Chapter 2 Solutions. Prob. 2.1 (a&b) Sketch a vacuum tube device. Graph photocurrent I versus retarding voltage V for several light intensities.

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1 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

2 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 o 1 o 1 h 8 h h 8ε h = = mq ( ) mq o 1 o ( ) (6.6 1 Js) = kg (1.6 1 C) 1 1 F 4 8 m m s = m = 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) YMAN IMIT 911Ǻ BAMER SERIES ^ ^4 4^/(^4) 9114^/(^4) PASCHEN IMIT 8199Ǻ BAMER IMIT 644Ǻ 15 Pearso Eduatio, I., Hoboke, NJ. All rights reserved. This material is proteted uder all opyright laws as they urretly

3 Prob..4 (a) Fid Δp x for Δx=1Ǻ. 4 h h Js Δp Δx = Δp = = = Δx 4π1 m 5 kgm x x s (b) Fid Δt for ΔE=1eV. 15 h h 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 Js p mv Em kg λ = = = = E = E J m For 1eV, J 19 1 λ = E J m = (1eV ) J m = 1. 1 m = 1. Å ev For 1keV, J λ = E J m = (1. 1 ev1.6 1 ) J m = 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

4 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 Ρ = (x) (x)dx = x dx + x1 dx 5 = (x) + (x1) = + = 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

5 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 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 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

6 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( kg) 1 x z 15 Pearso Eduatio, I., Hoboke, NJ. All rights reserved. This material is proteted uder all opyright laws as they urretly

7 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 = = = m (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