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

Transcription

1 Phyi 07 Problem 2.5 O. A. Pringle := h := := := h = Joule Note I had to et the zero tolerane here. e := ev <-> joule onverion ator ev := e ev =.776 ev Phyi 07 Problem 2.6 O. A. Pringle := h h := := h ev := ev e := := ev e h = Hz 22 := o := := = meter in nanometer, nm := 0 9 nm = Phyi 07 Problem 2.7 O. A. Pringle h := photon energy := h Power i energy per time. Photon per eond i energy per time divided by energy per photon. P:= 0 3 P n := n = photon per eond Note that a long a I keep everything in mk unit, the unit automatially work out OK.

2 Phyi 07 Problem 2.8 O. A. Pringle h := := 0 8 := := photon := h photon = N := photon N = 3.07 The eye an detet 3 o thee photon. Note again that unit work out OK i we tik with the mk ytem. Phyi 07 Problem 2.9 O. A. Pringle (a) How many photon all per eond on eah quare meter o the earth' urae diretly aing the un? h := photon := 50 4 on eah quare meter: P := a in problem 2.5, per eond i power divided by energy n = photon/ n := P h photon (b) What i the power output o the un, and how many photon per eond doe it emit? The power per area at a radiu o.5x0 m i given a.4x0 3 W/m 2. To anwer the irt part, jut multiply the power per area by the area o a phere o the given radiu. r :=.5 0 A:= 4 π r 2 P un := PA P un = watt N := P un h photon N = photon per eond emitted by the un () How many photon per ubi meter are there near the earth? := will need thi in a minute Thi i mainly a unit onverion problem. Look at the unit. := volume := volume denity := ρ := n 2 volume meter peed m 2 area ρ = photon per ubi meter 2

3 Phyi 07 Problem 2.0 O. A. Pringle t := P := 0.5 := h := := The energy o a pule i the power time the time. pule := Pt The o photon in the pule i the energy o the pule divided by the energy o a photon. N := 0 := pule h N = Phyi 07 Problem 2. O. A. Pringle The equation to ue i K max := h h 0 h := energie will ome out in ev i I ue h := in thee unit 0 := K max :=.5 Solve or K max + h 0 h = here' the requeny; not neeary, jut wanted to look at it := = Thi i a wavelength o 80 nm. Phyi 07 Problem 2.2 O. A. Pringle h := will give K in ev We are given the ollowing: 0 := Simply plug thee into equation 2. K max := h h 0 K max =.656 ev 3

4 The equation we will ue i Phyi 07 Problem 2.3 O. A. Pringle K max := h h 0 where h := will give u energie in ev Beaue =h and =/, longer wavelength o light have lower energie. Thi problem i equivalent to aking "what i the minimum requeny o light that will aue photoeletron to be emitted rom odium." That minimum requeny i jut the threhold requeny or odium, whih an be ound rom the work untion. φ := 2.3 ev, rom table 2. := Thi i jut the minimun energy needed to produe a photoeletron, o φ := h 0 or φ 0 := 0 = Hz h The maximum wavelength i jut, uing 0 := 0 0 = meter, or 540 nm (to get Beier' anwer, ue =2.998x0 8 and h=4.36x0-5 ) What will the maximum kineti energy o the photoeletron be i 200-nm light all on a odium urae? Thi i jut like problem 2.0, exept we are given wavelength intead o requenie. 0 = alulated above := h h K max := K max = 3.9eV 0 Phyi 07 Problem 2.4 O. A. Pringle Thi ound triky but really in't. Light inident on the ball will aue photoeletron to be emitted. The ball will aquire a poitive harge and thereore an eletrial potential a the eletron are emitted. When work untion plu the potential o the ball equal the energy o the inident light, no more eletron will be emitted. For ilver φ := 4.7 ev Plank' ontant h := ev* requeny o inident light: := := = Hz In word: inident energy=v+φ V:= h φ V =.5Volt 4

5 Phyi 07 Problem 2.5 O. A. Pringle The.5 mw (milliwatt) give the energy per unit time in the inident light beam. P := joule/ 400-nm tell u the photon energy. h := I'll work in mk unit here. := := h photon := photon = joule per photon Dividing the power (energy per time) by the energy o a photon (energy per photon) give u the o photon per eond inident on the ell. P N := photon N = photon per eond Only 0. perent o thee photon produe photoeletron, o the n produing photoeletron i N n := 000 n = eletron produed per eond Thi n i atually the urrent, but we hould expre it in the more amiliar unit o oulomb e := oulomb per eletron I:= n e I = oulomb per eond, or amp Phyi 07 Problem 2.6 O. A. Pringle a) Find the extintion voltage, that i, the retarding voltage at whih the photoeletron urrent diappear. The extintion voltage our when the retarding voltage plu the work untion equal the photoeletron energy. φ + ev ext := h Thi i an energy equation. The energy o an h := := φ := 2.50 e eletron aquire when it pae through a voltage equal to the extintion voltage i ev ext. := ( h φ) V ext := e V ext = volt b) Find the peed o the atet photoeletron. The mot energeti eletron will appear when V.ext=0, i.e., ( ) h φ K max := expre energy in ev or omparion with part a) e K max = eletron volt; ye, the ame a in part a xperiene hould tell u that 0.6 ev i nonrelativiti; we will do a nonrelativiti alulation, and i the peed i too great, go bak and do a relativiti alulation. K max 2 m 2 := eletron v max e :=

6 K max := K max e onvert to Joule! m eletron := K max v max := 2 m eletron v max = mall enough to be nonrelativiti Phyi 07 Problem 2.7 O. A. Pringle We are given light o requenie. and.2, and the maximum kineti energie K. and K.2 whih they produe. We want to ind the experimental value or h and φ. h := K + φ Remember, the little quare mean thee are ymboli equation only. h 2 := K 2 + φ It' very eay to olve thee imultaneouly or h. h := K + φ h 2 := K 2 φ ( ) h 2 := K K 2 h := K K 2 2 In a ouple o line, I'm going to opy thi ymboli equation (F2), pate it below (F4), and turn it "on" with "eq". Now plug in K :=.97 K 2 := 0.52 := := K K 2 Thi value i in unit o ev*. h := h = You an eaily onvert it to J*. 2 Now ue the value or h to olve either equation at the tart or φ. Ue Cut and Pate (F2 and F4) again. Now olve or φ. h := K + φ φ := h K φ = 3.00 Sine h i in ev*, unit o φ are ev Phyi 07 Problem 2.8 O. A. Pringle To olve thi, imply take the equation K max := h φ and olve it or h. K max :=.7 φ := 5.4 6

7 := := K max + φ h := h = ev* 5 The atual value o h in thee unit i 4.36x0-5. δ := Phyi 07 Problem 2.5 O. A. Pringle GM 2 R In mk unit, G := M := R := := := the wavelength o the light give the requeny o the light δ GM 2 R δ = The new requeny i le than the original requeny, o I'm going to work thi problem the traightorward brute trength way. Nothing triky, but it involve onverion bak and orth between and. You might ave ome time by doing the algebra irt on paper. prime := δ Now alulate the new wavelength: prime := prime prime = The red hit i the amount by whih the wavelength hanged: δ := prime δ = meter Or δ= nm. 7

8 Phyi 07 Problem 2.52 O. A. Pringle Thi i jut problem 2-5 with dierent. I will ue my 2-5 olution diretly here. In mk unit, G := M := R := := := the wavelength o the light give the requeny o the light GM δ 2 δ =.39 0 R The new requeny i le than the original requeny, o prime := δ Now alulate the new wavelength: prime := prime prime = The red hit i the amount by whih the wavelength hanged: δ := prime δ =.58 0 meter 0 Or δ=0.6 nm. Phyi 07 Problem 2.54 O. A. Pringle Find the Shwarzhild radiu o the un. G := M := := G M R S := 2 R S = meter I the un had le than thi radiu, it would be a blak hole. 8

9 Phyi 07 ( Gm M) U := R and m tart with a kineti energy o mv^2/2, o our initial total energy i ( Gm M) + R Ater m ha jut eaped rom M, m ha ued up all o it kineti energy in eaping, o it kineti energy i zero. I m had exatly enough energy to jut eape, it won't have eaped until it reahed a ditane o r=. At r= the gravitational potential energy i zero (/r=0 there), o the total energy i zero. Thereore ( Gm M) + R 2 m v2 = "Divide both ide by m, and olve or v to get v := 2G M R Problem m v2 O. A. Pringle The gravitational potential energy U relative to ininity o a body o ma m at a ditane R rom the enter o a body o ma M i U=-GmM/R. (a) I R i the radiu o the body o ma M, ind the eape peed v.e o the body (preumably Beier mean the body o ma m), whih i the minimum peed needed to leave it permanently. To eape, the "body o ma m" mut have enough kineti energy to overome the gravitational attration o the body o ma M. Let me all m the mall body and M the large body. The mall body tart at the urae o the large body, where the gravitational potential energy i (b) Obtain a ormula or the Shwarzhild radiu o the body by etting v.e=, the peed o light, and olving or r. := 2G M R Square both ide, olve or R to get R := 2G M 2 9