Test Bank for Global Business, 4th Edition

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. λ=2440Å=0.244µm Vo = hν – Φ = Φ =4.09eV 1.24eV ⋅ µm 1.24eV ⋅ µm -Φ = – 4.09eV = 5.08eV – 4.09eV ≈ 1eV λ(µm) 0.244µm Prob. 2.2 Show third Bohr postulate equates to integer number of DeBroglie waves fitting within circumference of a Bohr circular orbit. 4π∈o n 2 h2 q2 mv 2 rn = and = and pθ = mvr mq 2 4π∈o r 2 r rn = 4π∈o n 2 h2 n 2 h2 4π∈o rn 2 n 2 h2 rn n 2 h2 = ⋅ = ⋅ = 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 pθ = nh is the third Bohr postulate Prob. 2.3 (a) Find generic equation for Lyman, Balmer, and Paschen series. ΔE = mq 4 mq 4 hc = λ 32π 2∈o 2 n12 h2 32π 2∈o 2 n 2 2 h2 mq 4 (n 2 2 – n12 ) mq 4 (n 2 2 – n12 ) hc = = λ 32∈o 2 n12 n 2 2 h2 π 2 8∈o 2 n12 n 2 2 h 2 λ= 8∈o 2 n12 n 2 2 h 2 ⋅ hc 8ε o 2 h3c n12 n 2 2 ⋅ 2 2 = mq 4 (n 2 2 – n12 ) mq 4 n 2 – n1 λ= ⋅ ⋅ 10-12 mF ) 2 ⋅ (6.63 ⋅10−34 J⋅s)3⋅ 2.998 ⋅ 108 ms n12 n 2 2 8(8.85 ⋅ 2 2 9.11 ⋅ 10-31kg ⋅ (1.60 ⋅ 10-19 C)4 n 2 – n1 λ = 9.11 ⋅108 m ⋅ n12 n 2 2 n 12 n 2 2 ≈ ⋅ = 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 911Ǻ 911*4*n^2/(n^2-4) 6559 4859 4338 4100 3968 3644Ǻ 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 8199Ǻ Prob. 2.4 (a) Find Δpx for Δx=1Ǻ. h h 6.63 ⋅10-34 J ⋅ s Δpx ⋅ Δx = → Δp x = = = 5.03 ⋅10-25 kgs⋅m -10 4π 4π ⋅ Δx 4π ⋅10 m (b) Find Δt for ΔE=1eV. h h 4.14 ⋅10-15eV ⋅ s ΔE ⋅ Δt = → Δt = = = 3.30 ⋅10-16s 4π 4π ⋅ ΔE 4π ⋅1eV Prob. 2.5 Find wavelength of 100eV and 12keV electrons. Comment on electron microscopes compared to visible light microscopes. 2⋅E m h = 2⋅E⋅m E = 12 mv 2 → v = h h = = p mv For 100eV, -1 6.63 ⋅10-34 J ⋅ 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) λ= 2 ⋅ 9.11⋅10-31kg -1 -1 ⋅ E 2 = E 2 ⋅ 4.91⋅10-19 J 2 ⋅ m -1 1 λ = E 2 ⋅ 4.91⋅10-19 J 2 ⋅ m = (100eV ⋅1.602 ⋅10-19 eVJ ) 2 ⋅ 4.91⋅10-19J 2 ⋅ m = 1.23 ⋅10-10 m = 1.23≈ 1 For 12keV, -1 1 -1 λ = E 2 ⋅ 4.91⋅10-19 J 2 ⋅ m = (1.2 ⋅104eV ⋅1.602 ⋅10-19 eVJ ) 2 ⋅ 4.91⋅10-19J 2 ⋅ m = 1.12 ⋅10-11m = 0.112≈ 1 1 The resolution on a visible microscope is dependent on the wavelength of the light which is around 5000Ǻ; 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) Ψ (x) = C for all x. B) Ψ (x) = C for values of x between 2 and 8 cm, and Ψ (x) = 3.5 C for values of x between 5 and 10 cm. Ψ (x) is zero everywhere else. C) Ψ (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 ≤ 2 and x ≥ 10 is consistent with this? ∞ A) For a wavefunction Ψ (x) , we know Ρ = ∫ Ψ * (x)Ψ (x)dx = 1 -∞ ⎧ 0 c = 0 Ρ = ∫ Ψ * (x)Ψ (x)dx = c2 ∫ dx → Ρ= ⎨ ⇒ Ψ (x) cannot be a wave function ⎩∞ c ≠ 0 -∞ -∞ ∞ ∞ B) For 5 ≤ x ≤ 8 , Ψ (x) has two values, C and 3.5C. For c ≠ 0 , Ψ (x) is not a function ∞ and for c = 0 : Ρ = ∫ Ψ * (x)Ψ (x)dx = 0 ⇒ Ψ (x) cannot be a wave function. -∞ ⎧ iC ⎪⎪ 3 ( x-2 ) 2 ≤ x ≤ 5 C) Ψ (x)= ⎨ ⎪− iC ( x-10 ) 5 ≤ x ≤ 10 ⎪⎩ 5 ∞ 5 10 c2 c2 2 2 ( x-2 ) dx + ∫ ( x-10 ) dx 9 25 2 5 -∞ 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) Ρ = ∫ Ψ* (x)Ψ(x)dx = ∫ 5 10 c2 c2 (x-2)3 ⎤⎦ + (x-10)3 ⎤⎦ 2 5 3×9 3×25 2 ⎡ 27 125 ⎤ 8c = c 2 ⎢ + = ⎥ 3 ⎣ 27 3×25 ⎦ = Ρ=1 ⇒ 8c2 =1 ⇒ c=0.612 → Ψ (x) can be a wave function 3 Since Ψ (x) = 0 for x ≤ 2 and x ≥ 10 , the potential energy should be infinite in these two regions. Prob. 2.7 A particle is described in 1D by a wavefunction: Ψ = Be-2x for x ≥0 and Ce+4x for x<0, and B and C are real constants. Calculate B and C to make Ψ a valid wavefunction. Where is the particle most likely to be? A valid wavefunction must be continuous, and normalized. For Ψ (0) = C = B ∞ 2 To normalize Ψ , ∫ Ψ dx = 1 -∞ 0 ∞ 2 8x 2 -4x ∫ C e dx + ∫ C e dx = 1 -∞ 0 2 ∞ C 8x ⎛ −1 ⎞ ⎡⎣e ⎤⎦ + C2 ⎜ ⎟ ⎡⎣e-4x ⎤⎦ = 1 −∞ 0 8 ⎝ 4 ⎠ 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 ⇒ 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? Ψ = Ceikx 22 * 2 -1 1 cm 20 ∫ Ψ Ψdx = C (20) = 1 ⇒ C = 2 4 2 1 ⎛ 1 ⎞ Probability = ∫ Ψ dx = ⎜ ⎟ ( 2 ) = 10 ⎝ 20 ⎠ 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 (|ψ|2=0). However, it is possible for electrons to tunnel through the barrier at 5<x6 would be quantum mechanically greater than zero (|ψ|2>0) and classical mechanically zero. Prob. 2.10 Find 4 ⋅ px 2 + 2 ⋅ pz 2 + 7mE for Ψ( x, y, z, t ) = A ⋅ e j (10⋅x +3⋅ y -4⋅t ) . * px ∫ A ⋅e 2 = -∞ – j(10⋅x+3⋅ y-4⋅t) ⎛ h ∂ ⎞ ⎜ j ∂x ⎟ A ⋅ e ⎝ ⎠ ∞ 2 ∫A e – j(10⋅x+3⋅ y-4⋅t) -∞ ∞ * pz ∫ A ⋅e 2 = -∞ ∫A e – j(10⋅x+3⋅ y-4⋅t) -∞ * ∫ A ⋅e E = -∞ = 100 ⋅ h2 e j(10⋅ x+3⋅ y – 4⋅ t) j(10⋅ x+3⋅ y -4⋅t) dz =0 dz h ∂ ⎞ j(10⋅x+3⋅ y – 4⋅ t) dt ⎜ − j ∂t ⎟ A ⋅ e ⎝ ⎠ = 4⋅h – j(10⋅x+3⋅ y-4⋅t) ⎛ ∞ dx 2 ⎜ j ∂z ⎟ A ⋅ e ⎝ ⎠ 2 j(10⋅ x+3⋅ y – 4⋅ t) e j(10⋅x+3⋅y-4⋅t) dx – j(10⋅x+3⋅ y-4⋅t) ⎛ h ∂ ⎞ ∞ ∞ 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) ∞ 2 ∫A e – j(10⋅x+3⋅ y-4⋅t) e j(10⋅x+3⋅y-4⋅t) dt -∞ 4 ⋅ p x +2 ⋅ p z +7 mE = 400h2 + 28(9.11 ⋅10-31kg) h 2 2 Prob. 2.11 Find the uncertainty in position (Δx) and momentum (Δρ). L 2 ⎛ πx ⎞ ⋅ sin ⎜ ⎟ ⋅ e-2πjEt/h and ∫ Ψ* ⋅ Ψdx = 1 L ⎝ L ⎠ 0 Ψ(x,t) = L L x = ∫ Ψ * ⋅ x ⋅ Ψdx = 0 2 ⎛ π x ⎞ x ⋅ sin 2 ⎜ ⎟ dx = 0.5L (from problem note) ∫ L0 ⎝ L ⎠ L L 2 ⎛ π x ⎞ 2 = ∫ Ψ ⋅ x ⋅ Ψdx = ∫ x 2 ⋅ sin 2 ⎜ ⎟ dx = 0.28L (from problem note) L0 ⎝ L ⎠ 0 * 2 Δx = x2 – x Δp ≥ h h = 0.47 ⋅ 4π ⋅ Δx 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 10Ǻ quantum well with infinite walls. En = n 2 ⋅ π 2 ⋅ h2 (6.63 ⋅10-34 ) 2 = ⋅ n 2 = 6.03 ⋅10-20 ⋅ n 2 −31 −9 2 2 2⋅m⋅L 8 ⋅ 9.11⋅10 ⋅ (10 ) E1 = 6.03 ⋅10-20 J = 0.377eV E 2 = 4 ⋅ 0.377eV = 1.508eV E 3 = 9 ⋅ 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.