.......... CH00 CH00-page1 Quantum Physics.............. Robert Eisberg..Second edition.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ Quantum Physics of Atoms, molecules,solids, nuclei, and particles Second Edition Robert Eisberg CH01..Thermal radiation and Planck’s postulate CH02..Photons-particlelike properties of radiation CH03..De Broglie’s postulate-wavelike properties of particles CH04..Bohr’s model of the atom CH05..Schroedinger’s theory of quantum mechanics CH06..Solutions of time-independent Schroedinger equations CH07..One-electron atoms CH08..Magnetic dipole moments, spin and transition rates CH09..Multielectron atoms – ground states and x-ray excitations CH10..Multielectron atoms-optical exciations CH11..Quantum statistics CH12..Molecules CH13..Solids-conductors and semiconductors CH14..Solids-superconductors and magnetic properties CH15..Nuclear models CH16..Nuclear decay and nuclear reactions CH17..Introduction to elementary particles CH18..More elementary particles .......... CH00 CH00-page2 Solutions Supplement to Accompany Quantum Physics of Atoms, molecules,solids, nuclei, and particles Second Edition Robert Eisberg Prepared by Edward Derringh .......... CH01 - 1 - Quantum Physics.............. Robert Eisberg..Second edition.. CH 01..Thermal radiation and Planck’s postulate ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-1..At what wavelength does a cavity at 60000K radiated most per unit wavelength? ..60000K................................. ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-2..Show that the proportionality constant in (1-4) is 4 c . That is, show that the relation between spectral radiancy ( ) T R . and energy density ( ) T . . is ( ) ( ) T 4 T R . d. = c . . d. . ANS..(1-4) ( ) ( ) T T . . . R . 2 3 ( ) 8 1 T h kT d h d c e . p. . . . . = . - . . x h kT . = . 4 4 3 4 4 4 3 3 3 3 ( ) 8 8 T x 1 15 d k T x dx k T h c e h c p p p . . .= = - ..Hint : 3 4 x 1 15 x dx e p = - . .. .. 4 T R =s T , 5 4 2 3 2 15 k c h p s = ..Stefan’s law.. 4 4 4 5 4 4 3 3 3 2 ( ) 8 4 ( 2 ) 4 ( ) T 15 15 T d k T k T R d h c c h c c p p p . . . = = = . . .... ( ) ( ) T 4 T R . d. = c . . d. …………## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-3..Consider two cavities of arbitrary shape and material, each at the same temperature T, connected by a narrow tube in which can be placed color filters (assumed ideal) which will allow only radiation of a specified frequency . to pass through. (a) Suppose at a certain frequency . ', ( ) T . . ' d. for cavity 1 was greater than ( ) T . . ' d. for cavity 2. A color filter which passes only the frequency . ' is placed in the connecting tube. Discuss what will happen in terms of energy flow. (b) What will happened to their respective temperatures? (c) Show that this would violate the second law of thermodynamic; hence prove that all blackbodies at the same temperature must emit thermal radiation with the same spectrum independent of the details of their composition. ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-4..A cavity radiator at 60000K has a hole 10.0 mm in diameter drilled in its wall. Find the power radiated through the hole in the range 5500~5510Å...Hint : See Problem2.. ......60000K.................10.0mm...............................5500 ~5510Å ............. .......... CH01 - 2 - ANS.. 2 2 1 1 ( ) 1 ( ) 1 ( ) T 4 T 4 T av P A R d Ac d Ac . . . . = . . . = . . . . = . . .. 8 14 1 7 1 2.988 10 5.4509 10 5.50 10 . c Hz . - × = = = × × 8 14 2 7 2 2.988 10 5.4401 10 5.51 10 . c Hz . - × = = = × × Therefore, 14 1 2 1 ( ) 5.46 10 av 2 . = . +. = × Hz 11 2 1 .. =. -. = 9.9×10 Hz Since 3 3 ( ) 8 1 1 av av T av h kT h c e . p . . . = - Numerically 3 13 3 8 av 1.006 10 h c p . - = × .. av 4.37 h kT . = .. 1 78.04 h av e kT . - = 13 ( ) 1.006 10 1.289 10 15 T av 78.04 . . - × - = = × The aera of the hole is A =p r2 =p (5×10-3 )2 = 7.854×10-5m2 Hance, finally, 1 ( ) 1 (7.854 10 5 )(2.998 108 )(1.289 10 15 )(9.9 1011) 4 T av 4 P = Ac. . .. = × - × × - × P = 7.51W ............## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-5..(a) Assuming the surface temperature of the sun to be 57000K ,use Stefan’s law, (1-2), to determine the rest mass lost per second to radiation by the sun. Take the sun’s diameter to be 1.4×109m. (b) What fraction of the sun’s rest mass is lost each year from electromagnetic radiation? Take the sun’s rest mass to be 2.0×1030 kg . ANS..(a) L = 4p R2s T 4 = 4p (7×108 )2 (5.67×10-8 )(5700)4 = 3.685×1026W .. 7 108 sun R = × m.. L d (mc2 ) c2 dm dt dt = = 26 9 2 82 3.685 10 4.094 10 / (3 10 ) dm L kg s dt c × = = = × × (b) The mass lost in one year is M dm t (4.094 109 )(86400 365) 1.292 1017 kg dt . = = × × = × The desired fraction is , then, 17 14 30 1.292 10 6.5 10 2.0 10 f M M . × - = = = × × …………## .......... CH01 - 3 - ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-6..In a thermonuclear explosion the temperature in the fireball is momentarily 107 0K . Find the wavelength at which the radiation emitted is a maximum. .......................107 0K............................ ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-7..At a given temperature, max . = 6500Å for a blackbody cavity. What will max . be if the temperature of the cavity wall is increased so that the rest of emission of spectral radiation is double? .....................max . = 6500Å...................................... max . .......... ANS..Stefan’s law 4 T R =s T 4 4 R 2 T R T s s ' ' = = . T' = 4 2T Wien’s law : max max .' T' = . T . 0 4 max . ' ( 2T) = (6500 A)T . 0 0 max 4 6500 5466 2 .' = A = A……## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-8..At what wavelength does the human body emit its maximum temperature radiation? List assumptions you make in arriving at an answer. .............................................................. ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-9..Assuming that max . is in the near infrared for red heat and in the near ultraviolet for blue heat, approximately what temperature in Wien’s displacement law corresponds to red heat? To blue heat? ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-10..The average rate of solar radiation incident per unit area on the earth is 0.485cal / cm2 -min (or 338W /m2 ). (a) Explain the consistency of this number with the solar constant (the solar energy falling per unit time at normal incidence on a unit area) whose value is 1.94cal / cm2 -min (or 1353W /m2 ). (b) Consider the earth to be a blackbody radiating energy into space at this same rate. What surface temperature world the earth have under these circumstances? ANS..(a) The solar constant S is defined by 4 2 sun S L p r = r = Earth-sun distance......-........... sun L = rate of energy output of the sun...... .................R = radius of the earth..............The rate P at which energy impinges on the earth is .......... CH01 - 4 - 2 2 4 2 sun P L R R S r p p p = = The average rate, per m2 , of arrival of energy at the earth’s surface is 2 2 2 2 2 1 1(1353 / ) 338 / av 4 4 4 4 P P R S S W m W m R R p p p = = = = = (b) 338 =s T 4 = (5.67×10-8 )T 4 T = 277.860K …………## NOTE.......... Appendix S..S-1..(b) 2800K .. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-11..Attached to the roof of a house are three solar panels, each 1m× 2m . Assume the equivalent of 4 hrs of normally incident sunlight each day, and that all the incident light is absorbed and converted to heat. How many gallows of water can be heated from 40.. to 120.. each day? ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-12..Show that the Rayleugh-Jeans radiation law, (1-17), is not consistent with the Wien displacement law max . .T , (1-3a), or max. T = const , (1-3b). ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-13..We obtain max . in the blackbody spectrum by setting T ( ) 0 d d . . . = and max . by setting T ( ) 0 d d . . . = . Why is it not possible to get from max. T = const to max . = const ×T simply by using max max . c . = ? This is, why is it wrong to assume that max max . . = c , where c is the speed of light? ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-14..Consider the following number : 2,3,3,4,1,2,2,1,0 representing the number of hits garnered by each member of the Baltimore Orioles in a recent outing. (a) Calculate directly the average number of hits per man. (b) Let x be a variable signifying the number of hits obtained by a man, and let f (x) be the number of times the number s appears. 4 0 4 0 ( ) ( ) xf x x f x = S S . (c) Let p(x) be the probability of the number x being attained. Show that x is given by 4 0 x =Sxp(x) ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ .......... CH01 - 5 - 1-15..Consider the function ( ) 1 (10 )2 0 10 10 ( ) 0 f x x x f x all other x = - = = = (a) From ( ) ( ) xf x dx x f x dx 8 -8 8 -8 = . . find the average value of x. (b) Suppose the variable x were discrete rather then continuous. Assume .x =1 so that x takes on only integral values 0,1,2,……,10. Compute x and compare to the result of part(a). (Hint : It may be easier to compute the appropriate sum directly rather than working with general summation formulas.) (c) Compute x for .x = 5 , i.e. x = 0,5,10. Compare to the result of part (a). (d) Draw analogies between the results obtained in this problem and the discussion of Section 1-4. Be sure you understand the roles played by e ,.e , and P(e ) . ANS..(a) 10 2 0 10 2 0 1 (10 ) 10 1 (10 ) 10 x x dx x x dx - = - . . 10 2 0 10 2 0 (100 20 ) (100 20 ) x x x dx x x dx - + = - + . . 2 3 4 10 0 2 310 0 50 20 1 3 4 100 10 1 3 x x x x x x - + = - + 2 3 10 0 2 10 0 50 20 1 3 4 100 10 1 3 x x x x x - + = - + 1000 12 2.5 100 3 = = (b) 10 2 0 10 2 0 1 (10 ) 10 1 (10 ) 10 x x x x - = - S S 10 2 3 0 10 2 0 100 20 100 20 x x x x x - + = - + S S 100 10 11 20 10 11 21 (10 11)2 2 6 2 100 11 20 10 11 10 11 21 2 6 × × × × × - × + = × × × × - × + 825 2.143 385 = = (Hint : 1 ( 1) 2 n n n n + S = , 2 1 ( 1)(2 1) 6 n n n n n + + S = , 3 2 1 [ ( 1)] 2 n n n n + S = ) (c) Set x = 5n . 2 2 0 2 2 0 5 1 (10 5 ) 10 1 (10 5 ) 10 n n n n x n = = - = - S S 2 2 3 0 2 2 0 20 20 5 4 4 n n n n n n n = = - + = - + S S 20 2 3 20 2 3 5 5(2 3)2 2 6 2 4 3 4 2 3 2 3 5 2 6 × × × × × - × + = × × × × - × + 5 1 5 = = ……## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ .......... CH01 - 6 - 1-16..Using the relation ( ) e kT P kT e e - and 0 P(e )de 1 8 . = , evaluate the integral of (1-21) to deduce (1-22), e = kT . ....( ) e kT P kT e e - .. 0 P(e )de 1 8 . = .......1-21 ..............1-22 ....e = kT .. ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-17..Use the relation ( ) ( ) T 4 T R . d. = c . . d. between spectral radiancy and energy density, together with Planck’s radiation law, to derive Stefan’s law. That is, show that 3 4 2 0 2 1 T h kT R h d T c e . p . . s 8 = = - . where 5 4 2 3 2 15 k c h p s = . (Hint : 3 4 0 q 1 15 q dq e 8 p = - . ) ANS.. 3 2 0 0 ( ) 2 1 T T h kT R R d h d c e . p . . . . 8 8 = = - . . 4 4 2 3 0 2 x 1 k T xdx c h e p 8 = - . 4 4 4 2 3 2 15 k T c h p p = 5 4 4 4 2 3 2 15 k T T c h p = =s ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-18..Derive the Wien displacement law, max T 0.2014 hc k . = , by solving the equation T ( ) 0 d d . . . = . ..Hint : Set hc x .kT = and show that the equation quoted leads to 1 5 e- x + x = . Then show that x = 4.965 is the solution... ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-19..To verify experimentally that the 30K universal background radiation accurately fits a blackbody spectrum, it is decided to measure ( ) T R . from a wavelength below max . where its value is max 0.2 ( ) T R . to a wavelength above max . where its value is again max 0.2 ( ) T R . . Over what range of wavelength must the measurements be made? ANS.. 5 5 5 4 3 ( ) ( ) 2 T 4 T x 1 R c k t x h c e p . = . . = - With x hc .kT = . At max . = . , x = 4.965 , by problem 18. Thus 5 max 4 3 ( ) 42.403 ( ) T R kt h c . = p Now find x such that max ( ) 0.2 ( ) T T R . = R . : .......... CH01 - 7 - 5 5 5 5 4 3 4 3 2 (0.2)42.403 ( ) x 1 k t x kt h c e h c p = p - 5 4.2403 x 1 x e = - 1 x =1.882.. 2 x =10.136 Numerically, 34 8 23 1 (6.626 10 )(2.998 10 ) 1 (1.38 10 )(3) hc kT x x . - - × × = = × 4.798 10 3 x . × - = So that 3 1 4.798 10 2.55 1.882 . mm × - = = 3 2 4.798 10 0.473 10.136 . mm × - = = ……## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-20..Show that, at he wavelength max . , where ( ) T . . has its maximum 5 max 4 ( ) 170 ( ) T ( ) kT hc . . = p . ANS..If max x hc . kT = , then, by Problem 18, 1 5 e-x + x = . 1 5 ex x x - = - Hence, max 4 max ( ) 8 (5 ) T kT x p . . . = - . But, x = 4.965 , 4 4 max 1 (4.965 kT ) . hc = Upon substitution, there give 5 max 4 ( ) 170 ( ) T ( ) kT hc . . = p ……## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-21..Use the result of the preceding problem to find the two wavelengths at which ( ) T . . has a value one-half the value at max . . Give answers in terms of max . . ANS..By Problem 20, 5 max 4 ( ) 170 ( ) T ( ) kT hc . . = p . So that the wavelengths sought must satisfy 5 5 4 8 1 1 170 ( ) 2 ( ) 1 hc kT hc kT hc e. p p . = · - . Again let x hc .kT = . In terms of x, the preceding equation becomes 5 170 x 1 16 x e = - . Solutions are 1 x = 2.736 ; 2 x = 8.090 . Since, for max . , x = 4.965 , these solutions give 1 max . =1.815. , 2 max . = 0.614. …….## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-22..A tungsten sphere 2.30cm in diameter is heated to 2000... At this temperature tungsten .......... CH01 - 8 - radiates only about 30% of the energy radiated by a blackbody of the same size and temperature. (a) Calculate the temperature of a perfectly black spherical body of the same size that radiates at the same rate as the tungsten sphere. (b) Calculate the diameter of a perfectly black spherical body at the same temperature as the tungsten sphere that radiates at the same rate. ANS..(a) 0 4 4 30 0 tungsten black ×s T =s T . 0 4 30 0 ×(2000 + 273) = T T =16820K =14090C (b) 0 4 2 4 2 30 0 4 (2.3 ) 4 tungsten black ×s T × p cm =s T × p r tungsten black T = T . r =1.26cm……## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-23..(a) Show that about 25% of the radiant energy in a cavity is contained within wavelengths zero and max . ; i.e., show that max 0 0 ( ) 1 4 ( ) T T d d . . . . . . . 8 . . .. (Hint : max hc 4.965 . kT = ; hence Wien’s approximation is fairly accurate in evaluating the integral in the numerator above.) (b) By the percent does Wien’s approximation used over the entire wavelength range overestimate or underestimate the integrated energydensity? ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-24..Find the temperature of a cavity having a radiant energy density at 2000Å that is 3.82 times the energy density at 4000Å. ANS..Let . ' = 200nm , . '' = 400nm; then 5 5 1 1 3.82 1 1 1 1 hc hc e. kT e. kT . . ' '' = × ' '' - - . 1 3.82 ( )5 1 hc kT hc kT e e . . . . '' ' - ' = × '' - Numerically, 34 8 7 23 (6.626 10 )(2.988 10 ) 71734 (2 10 )(1.38 10 ) hc K . k - - - × × = = ' × × 34 8 7 23 (6.626 10 )(2.988 10 ) 35867 (4 10 )(1.38 10 ) hc K . k - - - × × = = '' × × so that 35867 5 71734 1 3.82 (1) 0.1194 2 1 T T e e - = × = - Let 35867 x = e T ; then 2 1 0.1194 1 1 1 x x x - = = - + . 35867 x = 7.375 = e T .......... CH01 - 9 - 35867 17950 ln 7.375 T = = K …………## NOTE.......... Appendix S..S-1..180200K.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ ............ ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-001..(a) ......................................................(b) .......... ....... 5 ( ) (8 ) 1 4 1 hc kT R c hc e. p . . = - ........ .0... .8......................... ...... ANS..(a) Blackbody.............................................................. .................................................................... ............ ....................................................... (b) ................. 5 ( ) (8 ) 1 4 1 hc kT R c hc e. p . . = - 5 0 5 ( 0) (8 ) 1 (8 ) 4 4 1 hc kT hc kT R c hc c hc e e . . . p p . . . - . . = = - …..Wien’s law.. 5 5 ( ) (8 ) 1 (8 ) 1 4 4 1 1 1 hc kT R c hc c hc hc e kT . . p p . . . . .8 . 8 = = - + - 4 (8 ) 4 c p kT . = …..Rayleigh-Jeans law..……## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-002..Use Planck’s radiation theory show (a) Wien displacement law : 2.898 10 3 m. T = × - mK . (b) Stefan’s law : R =s T 4 , 5 4 2 3 2 15 k c h p s = ANS..(a) .................. 5 ( ) 8 1 hc kT d hc d e. p . . . . . = - ........................ ( ) 0 m d d . . . . = ……(1) .......... CH01 - 10 - .... 5 ( ) [8 ] 1 hc kT d d hc d d e. . . p . . . = - 5 8 [ ] 1 hc kT hc d d e. . p . - = - 6 5 2 5( 1) 8 [ ] ( 1) hc hc kT kT hc kT e hc e hc kT e . . . . . p . - - - + - = - ……....(1).. ..[ 5( 1) ] 0 m hc hc kT kT e hc e kT . . . . - - + = ……(2) . x hc .kT = ....(2)........-5(ex -1) + xex = 0 . 5( x 1) x y e y xe . = - - . . = ........ .... Wien displacement law 2.898 10 3 m T hc mK xk . = = × - (b) ........ 5 ( ) ( ) (8 ) 1 4 4 1 hc kT R c c hc e. p . . . . = = - .......... 2 0 0 5 ( ) 8 4 ( 1) T hc kT R R d hc d e. p . . . . 8 8 = = - . . . x hc .kT = , .... 4 T R =s T , 5 4 2 3 2 15 k c h p s = ..Stefan’s law..……## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-003..A sphere of radius r is maintained at a surface temperature T by an internal heat source. The sphere is surrounded by a thin concentric shell of radius 2r. Both object emit and absorb as blackbodies. What is the temperature of the shell? ANS.. 4 T R =s T ..Stefan’s law.. ..................................................................... 4 2 4 2 4 aP = p r R = sp r T ............ ....................................... .......... CH01 - 11 - 2[4 (2 )2 ] 32 2 4 eP = p r R = sp r T' ........... ............................ a e P = P ................... 4 4 0.595 8 T' = T = T ……## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-004..Assuming that the probability distribution for occupation of the oscillator is given by Boltzmann law. (a) Show that the mean energy E of an oscillator for a given temperature is 1 hc kT h e. . - . (b) Show that for a sufficiently high temperature this expression becomes equal to the classical one. Give the other of magnitude of this temperature. ANS..(a) ( ) 1 ( ) ( ) 1 1 nh kT nh h kT kT P nh e h kT P e e kT . . . e e . . e e - - = = = - S S S S (b) .... T h hc k k . . > = ...... 1 1 1 h kT h h kT h e kT . . . e . = . = - + - ……..................……## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-005..Cosmic background radiation peaks at a wavelength of about 1 mm. What is the temperature of the universe? (Hint: use Wien’s law) ANS..According to the Wien’s law 3 0 max 0.2014 2.898 10 B T hc m K k . = = × - For 3 max . =1mm =10- m T = 2.8980K ˜ 30K ……## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 1-006..What is the wavelength of a photon whose energy is equal to the rest mass energy of an electron? ANS.. 2 0 m c E h. h c . = = = 0 2 0 hc 0.0243 A m c .= = ……## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ .......... CH02 - 1 - Quantum Physics.............. Robert Eisberg..Second edition.. CH 02..Photons-particlelike properties of radiation ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 2-1..(a) The energy required to remove an electron from sodium is 2.3eV. Does sodium show a photoelectric effect for yellow light, with . = 5890 Å? (b) What is the cutoff wavelength for photoelectric emission from sodium? ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 2-2..Light of a wavelength 2000Å falls on an aluminum surface. In aluminum 4.2eV are required to remove an electron. What is the kinetic energy of (a) the fastest and (b) the slowest emitted photoelectrons? (c) What is the stopping potential? (d) What is the cutoff wavelength for aluminum? (e) If the intensity of the incident light is 2.0W /m2 , what is the average number of photons per unit time per unit area that strike the surface? ANS..(2a)2.0eV (2b)zero (2c)2.0V (2d)2950Å (2e) 2.0×1014 / cm2 -sec ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 2-3..The work function for a clean lithium surface is 2.3eV. Make a rough plot of the stopping potential 0 V versus the frequency of the incident light for such a surface, indicating its important features. ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 2-4..The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength . = 4910Å is 0.71V. When the incident wavelength is changed the stopping potential is found to be 1.43V. What is the new wavelength? ANS..3820Å ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 2-5..In a photoelectric experiment in which monochromatic light and sodium photocathode are used, we find a stopping potential of 1.85V for . = 3000 Å and of 0.82V for . = 4000Å. From these data determine (a) a value for Planck’s constant, (b) the work function of sodium in electron volts, and (c) the threshold wavelength for sodium. ANS..The photoelectric equation is 0 0 hc = eV . + w . with 0 V =1.85V for . = 300nm , and 0 V = 0.82V for . = 400nm 26 7 0 hc = 8.891×10- + 3×10- w 26 7 0 hc = 5.255×10- + 4×10- w Hence, 26 7 26 7 0 0 8.891×10- + 3×10- w = 5.255×10- + 4×10- w 19 0 w = 3.636×10- J = 2.27eV .......... CH02 - 2 - (b) Therefore, hc = 8.891×10-26 + (3×10-7 )×(3.636×10-19 ) =19.799×10-26 J -m (a) 26 34 8 19.799 10 6.604 10 2.988 10 h Js - × - = = × - × (c) 0 0 w hc . = , 26 19 0 3.636 10 19.799 10 . - - × × = 7 0 . = 5.445×10- m = 544.5nm…………## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 2-6..Consider light shining on a photographic plate. The light will be recorded if it dissociates an AgBr molecule in the plate. The minimum energy to dissociate this molecule is of the order of 10-19 joule. Evaluate the cutoff wavelength greater than which light will not be recorded. ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 2-7..The relativistic expression for kinetic energy should be used for the electron in the photoelectric effect when 0.1 c . > , if errors greater than about 1% are to be avoied. For photoelectrons ejected from an aluminum surface ( 0 . = 4.2eV ) what is the amallest wavelength of an incident photo for which the classical expression may be used? ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 2-8..X rays with . = 0.71Å eject photoelectrons from a gold foil. The electrons from circular paths of radius r in a region of magnetic induction B. Experiment shows that rB =1.88×10-4 tesla -m. Find (a) the maximum kinetic energy of the photoelectrons and (b) the work done in removing the electron from the gold foil. ANS..In a magnetic field r mv eB = p = mv = erB = (1.602×10-19 )(1.88×10-4 ) = 3.012×10-23kg -m/ s 23 8 13 (3.012 10 )(2.988 10 ) 0.05637 (1.602 10 ) p MeV c c - - × × = = × Also, 2 2 2 2 0 E = p c + E E2 = (0.05637)2 + (0.511)2 E = 0.5141MeV Hence, (a) 0 K = E - E = 0.5141- 0.5110 = 0.0031MeV = 3.1keV (b) The photon energy is ( ) 1240 1240 0.0175 ph ( ) 0.071 E eV MeV . nm = = = 0 17.5 3.1 14.4 ph w = E - K = - = keV ……## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ .......... CH02 - 3 - 2-9..(a) Show that a free electron cannot absorb a photon and conserve both energy and momentum in the process. Hence, the photoelectric process requires a bund electron. (b) In the Compton effect, the electron can be free . Explain. ANS..(a) Assuming the process can operate, apply conservation of mass-energy and of momentum : h. + E0 = K + E0 .h. = K h 0 p c . + = These equations taken together imply that p K c = ……(1) But, for an electron, 2 2 2 2 0 E = p c + E 2 2 2 2 0 0 (K + E ) = p c + E . 2 0 K 2E K p c + = ……(2) (1) and (2) can be satisfied together only if 0 E . 0 ,which is not true for an electron. (b) In the Compton effect, a photon is present after the collision; this allows the conservation laws to hold without contradiction. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 2-10..Under ideal conditions the normal human eye will record a visual sensation at 5500Å if as few as 100 photons are absorbed per seconds. What power level does this correspond to? ANS..3.6×10-17W ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 2-11..An ultraviolet lightbulb, emitting at 4000Å, and an infrared lightbulb, emitting at 7000Å, each are rated at 40W. (a) Which bulb radiates photons at the greater rate, and (b) how many more photons does it produce each second over the other bulb? ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 2-12..Solar radiation falls on the earth at a rate of 1.94cal / cm2 -min on a surface normal to the incoming rays. Assuming an average wavelength of 5500Å, how many photons per cm2 - min is this? ANS..1.235×1020Hz , 2.427×10-2 Å, 2.731×10-22 kg -m/ sec ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 2-13..What are the frequency, wavelength, and momentum of a photon whose energy equals the rest mass energy of an electron? ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 2-14..In the photon picture of radiation, show that if beams of radiation of two different wavelengths are to have the same intensity (or energy density) then the numbers of the photons per unit cross-sectional areas per sec in the beams are in the same ratio as the wavelength. .......... CH02 - 4 - ANS..Let n = number of photons per unit volume. In time t, all photons initially a distance µ .. .......... CH04 - 8 - .. .. .. .. .. .. .. .. .. .... 2 2 2 2 1 { 1 1 } 4 { 1 1 } He H He f i f i R R . n n n n = - > - .. .. .. .. .. .. .. .. .. Compare to the hydrogen Ha line, the helium 6..4 line wavelength is a little shorter. .. smaller (c) Since . .µ -1 (the factor Z2 is combined with 2 2 1 1 f i n n - to give equal values for H and He) H He He H 1 H H He He . . µ µ µ . µ µ - - = = - 4 1 1 3 3 0.511 4.084 10 1 4 4 938.3 4 e p e H e p p m m m m m m . . - - . = - = = = × - 0 .. = (4.084×10-4 )×(656.3nm) = 0.268nm = 2.68 A……## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 4-39..In stars the Pickering series is found in the He+ spectrum. It is emitted when the electron in He+ jumps from higher levels into the level with n = 4 . (a) Show the exact formula for the wavelength of lines belonging to this series. (b) In what region of the spectrum is the series? (c) Find the wavelength of the series limit. (d) Find the ionization potential, if He+ is in the ground state, in electron volts. ANS..(a) 0 2 2 ( ) 3647 16 A n n . - , n=5,6,7,… (b) visible, infrared (c) 3647Å (d) 54.4eV ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 4-40..Assuming that an amount of hydrogen of mass number three (tritium) sufficient for spectroscopic examination can be put into a tube containing ordinary hydrogen, determine the separation from the normal hydrogen line of the first line of the Balmer series that should be observed. Express the result as a difference in wavelength. ANS..2.38 Å ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 4-41..A gas discharge tube contains H1 ,H2 , He3 , He4 , Li6 ,and Li7 ions and atoms (the superscript is the atomic mass),with the last four ionized so as to have only one electron. (a) As the potential across the tube is raised from zero, which spectral line should appear first? (b) Given, in order of increasing frequency, the origin of the lines corresponding to the first line of the Lyman series of H1 . ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 4-42 .. Consider a body rotating freely about a fixed axis. Apply the Wilson-Sommerfield quantization rules, and show that the possible values of the total energy are predicted to be .......... CH04 - 9 - 2 2 0,1,2,3... 2 E .. n n I = = ,where I is its rotational inertia, or moment of inertia, about the axis of rotation. ANS..The momentum associated with the angle . is L = I. . The total energy E is 2 1 2 2 2 E K I L I = = . = . L is independent of . for a freely rotating object. Hence, by the Willson-Sommerfeld rule, ... Ld. = nh L... d. = L(2p ) = 2IE(2p ) = nh 2IE = n.. 2 2 2 E n .. I = ……## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 4-43..Assume the angular momentum of the earth of mass 6.0×1024 kg due to its motion around the sun at radius 1.5×1011m to be quantized according to Bohr’s relation 2 L nh p = . What is the value of the quantum number n? Could such quantization be detected? ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ .......... CH05 - 1 - Quantum Physics.............. Robert Eisberg..Second edition.. CH 05..Schroedinger’s theory of quantum mechanics ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 5-01..If the wave function .1(x,t) , 2. (x,t) , and 3. (x,t) are three solutions to the Schroedinger equation for a particular potential V (x,t) , show that the arbitrary linear combination 1 1 2 2 3 3 .(x,t) = c . (x,t) + c . (x,t) + c . (x,t) is also a solution to that equation. ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 5-02..At a certain instant of time, the dependence of a wave function on position is as shown in Figure 5-20. (a) If a measurement that could locate the associated particle in an element dx of the x axis were made at that instant, where would it most likely be found? (b) Where would it least likely be found? (c) Are the chances better that it would be found at any positive value of x, or are they better that it would be found at any negative value of x? (d) Make a rough sketch of the potential V (x) which gives rise to the wave function. (e) To which allowed energy does the wave function correspond? Figure 5-20 The space dependence of a wave function considered in Problem 2, evaluated at a certain instant of time. ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 5-03..(a) Determine the frequency . of the time-dependent part of the wave function quoted in Example 5-3, for the lowest energy state of a simple harmonic oscillator. (b) Use this value of . , and the de Broglie-Einstein relation E = h. , to evaluate the total energy E of the oscillator. (c) Use this value of E to show that the limits of the classical motion of the oscillator, found in Example 5-6, can be written as 1/ 2 ( )1/ 4 x .. Cm = ± . ANS..(a) The time-dependent part of the wavefunction is 1 2 .. 2 it C iET i t m e e ep. - - - = = Therefore, 1 2 2 C m = p. . 1 4 C m . p = (b) Since E = h. = 2p... , 1 2 .. E C m = .......... CH05 - 2 - (c) The limiting x can be found from 1 2 2 Cx = E (2 )1/ 2 .. 1/ 2 ( ) 1/ 2 x E Cm C = ± = ± - ……## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 5-04..By evaluating the classical normalization integral in Example 5-6, determine the value of the constant B2 which satisfies the requirement that the total probability of finding the particle in the classical oscillator somewhere between its limits of motion must equal one. ANS..According to Example 5-6, the normalizing integral is 2 2 2 2 1 0 0 2 1 2 2 sin 2 2 E C E C B m dx B m x C E C E x C C = = - - . 1 B2 m C = p . 2 1/ 2 2 B ( C ) mp = ……## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 5-05..Use the results of Example 5-5, 5-6, and 5-7 to evaluate the probability of finding a particle, in the lowest energy state of a quantum mechanical simple harmonic oscillator, within the limits of the classical motion. (Hint : (i) The classical limits of motion are expressed in a convenient form in the statement of Problem 3c. (ii) The definite integral that will be obtained can be expressed as a normal probability integral, or an error function . It can then be evaluated immediately by consulting mathematical handbooks which tabulate these quantities. Or, the integral can easily be evaluated by expanding the exponential as an inifinite series before integrating, and then integrating the first few terms in the series. Alternatively, the definite integral can be evaluated by plotting the integrand on graph paper, and counting squares to find the area enclosed between the integrand, the axis, and the limits.) ANS..Problem 5-3(c) Provides the limits on x; the wavefunction is 1/8 2 2 1/ 4 ( ) ( ) .. .. Cm Cm x i t e e. p . = - - Hence, the desired probability is given by 1/ 2 1/ 4 1/ 4 ( ) 2 1/ 2 0 . 2 ( ) ( ) .. .. .. Cm Cm Cm x Prob e dx p - - = . If 1/ 4 1/ 2 (4 ) .. u = Cm x 2 2 2 0 . 2 1 2(0.42) 0.84 2 u Prob e du p - = . = = ……## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 5-06..At sufficiently low temperature, an atom of a vibrating diatomic molecule is a simple .......... CH05 - 3 - harmonic oscillator in its lowest energy state because it is bound to the other atom by a linear restoring force. (The restoring force is linear, at least approximately, because the molecular vibrations are very small.) The force constant C for a typical molecule has a value of about C~ 103nt /m. The mass of the atom is about m~ 10-26 kg . (a) Use these numbers to evaluate the limits of the classical motion from the formula quoted in Problem 3c. (b) Compare the distance between these limits to the dimensions of a typical diatomic molecule, and comment on what this comparison implies concerning the behavior of such a molecule at very low temperatures. ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 5-07..(a) Use the particle in a box wave function verified in Example 5-9, with the value of A determined in Example 5-10, to calculate the probability that the particle associated with the wave function would be found in a measurement within a distance of 3 a from the right-hand end of the box of length a. The particle is in its lowest energy state. (b) Compare with the probability that would be predicted classically from a very simple calculation related to the one in Example 5-6. ANS..(a) Since ( 2)1/ 2 cos .. x iEt e a a p - . = 2 2 2 2 6 6 . 2 cos ( ) 2 cos 1 3 0.1955 3 4 a a Prob x dx udu a a p p p p p = . = . = - = Independent of E. (b) Classically . 3 1 0.3333 3 Prob a a = = = ……## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 5-08..Use the results Example 5-9 to estimate the total energy of a neutron of mass about 10-27 kg which is assumed to move freely through a nucleus of linear dimensions of about 10-14m, but which is strictly confined to the nucleus. Express the estimate in MeV. It will be close to the actual energy of a neutron in the lowest energy state of a typical nucleus. ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 5-09..(a) Following the procedure of Example 5-9, verify that wave function sin 2 ( , ) 2 2 0 2 2 .. x iEt a a A e x x t a x a or x a p - - < < + . = < - > + is a solution to the schroedinger equation in the region 2 2 - a < x < + a for a particle which moves freely through the region but which is strictly confined to it. (b) Also determine the .......... CH05 - 4 - value of the total energy E of the particle in this first excited state of the system, and compare with the total energy of the ground state found in Example 5-9. (c) Plot the space dependence of this wave function. Compare with the ground state wave function of Figure 5-7, and give a qualitative argument relating the difference in the two wave functions to the difference in the total energies of the two states. ANS..(b) 2 2 2 0 2 .. 4E ma p = ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 5-10..(a) Normalize the wave function of Problem 9, by adjusting the value of the multiplicative constant A so that the total probability of finding the associated particle somewhere in the region of length a equals one. (b) Compare with the value of A obtained in Example 5-10 by normalizing the ground state wave function. Discuss the comparison. ANS..(a) To normalize the wavefunction, evaluate 2 2 1 a a * dx - = . . . (. = 0 outside this region ). With sin 2 .. x Et A e a p - - . = , this become 2 2 2 2 2 0 0 1 2 sin 2 sin 2 a A x dx a udu a A a p p p p p = . = . = A 2 a = ……## (b) This equals the value of A for the ground state wavefunction and, in fact, the normalization constant of all the excited states equals this also. Since all of the space wave functions are simple sines or cosines, this equality is understandable…….## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 5-11..Calculate the expectation value of x, and the expectation value of x2 , for the particle associated with the wave function of Problem 10. ANS..The wavefunction is 2 sin 2 .. x iEt e a a p . - = And therefore 2 2 2 2 sin 2 0 a a x x x dx a a p + - = . = ……## As for x2 : 2 2 2 2 2 2 2 2 2 3 2 0 2 2 sin 2 sin 1 (1 1 ) 0.07067 2 43 2 a a x x x dx a u udu a a a a p p p p + - = . = . = - = ……## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 5-12..Calculate the expectation value of p, and the expectation value of p2 , for the particle associated with the wave function of Problem 10. .......... CH05 - 5 - ANS..The linear momentum operator is i.. x . - . and therefore 2 0 2 2 sin 2 [ (sin 2 )] 4 .. sin cos 0 .. a a p x i x dx i u udu a a x a a p p p + - . = - = - = . . . ……## Similarly, 2 2 2 22 2 2 2 2 2 2 2 0 2 2 sin 2 [ (sin 2 )] 8 (.. ) sin 4 (.. ) ( ) .. a a p x i x dx i udu h a a x a a a a p p p p p + - . = =- = = . . . ……## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 5-13..(a) Use quantities calculated in the preceding two problems to calculate the product of the uncertainties in position and momentum of the particle in the first excited state of the system being considered. (b) Compare with the uncertainty product when the particle is in the lowest energy state of the system, obtained in Example 5-10, Explain why the uncertainty products differ. ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 5-14..(a) Calculate the expectation values of the kinetic energy and potential energy for a particle in the lowest energy state of a simple harmonic oscillator, using the wave function of Example 5-7. (b) Compare with the time-averaged kinetic and potential energies for a classical simple harmonic oscillator of the same total energy. ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 5-15..In calculating the expectation value of the product of position times momentum, an ambiguity arises because it is not approachwhich of the two expressions xp *x( i.. ) dx x 8 -8 . = . - . . . px *( i.. )x dx x 8 -8 . = . - . . . should be used. (In the first expression x . . operates on . ; in the second it operates on x. .) (a) Show that neither is acceptable because both violate the obvious requirement that xp should be real since it is measurable. (b) Then show that the expression * ( ) ( ) [ ] 2 x i.. i.. x xp x x dx 8 -8 . . - + - = . . . . . is acceptable because it does satisfy this requirement. (Hint : (i) A quantity is real if it equals .......... CH05 - 6 - its own complex conjugate. (ii) Try integrating by part. (iii) In any realistic case the wave function will always vanish at x = ±8 .) ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 5-16 .. Show by direct substitution into the Schroedinger equation that the wave function ( , ) ( ) .. iEt x t . x e- . = satisfies that equation if the eigenfunction . (x) satisfies the time-independent Schroedinger equation for a potential V (x) . ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 5-17..(a) Write the classical wave equation for a string of density per unit length which varies with x. (b) Then separate it into two ordinary differential equations, and show that the equation in x is very analogous to the time-independent Schroedinger equation. ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 5-18..By using an extension of the procedure leading to (5-31), obtain the Schroedinger equation for a particle of mass m moving in three diamensions (described by rectangular coordinates x,y,z) ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 5-19..(a) Separate the Schroedinger equation of Problem 18, for a time-independent potential, into a time- independent Schroedinger equation and an equation for the time dependence of the wave function. (b) Compare to the corresponding one-dimensional equations, (5-37) and (5-38), and explain the similarities and the differences. ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 5-20 .. (a) Separate the time-independent Schroedinger equation of Problem 19 into three time-independent Schroedinger equations, one in each of the coordinates. (b) Compare them with (5-37). (c) Explain clearly what must be assumed about the form of the potential energy in order to make the separation possible, and what the physical significance of this assumption is. (d) Give an example of a system that would have such a potential. ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 5-21..Starting with the relativistic expression for the energy, formulate a Schroedinger equation for photons, and solve it by separation of variables, assuming V = 0 . ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 5-22..Consider a particle moving under the influence of the potential V (x) = C x , where C is a constant, which is illustrated in Figure 5-21. (a) Use qualitative arguments, very similar to those of Example 5-12, to make a sketch of the first eigenfunction and of the tenth .......... CH05 - 7 - eigenfunction for system. (b) Sketch both of the corresponding probability density function. (c) Then use the classical mechanics to calculate, in the manner of Example 5-6, the probability density function predicted by that theory. (d) Plot the classical probability density functions with the quantum mechanical probability density functions, and discuss briefly their comparison. Figure 5-21 A potential function considered in Problem 22. ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 5-23..Consider a particle moving in the potential V(x) plotted in figure 5-22. For the following ranges of the total energy E, state whether there are any allowed values of E and if so, whether they are discretely separated or continuously distributed. (a) 0 E + a . ANS..Schroedinger’s equation is 2 2 2 2 ( ) 0 .. d mE V dx . + - . = In the region in questin, 0 V =V = constant , 0 E Hence, . = Ae-qx + Beqx , is the general solution. However, . (x = 8) = 0, requiring B = 0 . = Ae-qx as the wavefunction…….## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 5-28..Using the probability density corresponding to the eigenfunction of Problem 27, write an expression to estimate the distance D outside the binding region of the potential within which there would be an appreciable probability of finding the particle. (Hint : Take D to extend to the point at which .*. is smaller than its value at the edge of the binding region by a factor of e-1 . This e-1 criterion is similar to one often used in the study of electrical circuits.) ANS..Since . is real, the probability density P is P =. *. =. 2 = A2e-2qx Recalling that x is measured from the center of the binding region, the suggested criterion for D gives 2 ( 1 ) 2 ( 1 ) 2 2 1 2 2 q a D q a A e e A e - + = - - eqa-2qD = e-qa-1 1/ 2 0 1 2 2[2 ( )] D .. q mV E = = - ……## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 5-29..The potential illustrated in Figure 5-23 gives a good description of the forces acting on an .......... CH05 - 10 - electron moving through a block of metal. The energy difference 0 V - E , for the highest energy electron, is the work function for the metal. Typically, 0 V - E.. 5eV . (a) Use this value to estimate the distance D of Problem 28. (b) Comment on the results of the estimate. ANS..0.4Å ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 5-30..Consider the eigenfunction illustrated in the top part of Figure 5-26. (a) Which of the three potentials illustrated in the bottom part of the figure could lead to such an eigenfunction? Give qualitative arguments to justify your answer. (b) The eigenfunction shown is not the one corresponding to the lowest allowed energy for the potential. Sketch the form of the eigenfunction which does correspond to the lowest allowed energy 1 E . (c) Indicate on another sketch the range of energies where you would expect discretely separated allowed energy states, and the range of energies where you would expect the allowed energies to be continuously distributed. (d) Sketch the form of the eigenfunction which corresponds to the second allowed energy 2 E . (e) To which energy level does the eigenfunction presented in Figure 5-26 correspond? Figure 5-26 An eigenfunction (top curve) and three possible forms (bottom curves) of the potential energy function considered in Problem30. ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 5-31..Estimate the lowest energy level for a one-dimensional infinite square well of width a .......... CH05 - 11 - containing a cosine bump. That is the potential V is 0 cos 2 2 inf 2 2 V V x a x a a V inity x a or x a p = - < < + = <- > + where 2 2 0 2 2 V .. ma p << . ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 5-32..Using the first two normalized wave function 1. (x,t) and 2. (x,t) for a particle moving freely in a region of length a, but strictly confined to that region, construct the linear combination 1 1 2 2 .(x,t) = c . (x,t) + c . (x,t) . Then derive a relation involving the adjustable constants 1 c and 2 c which, when satisfied, will ensure that .(x,t) is also normalized. The normalized 1. (x,t) and 2. (x,t) are obtained in Example 5-10 and Problem 10. ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 5-33..(a) Using the normalized “mixed” wave function of Problem32, calculate the expectation value of the total energy E of the particle in terms of the energies 1 E and 2 E of the two states and the values 1 c and 2 c of the mixing parameters. (b) Interpret carefully the meaning of your result. ANS..(a) * * 1 1 1 2 2 2 c c E + c c E ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 5-34..If the particle described by the wave function of Problem 32 is a proton moving in a nucleus, it will give rise to a charge distribution which oscillates in time at the same frequency as the oscillations of its probability density. (a) Evaluate this frequency for values of 1 E and 2 E corresponding to a proton mass of 10-27 kg and a nuclear dimension of 10-14 m. (b) Also evaluate the frequency and energy of the photon that would be emitted by oscillating charge distribution as the proton drops from the excited state to the ground state. (c) In what region of the electromagnetic spectrum is such a proton? ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ .......... CH06 - 1 - Quantum Physics.............. Robert Eisberg..Second edition.. CH 06..Solutions of time-independent Schroedinger equations ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 6-01..Show that the step potential eigenfunction, for E 0 where 0 V (x) =V , and traveling in the direction of decreasing x towards the point x = 0 where the potential steps down to its value V (x) = 0 in the region x < 0 . Show that the transmission and reflection coefficients are the same as those obtained in Section 6-4. ANS..Assume that 1 1 . = Ce-ik x 2 2 2 . = Ae-ik x + Beik x Where A=amplitude of incident wave B=amplitude of reflected wave C=amplitude of transmitted wave There is no wave moving in the +x-direction in region I. Also, 1/ 2 1 (2 ) .. k = mE , 1/ 2 0 2 {2 ( )} .. k m E V - = Continuity of wavefunction and derivative at x = 0 imply A+ B = C , 2 2 1 -k A+ k B = -k C These equations may be solved to give the reflection and the transmission amplitudes in terms of the incident amplitude, the results being: 2 1 2 1 B k k A k k - = + ; 2 2 1 C 2k A k k = + The reflection coefficient R and transmission coefficient T now become 2 1 2 2 2 1 2 R B B B ( k k ) A A A k k * * - = = =+ 1 1 2 2 1 2 2 2 2 1 2 1 2 ( )( 2 ) 4 ( ) .. .. T v C C k k k k v A A k k k k k * * = = = + + These expressions for R and T are the same as those obtained if the incident wave came from the left…….## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 6-03..Prove (6-43) stating that the sum of the reflection and transmission coefficients equals one, for the case of a step potential with 0 E >V . ANS.. .......... CH06 - 2 - ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 6-04..Prove (6-44) which expresses the reflection and transmission coefficients in terms of the ratio 0 E V . ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 6-05..Consider a particle tunneling through a rectangular potential barrier. Write the general solutions presented in Section 6-5, which give the form of . in the different regions of the potential. (a) Then find four relations between the five arbitrary constants by matching . and d. dx at the boundaries between these regions. (b) Use these relations to evaluate the transmission coefficient T, thereby verifying (6-49). (Hint : First eliminate F and G, leaving relations between A, B, and C. Then eliminate B.) ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 6-06..Show that the expression of (6-49), for the transmission coefficient in tunneling through a rectangular potential barrier, reduces to the form quoted in (6-50) if the exponents are very large. ANS..If 2 k a >>1, then ek2a >> e-k2a and the transmission coefficient becomes, under these circumstances, 2 2 1 0 0 {1 } 16 (1 ) e k a T E E V V = + - - . Now 0 0 E 1 V < < and therefore 0 0 16 E (1 E ) 4 V V - = , the upper limit occurring at 0 1 2 E V = . Hence, if e2k2a > 4, 2 2 0 0 1 16 (1 ) e k a E E V V > - . Since, in fact, it is assumed that e2k2a >>1, 2 2 0 0 1 16 (1 ) e k a E E V V >> - , And therefore, under these conditions, 2 2 0 0 T 16 E (1 E )e k a V V = - - …….## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 6-07..Consider a particle passing over a rectangular potential barrier. Write the general solutions, presented in Section 6-5, which give the form of . in the different regions of the potential. (a) Then find four relations between the five arbitrary constants by matching . and d dx . at the boundaries between these regions. (b) Use these relations to evaluate the transmission coefficient T, thereby verifying (6-51). (Hint : Note that the four relations become exactly the same as those found in the fires part of Problem 5, if II k is replaced by III ik . Make .......... CH06 - 3 - this substitution in (6-49) to obtain directly (6-51).) ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 6-08..(a) Evaluate the transmission coefficient for an electron of total energy 2eV incident upon a rectangular potential barrier of height 4eV and thickness 10-10m, using (6-49) and then using (6-50). Repect the evaluation for a barrier thickness of (b) 9×10-9m and (c) 10-9m. ANS..(8a) 0.62 (8b) 1.07×10-56 (8c) 2.1×10-6 ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 6-09..A proton and a deuteron (a particle with the same charge as a proton, but twice the mass) attempt to penetrate a rectangular potential barrier of height 10MeV and thickness 10-14m. Both particle have total energies of 3MeV . (a) Use qualitative arguments to predict which particle has the highest probability of succeeding. (b) Evaluate quantitatively the probability of success for both particles. ANS..(a) The opacity of a barrier is proportional to 2 0 2 2 .. mV a and therefore the lower mass particle (proton) has the higher probability of getting through. (b) With 0 V =10MeV , E = 3MeV , a =10-14m, it follows that 0 0 16 E (1 E ) 3.36 V V - = . The required masses are 1.673 10 27 pm = × - kg , 2 d p m ˜ m . For the proton 2 k a = 5.803 and, using the approximate formula, 3.36 2(5.083) 3.06 10 5 pT = e- = × - . Since 2 d p m ˜ m , as noted above, 2 k a.. 2 ×5.803 = 8.207 . Hence, for the deuteron, 3.36 2(8.207) 2.5 10 7 dT = e- = × - ……## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 6-10..A fusion reaction important in solar energy production (see Question 16) involves capture of a proton by a carbon nucleus, which has six times the charge of a proton and a radius of r'.. 2×10-15m . (a) Estimate the Coulomb potential V experienced by the proton if it is at the nuclear surface. (b) The proton is incident upon the nucleus because of its thermal motion. Its total energy cannot realistically be assumed to be much higher than 10kT, where k is Boltzmann’s constant (see Chapter 1) and where T is the internal temperature of the sun of about 107 0K . Estimate this total energy, and compare it with the height of Coulomb barrier. (c) Calculate the probability that the proton can penetrate a rectangular barrier potential of height V extending from r' to r'' , the point at which the Coulomb barrier potential drops to 2 V . (d) IS the penetration through the actual Coulomb barrier potential greater or less than through the rectangular barrier potential of part (c)? .......... CH06 - 4 - ANS..(a) 19 2 9 0 15 0 1 (9 10 ) (6)(1)(1.6 10 ) 4 210 V qQ pe r - - × = = × ' × 13 0 13 6.912 10 4.32 1.6 10 / V J MeV J MeV - - × = = × (b) 23 7 15 3 0 E =10kT = (10)(1.38×10- )(10 ) =1.38×10- J = 8.625×10- MeV = 0.002V (c) Numerically, a = 2r' - r' = 2×10-15m; also, 0 0 16 E (1 E ) 0.032 V V - = ; 0 2 2 ( ) 0.91 .. m V E k a a - = = 2 {1 (2.484 0.403) } 1 0.0073 0.032 T - - = + = (d) The actual barrier can be considered as a series of barriers, each of constant height but the heights decreasing with r; hence 0 V - E diminishes with r and the probability of penetration is greater than for an equal width barrier of constant height 0 V ……## .... Appendix S ......(10a) 4.32MeV (10b) 3 0 2×10- V (10c) 0.0073 ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 6-11..Verify by substitution that the standing wave general solution, (6-62), satisfies the timeindependent Schroedinger equation, (6-2), for the finite square well potential in the region inside the well. ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 6-12..Verify by substitution that the exponential general solutions, (6-63) and (6-64), satisfy the time- independent Schroedinger equation (6-13) for the finite square well potential in the regions outside the well. ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 6-13..(a) From qualitative arguments, make a sketch of the form of a typical unbound standing wave eigenfunction for a finite square well potential. (b) Is the amplitude of the oscillation the same in all regions? (c) What does the behavior of the amplitude predict about the probabilities of finding the particle in a unit length of the x axis in various regions? (d) Does .......... CH06 - 5 - the prediction agree with what would be expected from classical mechanics? ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 6-14..Use the qualitative arguments of Problem 13 to develop a condition on the total energy of the particle, in an unbound state of a finite square well potential, which makes the probability of finding it in a unit length of the x axis the same inside the well as outside the well. (Hint : What counts is the relation between the de Broglie wavelength inside the well and the width of the well.) ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 6-15..(a) Make a quantitative calculation of the transmission coefficient for an unbound particle moving over a finite square well potential. (Hint : Use a trick similar to the one indicated in Problem 7.) (b) Find a condition on the total energy of the particle which makes the transmission coefficient equal to one. (c) Compare with the condition found in Problem 14, and explain why they are the same. (d) Give an example of an optical analogue to this system. ANS..(15a) 2 2 1 [1 (sin )] 4 ( 1) k a x x + - - , 0 x E V = (15b) 2 2 2 2 2 n .. ma p ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 6-16..(a) Consider a one-dimensional square well potential of finite depth 0 V and width a. What combination of these parameters determines the “strength” of the well-i.e., the number of energy levels the wells is capable of binding? In the limit that the strength of the well becomes small, will the number of bound levels become 1 or 0? Give convincing justification for your answers. ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 6-17..An atom of noble gas krypton exerts an attractive potential on an unbound electron, which has a very abrupt onset. Because of this it is a reasonable approximation to describe the potential as an attractive square well, of radius equal to the 4×10-10m radius of the atom. Experiments show that an electron of kinetic energy 0.7eV, in regions outside the atom, can travel through the atom with essentially no reflection. The phenomenon is called the Ramsaure effect. Use this information in the conditions of Problem 14 or 15 to determine the depth of the square well potential. (Hint : One de Broglie wavelength just fits into the width of the well. Why not one-half a de Broglie wavelength?) ANS..Numerically a = 2(4×10-10m) and K = 0.7eV . 0 E = K +V where 2 2 34 2 2 2 2 31 10 2 19 (6.626 10 ) (0.588 ) 8 8(9.11 10 )(8 10 ) (1.6 10 ) E n h n n eV ma - - - - × = = = × × × Set n =1; 1 E = 0.588eV < K , which is not possible. .......... CH06 - 6 - Using n = 2 gives 2 2 1 E = 2 E = 2.352eV 0 V = E - K =1.65eV The electron is too energetic for only half its wavelength to fit into the well; this may be verified by calculating the deBroglie wavelength of an electron with a kinetic energy over the well of 2.35eV……## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 6-18..A particle of total energy 0 9V is incident from the -x axis on a potential given by 0 0 8 0 0 0 5 V x V x a V x a < = < < > Find the probability that the particle will be transmitted on through to the positive side of the x axis, x > a . ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 6-19 ..Verify by substitution that the standing wave general solution, (6-67), satisfies the time-independent Schroedinger equation (6-2), for the infinite square well potential in the region inside the well. ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 6-20..Two possible eigenfunctions for a particle moving freely in a region of length a, but strictly confined to that region, are shown in Figure 6-37. When the particle is in the state corresponding to the eigenfunction I . , its total energy is 4eV. (a) What is its total energy in the state corresponding to II . ? (b) What is the lowest possible total energy for the particle in this system? Figure 6-37 Two eigenfunctions considered in Problem 20 ANS..(a) In the lowest energy state n =1, . has no nodes. Hence I . must correspond to n = 2 , II . to n = 3. Since 2 n E . n and 4 I E = eV , 2 2 3 2 II I E E = ; 9 II E = eV . (b) By the same analysis, 2 0 2 1 2 I E E = ; 0 E =1eV …….## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ .......... CH06 - 7 - 6-21..(a) Estimate the zero-point energy for a neutron in a nucleus, by treating it as if it were in an infinite square well of wide equal to a nuclear diameter of 10-14m. (b) Compare your answer with the electron zero-point energy of Example 6-6. ANS..(21a) 2.05MeV ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 6-22..(a) Solve the classical wave equation governing the vibrations of a stretched string, for a string fixed at both its ends. Thereby show that functions describing the possible shapes assumed by the string are essentially the same as the eigenfunctions for an infinite square well potential. (b) Also show that the possible frequencies of vibration of the string are essentially different from the frequencies of the wave functions for the potential. ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 6-23..(a) For a particle in a box, show that the fractional difference in the energy between adjacent eigenvalues is 2 2 1 n n E n E n . + = . (b) Use this formula to discuss the classical limit of the system. ANS..(a) The energy in question is 2 2 2 2 2 .. n E n ma p = , and therefore the energy of the adjacent level is 2 2 2 1 2 ( 1)2 .. nE n ma p + = + , so that 2 2 1 2 2 n n n ( 1) 2 1 n n E E E n n n E E n n + . - + - + = = = . (b) In the classical limit n . 8 ; but 2 lim n lim 2 1 0 n n n E n .8 E .8 n . + = = Meaning that the energy levels get so close together as to be indistinguishable. Hence, quantum effects are not apparent. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 6-24..Apply the normalization condition to show that the value of the multiplicative constant for the n = 3 eigenfunction of the infinite square well potential, (6-79), is 3 B 2 a = . ANS..The eigenfunctions for odd n are cos n n B n x a p . = . For normalization, 2 2 2 2 2 2 2 0 2 1 cos 2 cos a n n n n a dx B n x dx B a udu a n p p . p - = . = . = . 1 2 2 ( )( ) 2 n 4 2 n B a n a B n p p = = . 2 n B n = For all odd n and, therefore, for n = 3. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 6-25..Use the eigenfunction of Problem 24 to calculate the following expectation values, and comment on each result : (a) x , (b) p , (c) x2 , (d) p2 . .......... CH06 - 8 - ANS..(25a) zero (25b) zero (25c) 0.0777a2 (25d) 88.826(.. )2 a ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 6-26..(a) Use the results of Problem 25 to evaluate the product of the uncertainly in position times the uncertainty in momentum, for a particle in the n = 3 state of an infinite square well potential. (b) Compare with the results of Example 5-10 and Problem 13 of Chapter 5, and comment on the relative size of the uncertainty products for the n =1, n = 2 , and n = 3 state. (c) Find the limits of .x and .p as n approaches infinity. ANS..(a) Using the results of the previous problem, 2 1/ 2 2 2 (1 6 ) 12 x x a n p . = = - , 2 ( ) .. p p na . = = p Hence, for n = 3, 1/ 2 2 2 (1 6 ) 3 2.67 12 3 .. .. x p a a p p . . = - = . (b) The other results are n =1, .x.p = 0.57.. n = 2 , .x.p =1.67.. The increase with n is due mainly to the uncertainty in p: see Problem 6-25. (c) From (a), the limits as n . 8 are 12 .x. a ; .p.8…….## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 6-27..Form the product of the eigenfunction for the n =1 state of an infinite square well potential times the eigenfunction for n = 3 state of that potential. Then integrate it over all x, and show that the result is equal to zero. In other words, prove that 1 3 . (x). (x)dx 0 8 -8 . = . (Hint : Use the relation : cos cos cos( ) cos( ) 2 u v u v u v + + - = .) Students who have worked Problem 36 of Chapter 5 have already proved that the integral over all x of the n =1 eigenfunction times the n = 2 eigenfunction also equals zero. It can be proved that the integral over all x of any two different eigenfunctions of the potential equals zero. Furtherrmore, this is true for any two different eigenfunctions of any other potential. (If the eigenfunctions are complex, the complex conjugate of one is taken in the integrand.) This property is called orthogonality. ANS.. 2 2 1 3 2 2 2 cos cos 3 1 {cos 4 cos 2 } a a a a dx x x dx x x dx a a a a a a p p p p . . + + +8 -8 - - . = . = . - 1 3 1 (cos 2 cos ) 2 dx u u du p p . . p +8 -8 - . = . - The integrand being an even function of u…….## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ .......... CH06 - 9 - 6-28..Apply the results of Problem 20 of Chapter 5 to the case of a particle in a three-dimensional box. That is, solve the time-independent Schroedinger equation for a particle moving in a three-dimensional potential that is zero inside a cubical region of edge length a, and becomes infinitely large outside that region. Determine the eigenvalues and eigenfunctions for system. ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 6-29..Airline passengers frequently observe the wingtips of their planes oscillating up and down with periods of the order of 1 sec and amplitudes of about 0.1m. (a) Prove that this is definitely not due to the zero-point motion of the wings by comparing the zero-point energy with the energy obtained from the quoted values plus an estimated mass for the wings. (b) Calculate the order of magnitude of the quantum number n of the observed oscillation. ANS..(a) Let M = mass of wing. The zero-point energy is 0 (1 1) 1 2 2 2 .. E h h T = + . = . = , T = period of oscillation. The actual energy of oscillation is 2 2 2 2 2 2 1 1 2 2 2 E kA M A MA T p = = . = Thus, the value of M at which 0 E = E is 34 33 2 2 2 1 2 (6.626 10 )(1) 1.68 10 4 4 (10 ) M hT kg p A p - - - × = = = × This is less than the mass of an electron. Hence 0 E >> E and the observed vibration is not the zero-point motion. (b) Clearly then, n >>1 and therefore 2 2 2 2 2 E nh 2 MA n 2 MA T hT p p = . = . = As an example, take M = 2000kg : 2 12 35 34 2 (2000)(10 ) 6 10 (6.626 10 )(1) n p - - = = × × …….## .... Appendix S......(29b) .. 1036 ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 6-30..The restoring force constant C for the vibrations of the interatomic spacing of a typical diatomic molecule is about 103 joule /m2 . Use this value to estimate the zero-point energy of the molecular vibrations. The mass of the molecule is 4.1×10-26 kg . ANS..The zero-point energy is 1/ 2 0 1 1 ( ) 2 2 .. .. E C m = . = Therefore, 3 34 1/ 2 19 1 0 26 1 (1.055 10 )( 10 ) (1.6 10 ) 2 4.1 10 E - - - - = × × × 0 E = 0.051eV ……## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 6-31..(a) Estimate the difference in energy between the ground state and first excited state of the vibrating molecule considered in Problem 30. (b) From this estimate determine the energy of the photon emitted by the vibrations in the charge distribution when the system makes a .......... CH06 - 10 - transition between the first excited state and the ground state. (c) Determine also the frequency of the photon, and compare it with the classical oscillation frequency of the system. (d) In what range of the electromagnetic spectrum is it? ANS..(a) Using 0 E = 0.051eV , the level spacing will be 0 ( 1) 0.102 2 2 .E = . n + ... =... = eV = E . (b) The energy E of the proton = .E = 0.102eV . (c) For the proton, .. ph E = . But E = .E =... . ph . =. Where . = classical oscillation frequency. Thus, 19 13 34 (0.102)(1.6 10 ) 2.5 10 6.626 10 E Hz h . - - × = = = × × . (d) Photons of this frequency are in the infrared spectrum, . =12,000nm…….## ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 6-32..A pendulum, consisting of a weight of 1kg at the end of a light 1m rod, is oscillating with an amplitude of 0.1m. Evaluate the following quantities : (a) Frequency of oscillation, (b) energy of oscillation, (c) approximate value of quantum number for oscillation, (d) separation in energy between adjacent allowed energies, (e) separation in distance between adjacent bumps in the probability density function near the equilibrium point. ANS..(a) 9.8 3.13 / 1 g rad s L . = = = . 0.498 2 Hz . . p = = . (b) 1 2 1 2 2 2 E kA mg A L = = . E = 0.049J . (c) Since n >>1, 32 34 0.0490 1.5 10 (6.626 10 )(0.498) n E h. - = = = × × . (d) Since .n =1, .E = h. = 3.3×10-34 J . (e) A polynomial of degree n has n nodes; hence, the distance between “bumps”=distance adjacent nodes 33 32 2 2(0.1) 1.3 10 1.5 10 A m n = = = × - × ……## .... Appendix S......(32a) 0.5Hz (32b) 0.049 joule (32c) 1.5×1032 (32d) 3.3×10-34 joule (32e) 1.3×10-33m ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 6-33..Devise a simple argument verifying that the exponent in the decreasing exponential, which governs the behavior of simple harmonic oscillator eigenfunctions in the classically excluded region, is proportional to x2 . (Hint : Take the finite square well eigenfunctions of (6-63) and (6-64), and treat the quantity ( 0 V - E ) as if it increased with increasing x in proportion to x2 .) ANS.. .......... CH06 - 11 - ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ 6-34..Verify the eigenfunction and eigenvalue for the n = 2 state of a simple harmonic oscillator by direct substitution into the time-independent Schroedinger equation, as in Example 6-7. ANS.. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~