# Test Bank for Global Business, 4th Edition

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Chapter 2 Solutions Prob. 2.1 (a&b) Sketch a vacuum tube device. Graph photocurrent I versus retarding voltage V for several light intensities. I light intensity Vo V T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b) Note that Vo remains same for all intensities. (c) Find retarding potential. ฮป=2440ร=0.244ยตm Vo = hฮฝ – ฮฆ = ฮฆ =4.09eV 1.24eV โ ยตm 1.24eV โ ยตm -ฮฆ = – 4.09eV = 5.08eV – 4.09eV โ 1eV ฮป(ยตm) 0.244ยตm Prob. 2.2 Show third Bohr postulate equates to integer number of DeBroglie waves fitting within circumference of a Bohr circular orbit. 4ฯโo n 2 h2 q2 mv 2 rn = and = and pฮธ = mvr mq 2 4ฯโo r 2 r rn = 4ฯโo n 2 h2 n 2 h2 4ฯโo rn 2 n 2 h2 rn n 2 h2 = โ = โ = mq 2 mrB 2 q 2 mrn 2 mv 2 m 2 v 2 rn m 2 v 2 rn 2 = n 2 h2 mvrn = nh pฮธ = nh is the third Bohr postulate Prob. 2.3 (a) Find generic equation for Lyman, Balmer, and Paschen series. ฮE = mq 4 mq 4 hc = ฮป 32ฯ 2โo 2 n12 h2 32ฯ 2โo 2 n 2 2 h2 mq 4 (n 2 2 – n12 ) mq 4 (n 2 2 – n12 ) hc = = ฮป 32โo 2 n12 n 2 2 h2 ฯ 2 8โo 2 n12 n 2 2 h 2 ฮป= 8โo 2 n12 n 2 2 h 2 โ hc 8ฮต o 2 h3c n12 n 2 2 โ 2 2 = mq 4 (n 2 2 – n12 ) mq 4 n 2 – n1 ฮป= โ โ 10-12 mF ) 2 โ (6.63 โ10โ34 Jโs)3โ 2.998 โ 108 ms n12 n 2 2 8(8.85 โ 2 2 9.11 โ 10-31kg โ (1.60 โ 10-19 C)4 n 2 – n1 ฮป = 9.11 โ108 m โ n12 n 2 2 n 12 n 2 2 โ โ = 9.11 n 2 2 – n12 n 2 2 – n12 T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b) n1 =1 for Lyman, 2 for Balmer, and 3 for Paschen (b) Plot wavelength versus n for Lyman, Balmer, and Paschen series. n 2 3 4 5 n^2 4 9 16 25 LYMAN SERIES n^2-1 n^2/(n^2-1) 3 1.33 8 1.13 15 1.07 24 1.04 LYMAN LIMIT n 3 4 5 6 7 n^2 9 16 25 36 49 BALMER SERIES n^2-4 4n^2/(n^2-4) 5 7.20 12 5.33 21 4.76 32 4.50 45 4.36 BALMER LIMIT 911*n^2/(n^2-1) 1215 1025 972 949 911วบ 911*4*n^2/(n^2-4) 6559 4859 4338 4100 3968 3644วบ n 4 5 6 7 8 9 10 n^2 16 25 36 49 64 81 100 PASCHEN SERIES n^2-9 9*n^2/(n^2-9) 7 20.57 16 14.06 27 12.00 40 11.03 55 10.47 72 10.13 91 9.89 PASCHEN LIMIT 911*9*n^2/(n^2-9) 18741 12811 10932 10044 9541 9224 9010 8199วบ Prob. 2.4 (a) Find ฮpx for ฮx=1วบ. h h 6.63 โ10-34 J โ s ฮpx โ ฮx = โ ฮp x = = = 5.03 โ10-25 kgsโm -10 4ฯ 4ฯ โ ฮx 4ฯ โ10 m (b) Find ฮt for ฮE=1eV. h h 4.14 โ10-15eV โ s ฮE โ ฮt = โ ฮt = = = 3.30 โ10-16s 4ฯ 4ฯ โ ฮE 4ฯ โ1eV Prob. 2.5 Find wavelength of 100eV and 12keV electrons. Comment on electron microscopes compared to visible light microscopes. 2โE m h = 2โEโm E = 12 mv 2 โ v = h h = = p mv For 100eV, -1 6.63 โ10-34 J โ s T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b) ฮป= 2 โ 9.11โ10-31kg -1 -1 โ E 2 = E 2 โ 4.91โ10-19 J 2 โ m -1 1 ฮป = E 2 โ 4.91โ10-19 J 2 โ m = (100eV โ1.602 โ10-19 eVJ ) 2 โ 4.91โ10-19J 2 โ m = 1.23 โ10-10 m = 1.23โ 1 For 12keV, -1 1 -1 ฮป = E 2 โ 4.91โ10-19 J 2 โ m = (1.2 โ104eV โ1.602 โ10-19 eVJ ) 2 โ 4.91โ10-19J 2 โ m = 1.12 โ10-11m = 0.112โ 1 1 The resolution on a visible microscope is dependent on the wavelength of the light which is around 5000วบ; so, the much smaller electron wavelengths provide much better resolution. Prob. 2.6 Which of the following could NOT possibly be wave functions and why? Assume 1-D in each case. (Here i= imaginary number, C is a normalization constant) A) ฮจ (x) = C for all x. B) ฮจ (x) = C for values of x between 2 and 8 cm, and ฮจ (x) = 3.5 C for values of x between 5 and 10 cm. ฮจ (x) is zero everywhere else. C) ฮจ (x) = i C for x= 5 cm, and linearly goes down to zero at x= 2 and x = 10 cm from this peak value, and is zero for all other x. If any of these are valid wavefunctions, calculate C for those case(s). What potential energy for x โค 2 and x โฅ 10 is consistent with this? โ A) For a wavefunction ฮจ (x) , we know ฮก = โซ ฮจ * (x)ฮจ (x)dx = 1 -โ โง๏ฃฑ 0 c = 0 ฮก = โซ ฮจ * (x)ฮจ (x)dx = c2 โซ dx โ ฮก= โจ๏ฃฒ โ ฮจ (x) cannot be a wave function โฉ๏ฃณโ c โ  0 -โ -โ โ โ B) For 5 โค x โค 8 , ฮจ (x) has two values, C and 3.5C. For c โ  0 , ฮจ (x) is not a function โ and for c = 0 : ฮก = โซ ฮจ * (x)ฮจ (x)dx = 0 โ ฮจ (x) cannot be a wave function. -โ โง๏ฃฑ iC โช๏ฃดโช๏ฃด 3 ( x-2 ) 2 โค x โค 5 C) ฮจ (x)= โจ๏ฃฒ โช๏ฃดโ iC ( x-10 ) 5 โค x โค 10 โช๏ฃดโฉ๏ฃณ 5 โ 5 10 c2 c2 2 2 ( x-2 ) dx + โซ ( x-10 ) dx 9 25 2 5 -โ T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b) ฮก = โซ ฮจ* (x)ฮจ(x)dx = โซ 5 10 c2 c2 (x-2)3 โค๏ฃนโฆ๏ฃป + (x-10)3 โค๏ฃนโฆ๏ฃป 2 5 3ร9 3ร25 2 โก๏ฃฎ 27 125 โค๏ฃน 8c = c 2 โข๏ฃฏ + = โฅ๏ฃบ 3 โฃ๏ฃฐ 27 3ร25 โฆ๏ฃป = ฮก=1 โ 8c2 =1 โ c=0.612 โ ฮจ (x) can be a wave function 3 Since ฮจ (x) = 0 for x โค 2 and x โฅ 10 , the potential energy should be infinite in these two regions. Prob. 2.7 A particle is described in 1D by a wavefunction: ฮจ = Be-2x for x โฅ0 and Ce+4x for x<0, and B and C are real constants. Calculate B and C to make ฮจ a valid wavefunction. Where is the particle most likely to be? A valid wavefunction must be continuous, and normalized. For ฮจ (0) = C = B โ 2 To normalize ฮจ , โซ ฮจ dx = 1 -โ 0 โ 2 8x 2 -4x โซ C e dx + โซ C e dx = 1 -โ 0 2 โ C 8x โ๏ฃซ โ1 โ๏ฃถ โก๏ฃฎโฃ๏ฃฐe โค๏ฃนโฆ๏ฃป + C2 โ๏ฃฌ โ๏ฃท โก๏ฃฎโฃ๏ฃฐe-4x โค๏ฃนโฆ๏ฃป = 1 โโ 0 8 โ๏ฃญ 4 โ ๏ฃธ 0 T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b) C2 C2 8 + =1 โ C= 8 4 3 Prob. 2.8 The electron wavefunction is Ceikx between x=2 and 22 cm, and zero everywhere else. What is the value of C? What is the probability of finding the electron between x=0 and 4 cm? ฮจ = Ceikx 22 * 2 -1 1 cm 20 โซ ฮจ ฮจdx = C (20) = 1 โ C = 2 4 2 1 โ๏ฃซ 1 โ๏ฃถ Probability = โซ ฮจ dx = โ๏ฃฌ โ๏ฃท ( 2 ) = 10 โ๏ฃญ 20 โ ๏ฃธ 0 2 Prob. 2.9 Find the probability of finding an electron at x0 zero or non-zero? Is the classical probability of finding an electron at x>6 zero or non? The energy barrier at x=0 is infinite; so, there is zero probability of finding an electron at x<0 (|ฯ|2=0). However, it is possible for electrons to tunnel through the barrier at 5<x6 would be quantum mechanically greater than zero (|ฯ|2>0) and classical mechanically zero. Prob. 2.10 Find 4 โ px 2 + 2 โ pz 2 + 7mE for ฮจ( x, y, z, t ) = A โ e j (10โx +3โ y -4โt ) . * px โซ A โe 2 = -โ – j(10โx+3โ y-4โt) โ๏ฃซ h โ โ๏ฃถ โ๏ฃฌ j โx โ๏ฃท A โ e โ๏ฃญ โ ๏ฃธ โ 2 โซA e – j(10โx+3โ y-4โt) -โ โ * pz โซ A โe 2 = -โ โซA e – j(10โx+3โ y-4โt) -โ * โซ A โe E = -โ = 100 โ h2 e j(10โ x+3โ y – 4โ t) j(10โ x+3โ y -4โt) dz =0 dz h โ โ๏ฃถ j(10โx+3โ y – 4โ t) dt โ๏ฃฌ โ j โt โ๏ฃท A โ e โ๏ฃญ โ ๏ฃธ = 4โh – j(10โx+3โ y-4โt) โ๏ฃซ โ dx 2 โ๏ฃฌ j โz โ๏ฃท A โ e โ๏ฃญ โ ๏ฃธ 2 j(10โ x+3โ y – 4โ t) e j(10โx+3โy-4โt) dx – j(10โx+3โ y-4โt) โ๏ฃซ h โ โ๏ฃถ โ โ 2 T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b) โ 2 โซA e – j(10โx+3โ y-4โt) e j(10โx+3โy-4โt) dt -โ 4 โ p x +2 โ p z +7 mE = 400h2 + 28(9.11 โ10-31kg) h 2 2 Prob. 2.11 Find the uncertainty in position (ฮx) and momentum (ฮฯ). L 2 โ๏ฃซ ฯx โ๏ฃถ โ sin โ๏ฃฌ โ๏ฃท โ e-2ฯjEt/h and โซ ฮจ* โ ฮจdx = 1 L โ๏ฃญ L โ ๏ฃธ 0 ฮจ(x,t) = L L x = โซ ฮจ * โ x โ ฮจdx = 0 2 โ๏ฃซ ฯ x โ๏ฃถ x โ sin 2 โ๏ฃฌ โ๏ฃท dx = 0.5L (from problem note) โซ L0 โ๏ฃญ L โ ๏ฃธ L L 2 โ๏ฃซ ฯ x โ๏ฃถ 2 = โซ ฮจ โ x โ ฮจdx = โซ x 2 โ sin 2 โ๏ฃฌ โ๏ฃท dx = 0.28L (from problem note) L0 โ๏ฃญ L โ ๏ฃธ 0 * 2 ฮx = x2 – x ฮp โฅ h h = 0.47 โ 4ฯ โ ฮx L Prob. 2.12 = 0.28L2 – (0.5L) 2 = 0.17L T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g. in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n . We or b) x 2 Calculate the first three energy levels for a 10วบ quantum well with infinite walls. En = n 2 โ ฯ 2 โ h2 (6.63 โ10-34 ) 2 = โ n 2 = 6.03 โ10-20 โ n 2 โ31 โ9 2 2 2โmโL 8 โ 9.11โ10 โ (10 ) E1 = 6.03 โ10-20 J = 0.377eV E 2 = 4 โ 0.377eV = 1.508eV E 3 = 9 โ 0.377eV = 3.393eV Prob. 2.13 Show schematic of atom with 1s22s22p4 and atomic weight 21. Comment on its reactivity. nucleus with 8 protons and 13 neutrons 2 electrons in 1s 2 electrons in 2s 4 electrons in 2p = proton = neturon = electron This atom is chemically reactive because the outer 2p shell is not full. It will tend to try to add two electrons to that outer shell.

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### Test Bank for Global Business, 4th Edition

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