Chapter Solutios Prob..1 (a&b) Sketh a vauum tube devie. Graph photourret I versus retardig voltage V for several light itesities. I light itesity V o V Note that V o remais same for all itesities. () Fid retardig potetial. λ=44å=.44μm =4.9eV 1.4eV μm 1.4eV μm V o = hν Φ = = 4.9eV = 5.8eV 4.9eV 1eV λ(μm).44μm Prob.. Show third Bohr postulate equates to iteger umber of DeBroglie waves fittig withi irumferee of a Bohr irular orbit. 4 q mv r = ad = ad p = mvr o mq 4π or r 4 4πr r r = = = = o o mq mrb q mr mv m v r m v r = mvr = p = is the third Bohr postulate 15 Pearso Eduatio, I., Hoboke, NJ. All rights reserved. This material is proteted uder all opyright laws as they urretly
Prob.. (a) Fid geeri equatio for yma, Balmer, ad Pashe series. 4 4 h mq mq ΔE = = λ π π o 1 o h mq ( ) mq ( ) = = λ π 8 4 4 1 1 o 1 o 1 h 8 h h 8ε h = = mq ( ) mq o 1 o 1 4 4 1 1 8(8.85 1 ) (6.6 1 Js).998 1 = 9.11 1 kg (1.6 1 C) 1 1 F 4 8 m m s 1 1 19 4 1 = 9.111 m = 9.11 8 1 1 Å 1 1 =1 for yma, for Balmer, ad for Pashe (b) Plot wavelegth versus for yma, Balmer, ad Pashe series. YMAN SERIES PASCHEN SERIES ^ ^1 ^/(^1) 911^/(^1) ^ ^9 9^/(^9) 9119^/(^9) 4 1. 115 4 16 7.57 18741 9 8 1.1 15 5 5 16 14.6 1811 4 16 15 1.7 97 6 6 7 1. 19 5 5 4 1.4 949 7 49 4 11. 144 YMAN IMIT 911Ǻ 8 64 55 1.47 9541 9 81 7 1.1 94 1 1 91 9.89 91 BAMER SERIES ^ ^4 4^/(^4) 9114^/(^4) PASCHEN IMIT 8199Ǻ 9 5 7. 6559 4 16 1 5. 4859 5 5 1 4.76 48 6 6 4.5 41 7 49 45 4.6 968 BAMER IMIT 644Ǻ 15 Pearso Eduatio, I., Hoboke, NJ. All rights reserved. This material is proteted uder all opyright laws as they urretly
Prob..4 (a) Fid Δp x for Δx=1Ǻ. 4 h h 6.6 1 Js Δp Δx = Δp = = = 5. 1 1 4 4Δx 4π1 m 5 kgm x x s (b) Fid Δt for ΔE=1eV. 15 h h 4.14 1 evs 16 ΔE Δt = Δt = = =. 1 s 4 4ΔE 4π 1eV Prob..5 Fid wavelegth of 1eV ad 1keV eletros. Commet o eletro mirosopes ompared to visible light mirosopes. 1 E = mv v = E m h h h 4 6.61 Js p mv Em 1 9.111 kg 1 1 1 19 λ = = = = E = E 4.911 J m For 1eV, 1 1 1 1 19 19 J 19 1 λ = E 4.91 1 J m = (1eV 1.6 1 ) 4.91 1 J m = 1. 1 m = 1. Å ev For 1keV, 1 1 1 1 19 4 19 J 19 11 λ = E 4.91 1 J m = (1. 1 ev1.6 1 ) 4.91 1 J m = 1.1 1 m =.11 Å The resolutio o a visible mirosope is depedet o the wavelegth of the light whih is aroud 5Ǻ; so, the muh smaller eletro wavelegths provide muh better resolutio. Prob..6 Whih of the followig ould NOT possibly be wave futios ad why? Assume 1D i eah ase. (Here i= imagiary umber, C is a ormalizatio ostat) A) Ψ (x) = C for all x. B) Ψ (x) = C for values of x betwee ad 8 m, ad Ψ (x) =.5 C for values of x betwee 5 ad 1 m. Ψ (x) is zero everywhere else. C) Ψ (x) = i C for x= 5 m, ad liearly goes dow to zero at x= ad x = 1 m from this peak value, ad is zero for all other x. If ay of these are valid wavefutios, alulate C for those ase(s). What potetial eergy for x ad x 1 is osistet with this? ev 15 Pearso Eduatio, I., Hoboke, NJ. All rights reserved. This material is proteted uder all opyright laws as they urretly
A) For a wavefutio (x), we kow Ρ = (x) (x)dx = 1 Ρ = (x) (x)dx = dx Ρ= (x) aot be a wave futio = B) For 5x 8, (x) has two values, C ad.5c. For, (x) is ot a futio ad for = : Ρ = (x) (x)dx = (x) aot be a wave futio. C) ic x x 5 (x)= ic x1 5 x 1 5 9 5 5 1 Ρ = (x) (x)dx = x dx + x1 dx 5 = (x) + (x1) 9 5 7 15 8 = + = 7 5 5 1 5 8 Ρ = 1 =1 =.61 (x) a be a wave futio Sie (x) = for x ad x 1, the potetial eergy should be ifiite i these two regios. 15 Pearso Eduatio, I., Hoboke, NJ. All rights reserved. This material is proteted uder all opyright laws as they urretly
Prob..7 A partile is desribed i 1D by a wavefutio: Ψ = Be x for x ad Ce +4x for x<, ad B ad C are real ostats. Calulate B ad C to make Ψ a valid wavefutio. Where is the partile most likely to be? A valid wavefutio must be otiuous, ad ormalized. For () = C = B To ormalize, dx = 1 8x 4x C e dx + C e dx = 1 C 8x 1 4x e + C e 1 8 4 C C 8 + = 1 C = 8 4 Prob..8 The eletro wavefutio is Ce ikx betwee x= ad m, ad zero everywhere else. What is the value of C? What is the probability of fidig the eletro betwee x= ad 4 m? ikx = Ce 1 dx = C () = 1 C = m 4 1 1 Probability = dx = = 1 1 Prob..9 Fid the probability of fidig a eletro at x<. Is the probability of fidig a eletro at x> zero or ozero? Is the lassial probability of fidig a eletro at x>6 zero or o? The eergy barrier at x= is ifiite; so, there is zero probability of fidig a eletro at x< ( ψ =). However, it is possible for eletros to tuel through the barrier at 5<x<6; so, the probability of fidig a eletro at x>6 would be quatum mehaially greater 15 Pearso Eduatio, I., Hoboke, NJ. All rights reserved. This material is proteted uder all opyright laws as they urretly
tha zero ( ψ >) ad lassial mehaially zero. Prob..1 Fid 4 x z 7 p p me for j(1 x y4 t ) ( x, y, z, t) A e. j(1x+ y 4t) j(1x+ y 4t) x j(1x+ y 4t) j(1x+ y4 t) z A e A e dx j x p = = 1 A e e dx p = j(1x+ y 4t) j(1x+ y 4t) A e A e dz j z = j(1x+ y4t) j(1x+ y4t) A e e dz j(1x+ y4t) j(1 x+ y4t) A e A e dt j t E = = 4 j(1x+ y 4t) j(1x+ y 4t) A e e dt 4p + p +7 me = 4 8(9.11 1 kg) 1 x z 15 Pearso Eduatio, I., Hoboke, NJ. All rights reserved. This material is proteted uder all opyright laws as they urretly
Prob..11 Fid the uertaity i positio (Δx) ad mometum (Δρ). πx πjet/ h Ψ(x,t) = si e ad dx = 1 x x = xdx = xsi dx =.5 (from problem ote) x x = xdx = x si dx =.8 (from problem ote) Δx = x h h p =.47 4π Δx x =.8 (.5) =.17 Prob..1 Calulate the first three eergy levels for a 1Ǻ quatum well with ifiite walls. 4 π (6.6 1 ) E = = = 6.1 1 9 m 89.111 (1 ) E 1 = 6.1 J =.77eV E = 4.77eV = 1.58eV E = 9.77eV =.9eV Prob..1 Show shemati of atom with 1s s p 4 ad atomi weight 1. Commet o its reativity. uleus with 8 protos ad 1 eutros eletros i 1s eletros i s This atom is hemially reative beause the outer p shell is ot full. It will ted to try to add two eletros to that outer shell. 4 eletros i p = proto = eturo = eletro 15 Pearso Eduatio, I., Hoboke, NJ. All rights reserved. This material is proteted uder all opyright laws as they urretly