnt can b takn. T Hisnbrg uncrtainty principl corrlats tis maximum uncrtainty in t tim masurmnt wit a minimum uncrtainty in t nrgy masurmnt. Stratgy: Solv quation 30-0 or t minimum uncrtainty in t particl s nrgy. Solution: Calculat t minimum uncrtainty in nrgy: Et E J s min 10 tmax.510 s J Insigt: Bcaus t minimum uncrtainty in nrgy is invrsly proportional to t man litim o t particl, particls wit smallr litims will ncssarily av gratr uncrtaintis in nrgy

18 78. Pictur t Problm: Wn a proton s location is known to witin an uncrtainty o 0.15 nm, it will av a minimum uncrtainty in momntum as rquird by t uncrtainty principl. Tat minimum uncrtainty rquirs t proton to av a nonzro kintic nrgy. Stratgy: Us quation to calculat t minimum uncrtainty p in t momntum, and tn us p = p to ind t kintic nrgy o t proton according to t kintic nrgy quation ( K p m). Solution: 1. (a) Solv quation or t minimum uncrtainty in momntum:. (b) Calculat t minimum kintic nrgy: px J s pmin 9 x m kg m/s kg m/s p K m kg J/V 0.9 mv Insigt: Wn w compar t rsults o tis problm wit problm 77 (wic concrnd an lctron conind wit t sam x), w s tat t uncrtainty in momntum or t lctron and proton ar t sam. T minimum kintic nrgy o t lctron, owvr, is 1840 tims gratr tan t minimum kintic nrgy o t proton. 80. Pictur t Problm: You prorm an xprimnt on t potolctric ct using ligt wit a rquncy ig noug to jct lctrons. T intnsity o t ligt is incrasd wil t rquncy is ld constant. Stratgy: Rcall t principls tat govrn t potolctric ct wn answring t concptual qustions. Not tat incrasing t intnsity wil olding t rquncy constant simply mans tat mor potons (ac wit t sam original nrgy) strik t mtal surac pr scond. Solution: 1. (a) Bcaus t nrgis o t potons stay t sam wn t rquncy is ld constant, t maximum kintic nrgy o an jctd lctron will stay t sam.. (b) Bcaus t nrgis o t potons stay t sam wn t rquncy is ld constant, t maximum kintic nrgy o an jctd lctron and tror its minimum d Brogli wavlngt will stay t sam. 3. (c) Bcaus t numbr o potons will incras wn t intnsity o t ligt is incrasd, t numbr o lctrons jctd pr scond will incras. 4. (d) Bcaus t numbr o potons will incras wn t intnsity o t ligt is incrasd, t numbr o lctrons jctd pr scond and tror t currnt in t pototub will incras. Insigt: T maximum kintic nrgy o an jctd lctron will incras only i t rquncy o t ligt is incrasd. 8. Pictur t Problm: An lctron tat is acclratd rom rst troug a potntial dirnc V 0 as a d Brogli wavlngt 0. Stratgy: Us t xprssions p (quation 30-16), K p m (Concptual Cckpoint 9-3), and K U qv (quation 0-) to answr t concptual qustion. Solution: To doubl t lctron s wavlngt w must alv its momntum. Tis mans, in turn, tat w must dcras its kintic nrgy by a actor o our (rcall tat K = p /m). T kintic nrgy is linarly proportional to t potntial dirnc. Tror, t lctron sould b acclratd rom rst troug t potntial dirnc V 0 /4. Insigt: W can also com to tis conclusion using a ratio: V K q K p m p V0 V V0 K0 q K0 p0 m p

19 84. Pictur t Problm: In ordr to construct a potocll tat works on visibl ligt, t nrgy rom a poton in t visibl spctrum must b suicint to jct an lctron rom t mtal. Stratgy: Us quations 30-6 and 14-1 to calculat t maximum wavlngt tat a poton can av so tat it is abl to jct lctrons rom ac mtal. Compar t rsults wit t sortst wavlngt o visibl ligt (400 nm). Solution: 1. Dtrmin t max rquird to jct lctrons rom ac mtal surac: 34 8 W J/V c c J s m/s 140 V nm max 19 0 W0 W0 140 V nm. Calculat t max or aluminum: max 90 nm 4.8 V 140 V nm 3. Calculat t max or lad: max 9 nm 4.5 V 140 V nm 4. Calculat t max or csium: max 579 nm.14 V 5. O t tr matrials, csium is t only matrial tat will jct lctrons rom visibl ligt. Insigt: Rd ligt will not jct lctrons rom t surac bcaus t maximum wavlngt or csium is 579 nm, wic is in t yllow portion o t spctrum. T potocll will work or visibl ligt o wavlngts btwn 579 nm (yllow) and 400 nm (violt). 86. Pictur t Problm: T lngt o a pndulum dtrmins t rquncy at wic it oscillats. T nrgy o a typical pndulum oscillation can b considrd as a multipl o t quantum oscillation. Stratgy: Calculat t rquncy o t oscillation using quations 13-1 and Multiply t rquncy by Planck s constant and t quantum numbr n to calculat t nrgy o t oscillation. St t nrgy qual to t kintic nrgy and solv or t vlocity. Solution: 1. (a) Calculat t oscillation rquncy:. (b) St t oscillation nrgy qual to t kintic nrgy and 1 1 g m/s 0.56 Hz T L 0.78 m E n mv solv or t spd: n J s 0.56 Hz 1 v m 0.15 kg. m/s Insigt: T n = 1 quantum stat corrsponds to t minimum nrgy stat or tis pndulum oscillator. T maximum spd o t pndulum bob in tis stat would b m/s. 88. Pictur t Problm: As ic absorbs potons o ligt, t nrgy is convrtd into at and mlts t ic. Stratgy: Us quation 30-4 to calculat t nrgy o on poton. Divid t nrgy ndd to mlt 1.0-kg o ic ( Q ml) by t nrgy o on poton to calculat t numbr o potons ndd. To calculat t numbr o ic molculs tat on poton will mlt, divid t numbr o ic molculs in on kilogram by t numbr o potons to mlt t kilogram o ic. T numbr o watr molculs is qual to Avogadro s numbr tims 1 kilogram dividd by t atomic mass o watr (18 g/mol). Solution: 1. (a) Calculat E t nrgy in on poton:. Divid t at by t poton nrgy: poton J s Hz J 1.0 kg80.0 kcal/kg4186 J/kcal ml N E poton potons J 30 19

20 3. (b) Divid t numbr o molculs by t numbr o potons: NAm M N NAm NM molculs/mol1.0 kg potons0.018 kg/mol 40 molculs/poton Insigt: Wn a poton is absorbd by a watr molcul, t intrmolcular bonds tat attac t molcul to t otr molculs in t lattic can b brokn. T molcul will tn av suicint rmaining kintic nrgy tat it can collid wit 39 otr molculs and brak tir bonds as wll. 30 0

21 90. Pictur t Problm: As watr absorbs microwav potons, t nrgy is convrtd into at and raiss t tmpratur o t watr. Stratgy: Us quations 30-4 and 14-1 to calculat t nrgy o ac poton. Us quation to ind t nrgy rquird to at t watr, and divid tis nrgy by t nrgy o on poton in ordr to calculat t numbr o potons ndd to at t watr. Solution: 1. Calculat t poton nrgy: E poton J s m/s c 0.1 m. Calculat t at addd to t watr: Q mct 0.05 kg4186 J/kg C90.0C 0.0C 3. Divid t at by t nrgy pr poton: J 4 Q J N E J poton potons J Insigt: I a magniying glass tat is 3.00 incs in diamtr capturs t ntir solar spctrum on a day in wic t solar intnsity is 1000 W/m, it would tak 3.66 ours to cang t tmpratur o t sam mass o watr by t sam amount. 9. Pictur t Problm: T d Brogli wavlngt o t lctron and t wavlngt o t poton ar bot invrsly proportional to tir momnta. Stratgy: Us quation to calculat t d Brogli wavlngt o t lctron, wr t momntum is givn by quation 9-1. Calculat t wavlngt o t poton using quation 30-11, stting t momntum o t poton qual to t momntum o t lctron. Solution: 1. (a) Find t d Brogli wavlngt: J s p m v kg.710 m/s 0.7 nm. (b) Calculat t wavlngt o t poton: 0.7 nm p p Insigt: Any objct (lctron, poton, proton, prson, tc.) wit t sam momntum will av t sam wavlngt. 94. Pictur t Problm: Ligt o a givn rquncy and nrgy will jct lctrons rom t surac o t mtal. Part o t nrgy rom t poton ovrcoms t work unction o t mtal. T rmaindr o t nrgy gos into t kintic nrgy o t lctron. 1 Stratgy: (a) It is not ncssary to us rlativistic mcanics bcaus v << c. Substitut mv or max K max in t quation Kmax c W0 and solv or W. Tn calculat t cuto rquncy using 0 0 W0. Solution: 1. (b) Solv q.30-7 or t work unction:. (b) Us quation 30-6 to calculat t cuto rquncy: c Kmax W0 c c 1 W0 Kmax mvmax J s m/s m kg m/s W J 1 V J.01 V W J s J Hz

22 Insigt: Tis cuto rquncy alls in t orang part o t visibl spctrum. Tror, ssntially t ntir visibl spctrum (xcpt rd ligt) will jct lctrons rom t mtal. 30

23 96. Pictur t Problm: As t atoms mrg rom an ovn, tir most probabl spd is proportional to tir most probabl momntum, wic is invrsly proportional to tir most probabl wavlngt. Stratgy: Writ t most probabl wavlngt o t atoms in trms o t momntum using quation and quation 9-1. Solv t rsulting quation or t most probabl spd. Solution: 1. Writ quation in trms o t vlocity: mp p mv mkt mp mp J/K450 K. Solv or t most probabl spd: 5kT vmp 7 m kg Insigt: T most probabl spd incrass in proportion to t squar root o t Klvin tmpratur km/s 98. Pictur t Problm: Hlium atoms in a jar at room tmpratur and atmospric prssur ar otn considrd to bav lik point particls. Quantum cts btwn t particls ar gnrally ignord. Tis problm tsts t assumption tat quantum cts ar small or t intraction btwn lium atoms. Stratgy: Us quation 17-1 to calculat t kintic nrgy o t lium atoms. Writ t kintic nrgy in trms o t momntum. Tn using quation writ t momntum in trms o t wavlngt. Solv t rsulting quation or t avrag d Brogli wavlngt o t atoms. Solv t idal gas law (quation 17-) or t volum pr atom, and tn tak t cub root o t volum pr atom to calculat t avrag sparation distanc. Solution: 1. (a) Writ t kintic nrgy p in trms o and st it qual to 3 kt : K. Solv or t wavlngt st 3 av kt m m m J s av 3mkT kg J/K K in trms o t tmpratur: 3. (b) Us t idal gas law to calculat t volum occupid pr atom: nm PV NkT V kt J/K95.15 K m 5 N P Pa 4. Tak t cub root o t volum to V davg m 3.44 nm calculat t avrag sparation: N Insigt: T d Brogli wavlngt o t lium atom is 47 tims smallr tan t sparation distanc btwn atoms. Tror, quantum cts ar not a signiicant actor in dtrmining t bavior o t atoms in tis lium gas Pictur t Problm: Wn a poton is absorbd by t surac o a mtal it will jct an lctron. T kintic nrgy o t jctd lctron will qual t dirnc btwn t poton nrgy and t work unction o t mtal. Wn potons o t sam rquncy ar incidnt on two mtals o dirnt work unctions, t jctd lctrons will av dirnt maximum kintic nrgis. Stratgy: Solv quation 30-7 or t work unction. Tn subtract t two work unctions to sow tat t dirnc in kintic nrgis o t jctd lctrons is qual to t dirnc in work unctions or any incidnt rquncy. Solution: 1. (a) T dirnc in maximum kintic nrgy obsrvd rom t two suracs is du to t dirnc in t work unctions o t two matrials. Bcaus t maximum kintic nrgy dpnds linarly upon t work unction, and bcaus t work unction dos not dpnd upon t rquncy o t ligt, t dirnc in maximum kintic nrgy will stay t sam.. (b) Solv quation 30-8 Kmax W0 or t work unction: W K 0 max

24 3. Subtract t work unctions o t two mtals: W W K K 0B 0A max,b max,a J Kmax,A Kmax,B 1.5 V J/V Insigt: As long as bot work unctions ar blow 3.41 V, t ligt will jct lctrons rom bot suracs wos kintic nrgis dir by1.5 V. I on o t work unctions wr gratr tan 3.41 V, tat surac would not jct potolctrons. 10. Pictur t Problm: Wn potons o dirnt wavlngts sin on a litium surac, ty jct lctrons wit kintic nrgis qual to t dirnc btwn t nrgy o t poton and t work unction o litium. Stratgy: Insrt t data rom on o t xprimnts (and t calculatd work unction rom t prvious qustion) into t quation or c and solv or Planck s constant m Solution: Solv or Planck s constant: W0 Kmax 8.89 V V c m/s V s J/V J s Insigt: Millikan was abl to calculat Planck s constant to witin 1% o its prsntly accptd valu using ts two xprimnts Pictur t Problm: T imag sows a nm poton scattring o o an lctron. Atr t collision, t lctron as a kintic nrgy o V and t poton scattrs at an angl. Stratgy: Us quations 0-4 and 14-1 to calculat t initial nrgy o t poton. Subtract t kintic nrgy o t lctron rom t poton s nrgy to calculat t inal nrgy o t poton, and tn us quation 0-4 to ind t inal wavlngt. Tn us quation to calculat t scattring angl. A rcurring valu in similar Compton scattring problms is 1 mc.4610 m nm. Solution: 1. (a) In tis problm t poton as transrrd lss nrgy to t lctron tan it did in Exampl 30-4, so t scattring angl will b lss tan 15.. (b) Calculat t initial nrgy o t poton: J s3.010 m/s nm J/V E c V 3. Calculat t inal nrgy o t poton: E EK V V V 4. Calculat t inal wavlngt o t poton: 5. Solv quation or t scattring angl: J s3.010 m/s V1.610 J/V c E nm 1 cos 1 mc nm nm cos nm Insigt: T scattring angl o t poton incrass as t momntum o t lctron incrass. Tror, i t kintic nrgy o t lctron dcrass, t momntum o t lctron also dcrass, and t scattring angl o t poton also dcrass. 30 4