Physics 107 Problem 2.5 O. A. Pringle h Physics 107 Problem 2.6 O. A. Pringle

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1 Pysis 07 Problem 25 O A Pringle = Joules ote I ad to set te zero tolerane ere e ev <-> joules onversion ator ev e ev = 776 ev Pysis 07 Problem 26 O A Pringle ev ev e ev e = Hz so = meters in nanometers, nm nm = Pysis 07 Problem 27 O A Pringle poton energy Power is energy per time Potons per seond is energy per time divided by energy per poton P 0 3 P n n = potons per seond ote tat as long as I keep everyting in mks units, te units automatially work out OK

2 Pysis 07 Problem 28 O A Pringle poton poton = poton = 307 Te eye an detet 3 o tese potons ote again tat units work out OK i we stik wit te mks system Pysis 07 Problem 29 O A Pringle (a) How many potons all per seond on ea square meter o te eart's surae diretly aing te sun? poton 50 4 on ea square meter: P P as in problem 25, number per seond is power divided by energy n n = potons/s poton (b) Wat is te power output o te sun, and ow many potons per seond does it emit? Te power per area at a radius o 5*0^ m is given as 4x0^3 W/m^2 To answer te irst part, just multiply te power per area by te area o a spere o te given radius r 5 0 A 4 π r 2 P sun PA P sun = watts P sun poton = potons per seond emitted by te sun () How many potons per ubi meter are tere near te eart? 3 will need tis in a minute Tis is mainly a unit onversion problem Look at te units number number s = volume s volume number number s = volume s m m 2 number density = s speed area ρ n 2 ρ = potons per ubi meter 2

3 Pysis 07 Problem 20 O A Pringle t P Te energy o a pulse is te power times te time pulse Pt Te number o potons in te pulse is te energy o te pulse divided by te energy o a poton pulse = Pysis 07 Problem 2 O A Pringle Te equation to use is 0 energies will ome out in ev i I use in tese units Solve or 0 = ere's te requeny; not neessary, just wanted to look at it = Tis is a wavelengt o 80 nm Pysis 07 Problem 22 O A Pringle will give K in ev We are given te ollowing: Simply plug tese into equation 2 0 = 656 ev 3

4 Pysis 07 Problem 23 O A Pringle Te equation we will use is 0 were will give us energies in ev Beause = and =/, longer wavelengts o ligt ave lower energies Tis problem is equivalent to asking "wat is te minimum requeny o ligt tat will ause potoeletrons to be emitted rom sodium" Tat minimum requeny is just te tresold requeny or sodium, wi an be ound rom te work untion φ 23 ev, rom table 2 3 Tis is just te minimun energy needed to produe a potoeletron, so φ 0 or φ 0 0 = Hz Te maximum wavelengt is just, using = meters, or 540 nm (to get Beiser's answer, use =2998*0^8 and =436*0^-5) Wat will te maximum kineti energy o te potoeletrons be i 200-nm ligt alls on a sodium surae? Tis is just like problem 20, exept we are given wavelengts instead o requenies 0 = alulated above 200 = 39eV 0 Pysis 07 Problem 24 O A Pringle Tis sounds triky but really isn't Ligt inident on te ball will ause potoeletrons to be emitted Te ball will aquire a positive arge and tereore an eletrial potential as te eletrons are emitted Wen work untion plus te potential o te ball equals te energy o te inident ligt, no more eletrons will be emitted For silver φ 47 ev Plank's onstant ev*s requeny o inident ligt: = Hz In words: inident energy=v+φ V φ V = 5Volts 4

5 Pysis 07 Problem 25 O A Pringle Te 5 mw (milliwatts) gives te energy per unit time in te inident ligt beam P joules/s 400-nm tells us te poton energy I'll work in mks units ere poton poton = joules per poton Dividing te power (energy per time) by te energy o a poton (energy per poton) gives us te number o potons per seond inident on te ell P poton = potons per seond Only 0 perent o tese potons produe potoeletrons, so te number n produing potoeletrons is n 000 n = eletrons produed per seond Tis n is atually te urrent, but we sould express it in te more amiliar units o oulombs e oulombs per eletron I n e I = oulombs per seond, or amps Pysis 07 Problem 26 O A Pringle a) Find te extintion voltage, tat is, te retarding voltage at wi te potoeletron urrent disappears Te extintion voltage ours wen te retarding voltage plus te work untion equal te potoeletron energy φ V ext φ V ext φ V ext = 0605eletron volts b) Find te speed o te astest potoeletrons Te most energeti eletrons will appear wen Vext=0, ie, φ = 0605 eletron volts; yes, te same number as in part a xperiene sould tell us tat 06 ev is nonrelativisti; we will do a nonrelativisti alulation, and i te speed is too great, go bak and do a relativisti alulation K max 2 m 2 eletron v max e

6 K max e onvert to Joules! m eletron v max 2 m eletron v max = small enoug to be nonrelativisti Pysis 07 Problem 27 O A Pringle We are given ligt o requenies and 2, and te maximum kineti energies K and K2 wi tey produe We want to ind te experimental value or and φ K φ Remember, te little square means tese are symboli equations only 2 K 2 φ It's very easy to solve tese simultaneously or K φ 2 K 2 φ K K 2 K K 2 2 In a ouple o lines, I'm going to opy tis symboli equation (F2), paste it below (F4), and turn it "on" wit "eq" ow plug in numbers K 97 K K K 2 = Tis value is in units o ev*s You an easily onvert it to J*s ow use te value or to solve eiter equation at te start or φ Use Cut and Paste (F2 and F4) again ow solve or φ K φ φ K φ = 300 Sine is in ev*s, units o φ are ev Pysis 07 Problem 28 O A Pringle To solve tis, simply take te equation φ and solve it or 7 φ 54 6

7 75 3 φ = ev*s Te atual value o in tese units is 436x0-5 δ Pysis 07 Problem 25 O A Pringle GM 2 R In mks units, G M R te wavelengt o te ligt δ gives te requeny o te ligt GM 2 R I'm going to work tis problem te straigtorward brute strengt way oting triky, but it involves onversions bak and ort between and You migt save some time by doing te algebra irst on paper δ = 27 Te new requeny is less tan te original requeny, so prime δ ow alulate te new wavelengt: prime prime prime = Te red sit is te amount by wi te wavelengt anged: δ prime δ = meters Or δ=00006 nm 7

8 Pysis 07 Problem 252 O A Pringle Tis is just problem 2-5 wit dierent numbers I will use my 2-5 solution diretly ere In mks units, G M R te wavelengt o te ligt gives te requeny o te ligt δ GM δ = R Te new requeny is less tan te original requeny, so prime δ ow alulate te new wavelengt: prime prime = prime Te red sit is te amount by wi te wavelengt anged: δ prime δ = meters Or δ=06 nm Pysis 07 Problem 254 O A Pringle Find te Swarzsild radius o te sun G M R S 2G M 2 R S = meters I te sun ad less tan tis radius, it would be a blak ole 8

9 Pysis 07 Problem 255 O A Pringle Te gravitational potential energy U relative to ininity o a body o mass m at a distane R rom te enter o a body o mass M is U=-GmM/R (a) I R is te radius o te body o mass M, ind te esape speed ve o te body (presumably Beiser means te body o mass m), wi is te minimum speed needed to leave it permanently To esape, te "body o mass m" must ave enoug kineti energy to overome te gravitational attration o te body o mass M Let me all m te small body and M te large body Te small body starts at te surae o te large body, were te gravitational potential energy is ( Gm M) U R and m starts wit a kineti energy o mv^2/2, so our initial total energy is ( Gm M) R 2 m v 2 Ater m as just esaped rom M, m as used up all o its kineti energy in esaping, so its kineti energy is zero I m ad exatly enoug energy to just esape, it won't ave esaped until it reaed a distane o r= At r= te gravitational potential energy is zero (/r=0 tere), so te total energy is zero Tereore ( Gm M) R 2 m v 2 = 0 "Divide bot sides by m, and solve or v to get v 2G M R (b) Obtain a ormula or te Swarzsild radius o te body by setting ve=, te speed o ligt, and solving or r 2G M R Square bot sides, solve or R to get R 2G M 2 9