CYL Practice Harmonic Oscillat
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"-v~ •.• PC1f U",",J..C IU cue~(a[e 8.14(b) t .onfirrn that the wavefuncuon for the first excited state of a n = 2 in a square-well potential. one-dimensional linear harmonic oscillator given in Table 8,1 is a solution of the Schrodinger equation for the oscillator and that its energy is t1iro. ~) An electron is confined to a a square well oflength L. What would be the length of the box such that the zero-point energy of the electron is equal 8.1S(a) Locate the nodes of the harmonic oscillator wave function with to its rest mass energy, m.c2? Express your answer in terms of the parameter v=4. A.c = hlm.c, the 'Compton wavelength' of the electron. 8.1S(b) Locate the nodes of the harmonic oscillator wavefunction with 8.S(b) Repeat Exercise 8.5a for a general particle of mass m in a cubic box. v=5.
8.6(a) What are the most likely locations of a particle in a box oflength L in 8.16(a) What are the most probable displacements of a harmonic oscillator the state n = 3? withv=l?
8.6(b) What are the most likely locations of a particle in a box oflength L in 8.16(b) What are the most probable displacements of a harmonic oscillator the state n = 5? with v=3?
8.7(a) Calculate the percentage change in a given energy level of a particle in a 8.17(a) Assuming that the vibrations of a ,sCl2 molecule are equivalent to one-dimensional box when the length of the box is increased by 10 per cent. those of a harmonic oscillator with a force constant k = 329 N m-I, what is the zero-point energy of vibration of this molecule? The effective mass 8.7(b) Calculate the percentage change in a given energy level of a particle in a of a homonuclear diatomic molecule is half its total mass, and cubic box when the length of the edge of the cube is decreased by 10 per cent mesCl) = 34.9688711u' in each direction. 8.17(b) Assuming that the vibrations of a 14N2molecule are equivalent to 8.8(a) What is the value of n of a particle in a one-dimensional box such that those of a harmonic oscillator with a force constant k= 2293.8 N m ", the separation between neighbouring levels is equal to the energy of thermal what is the zero-point energy of vibration of this molecule? The effective motion (ikT). mass of a homonuclear diatomic molecule is half its total mass, and m(l4N) = 14.0031m ' 8.8(b) A nitrogen molecule is confined in a cubic box of volume 1.00 m'. u Assuming that the molecule has an energy equal to tkTat T= 300 K, what is 8.18(a). The wavefunction, 1If(¢), for the motion of a particle in a ring is of the the value of n = (n; + n: + n:)I12 for this molecule? What is the energy form 1If=Neim;. Determine the normalization constant, N. separation between the levels n and n + I? What is its de Broglie wavelength? 8.18(b) Confirm that wavefunctions for a particle in a ring with different 8.9{a) Calculate the zero-point energy of a harmonic oscillator consisting values of the quantum number ml are mutually orthogonal.
of a particle of mass 2.33 x 10-26 kg and force constant 155 N m-I. 8.19(a) Calculate the minimum excitation energy of a proton constrained to 8.9(b) Calculate the zero-point energy of a harmonic oscillator consisting rotate in a circle of radius 100 pm around a fixed point. of a particle of mass 5.16 X 10-26 kg and force constant 285 N m-I. 8.19(b) Calculate the value of 1mII for the system described in the preceding 8.10(a) For a certain harmonic oscillator of effective mass 1.33 X 10-25kg, the exercise corresponding to a rotational energy equal to the classical average difference in adjacent energy levels is 4.82 z]. Calculate the force constant of energy at 25°C (which is equal to ikT). the oscillator. 8.2O(a) Estimate the rotational quantum number of a bicycle wheel of 8.10(b) For a certain harmonic oscillator of effective mass 2.88 x 10-25 kg, diameter 60 cm and mass 1.0 kg when the bicycle is travelling at 20 km h-I. the difference in adjacent energy levels is 3.17 z], Calculate the force constant 8.20(b) The mass of a vinyl gramophone record is 130 g and its diameter is of the oscillator. 30 ern. Given that the moment of inertia of a solid uniform disc of mass m 8.11 (a) Calculate the wavelength of a photon needed to excite a transition and radius r is I =imr2, estimate the rotational quantum number when between neighbouring energy levels of a harmonic oscillator of effective the disc is rotating at 33 r.p.m. I mass equal to that of a proton (1.0078mu) and force constant 855 N m- . 8.21(a) The moment of inertia of a CH4 molecule is 5.27 x 10-47kg m2. 8.11(b) Calculate the wavelength of a photon needed to excite a transition What is the minimum energy needed to start it rotating? between neighbouring energy levels of a harmonic oscillator of effective mass 8.21(b) The moment of inertia ofan SF6 molecule is 3.07 X 10-45kg nr'. equal to that ofan oxygen atom (15.9949m ) and force constant 544 N m-I. u What is the minimum energy needed to start it rotating? 8.12(a) The vibrational frequency of H, is 131.9 THz. What is the vibrational 8.22(a) Use the data in Exercise 8.21a to calculate the energy needed to frequency of O2 (0 = 2H)? excite a CH4 molecule from a state with 1= I to a state with 1= 2. 8.12(b) The vibrational frequency of H, is 131.9 THz. What is the vibrational 8.22(b) Use the data in Exercise 8.21b to calculate the energy needed to frequency ofT2 (T = 'H)? excite an SF6 molecule from a state with 1= 2 to a state with 1= 3. 8.13(a) Calculate the minimum excitation energies of (a) a pendulum of 8.23(a) What is the magnitude of the angular momentum of a CH. molecule length 1.0 m on the surface of the Earth, (b) the balance-wheel of a clockwork when it is rotating with its minimum energy? watch (v= 5 Hz).
8.23(b) What is the magnitude of the angular momentum of an SF6 molecule 8.13(b) Calculate the minimum excitation energies of (a) the 33 kHz quartz when it is rotating with its minimum energy? crystal of a watch, (b) the bond between two °atoms in 02' for which • I 1 kr= 1177 N m-I. 8.24(a) Draw scale vector diagrams to represent the states (a) 5=2' m'=+2' (b) 1= 1, ml=+I, (c) 1=2, ml=O. 8.14(a) Confirm that the wavefunction for the ground state of a one- dimensional linear harmonic oscillator given in Table 8.1 is a solution of the 8.24(b) Draw the vector diagram for all the permitted states of a particle Schrodinger equation for the oscillator and that its energy is inca with 1=6.
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QUANTUM THEORY: TECHNIQUES AND APPLICATIONS 183
Hence, we require the values of n I, nz, and n3 that make
2 2 2 9 nl + n2 + n3 = . Therefore, (nl, 112,n3) = (1, 2, 2), (2, 1,2), and (2, 2, 1) and the degeneracy is [D
II n2 3 Question. What is the smallest multiple of the lowest energy, EI for which E,,, n does not exist?
E9.7(a) h2) K 2 2 2 (~) E = (nT + n~ + n5) x ( 8mL2 = L2' K = (nl + n2 + n3) x 8m .
6.E _ (KI(0.9L)2) - (KIL2) = _1_ -1 =10.23~ or 123percentl. IF - KIL2 0.81
1) 1/2 ~ E = v + 2: lu», co =;;;(k) [9.25]. ~~'t[~) ( The zero-point energy corresponds to v =·0; hence
1/2 11k 1/2 1 155Nm-1 Eo = =tu» = -Ii (-) = (-) x (1.055 x 10-34 J s) x 26 22m 2 ( 2.33 x 10- kg )
= 14.30 x 10-21 J I.
1/2 ~ 6.E = Ev+1 - E; = tuo [9.26] = Ii;;; k) [9.25]. '6 "(OUt-') (
6.E)2 ( 4.82 x 10-21 Hence k = m ( - = (1.33 x 10-25 kg) x 34 J)2 = I278 N m-I I[11 = 1 N m]. Ii l.055 x 10- J s
The requirement for a transition to occur is that 6.E(system) = E(photon), so J) ~~t\t.~ 6.E(system) = tuo [9.26], he E(photon) = h» = -. A
Therefore, -he = -hco = (h-) x ( -k ) 1/2 A 2IT 2IT m
27 In) 1/2 8 (1.673 x 10- kg) 1/2 A = 2ITe ( - = (2n) x (2.998 x 10 ms-I) x I k 855Nm- = 2.63 x 10-6 m = 12.63 Il-m,. ,/2.0....
Since A $S'·C3a... 1/2 J ) (a) co= (T) [elementary physics]. 6.E = Iua [harmonic oscillator level separations, 9.26] 981 -2) 1/2 = (l.055 x 10-34 J s) x . m s = 13.3 x 10-34 J I. ( l.Om 184 STUDENT'S SOLUTIONS MANUAL (b) !)'E =; h» = (6.626 X 10-34 JI1z ~)':i<:~?:'~)= 13.3 x 10-33 J I· ~) The Schrodinger equation for the linear harmonic oscillator is glttt~) !i2 d21/f 1 - 2m d.x2 + 2kx21/f = E1/f [9.24]. The ground-state wavefunction is 2 1/fo = Noe-x2/2a [9.29aJ with a = (If) 1/4 = (~) 1/4 k=£ (a) mk m2{j)2 ma4 . Performing the operations d!O = ( _a~x) 1/fo, 2 --d 1/fo = ( --x1) x ( --x1 ) x 1/fo + (1)--1/fo = -x21/fo --~"'1f{io = (X~- 1/fo, d.x2 a2 a2 a2 a4 a2 a4 - :2) and substituting into the Schrodinger equation, 2 -- n (X2- - - 1 ) 1/fo + =1kx 2 1/fo = E01/l0 2m a4 a2 2 which implies 2 Eo = _n (X2 _~) + ~ (b) 2m a4 a2 - But Eo is a constant, independent of x; therefore the terms that contain x must drop out, which is possible only if n2 1 ---+-k=O 2ma4 2 which is consistent with k = n2/ma4 as in (a). What is left in (b) is 2 2 n 1 k) 1/2 n ] Eo = -- =-nw using (j) = -; and k = ma . 2ma2 2 [ ( 4 1 Therefore, 1/fo is a solution of the Schrodinger equation with energy 2nw. gllS-(~) ~) The harmonic oscillator wavefunctions have the form 1 ) (n2) 1/4 1/fv(x) = NvHv(y) exp ( -2i with y = ~ and a = mk [9.28J. The exponential function approaches zero only as x approaches ±oo, so the nodes of the wavefunction are the nodes of the Hermite polynomials. H4(y) = 16i - 48i + 12 = 0 [Table 9.1J. QUANTUM THEORY: TECHNIQUES AND APPLICATIONS 185 Dividing through by 4 and letting z = i,we have a quadratic equation 2 4z - 12z + 3 = 0 -b ±.Jb2 - 4ac 12 ± Jl22 - 4 x 4 x 3 3 ±.J6 so z = 2a 2x4 2 Evaluating the result numerically yields z = 0.275 or 2.72, so y = ±0.525 or ±1.65. Therefore x =1 ± 0.525a or ± 1.65a I· COMMENT. Numerical values could also be obtained graphically by plotting H4(y). ~'I1-COv) ~) For a two-particle oscillator meff replaces m in the expression for co, where I I 1 -=-+-. meff m\ rn (See Chapter 13 for a more complete discussion of the vibration of a diatomic molecule.) Here ml = mo ; so meff = m12. Thus Eo = ~1Uu = ~n(~)1/2 = ~n(2k)1/2 2 2 meff 2 m mcJ5Cl) = 34.96888 u = (34.96888 u) x (1.66054 x 10-27 kg/u) = 5.807 x 10-26 kg. 34 1 _ (1.05457 x 1O- JS) ((2) x (329Nm- ») \/2 _I 6 -21 I Eo - x 26 - 5. 1 x 10 J. 2 5.807 x 10- kg ~.t<6~ E9::t6(a) We require ) f 1{1*1{1 d. = 1, that is, 1r0 :N2e-im'¢eim{¢d¢ = 1r0 :N2d¢ = 2nN2 = I. 2 N2_-~ , N-_~1 )\/ . 2n 2n