Chemistry 36 Dr Jen M Stndrd Problem Set 3 Solutions 1 Verify for the prticle in one-dimensionl box by explicit integrtion tht the wvefunction ψ ( x) π x is normlized To verify tht ψ ( x) is normlized, we hve to show tht ψ (x) dx 1 Note tht the integrtion rnge is from x to x since the prticle in box wvefunctions vnish outside tht rnge Substituting, the integrl becomes ψ (x) dx π x sin π x dx π x dx The integrl tht we need cn be found, for exmple, in the CRC or in most ny "useful" tble of integrls: sin bx dx x sinbx 4b Setting b π nd substituting, the normliztion integrl becomes ψ (x)dx L sin π x dx x 1 4 π sin 4π x
1 continued Evluting the expression t the limits, ψ (x) dx ψ (x) dx 1 x 1 4 π Therefore, the wvefunction ψ ( x) is normlized sin 4π x sin( 4π ) 8π 8π sin ( ) [ ]
3 Clculte the energy level spcings in Joules between the ground (n1) nd first excited (n) levels for the following cses ) An electron confined to one-dimensionl box of width 5 Å The energy level spcing ΔE between the ground nd first excited stte is ΔE E E 1 Using the prticle in box energy expression, we cn obtin n eqution for the energy level spcing, ΔE E E 1 ΔE 3h 8m h 8m 1 h 8m Substituting, ΔE E E 1 3( 66618 1 34 Js) ( )( 5 1 1 m) 8 91953 1 31 kg ΔE E E 1 73 1 19 J b) A bsebll with mss of 14 g confined to one-dimensionl box of width 1 meter Using the sme expression from prt () for the energy level spcing, we hve ΔE E E 1 ΔE 3h 8m h 8m 1 h 8m Substituting, ( ) ΔE E E 1 3 66618 134 Js ( )( 1 m) 8 1 kg ΔE E E 1 1646 1 66 J Notice how much smller the energy level spcing is for the mcroscopic object Energy level spcings this smll re not mesurble, nd therefore the energy level spcings re so smll s to be effectively continuous for mcroscopic objects
4 3 An electron in box of width undergoes trnsition from the lowest energy level (n1) to the first excited level (n) The wvelength of light bsorbed in this trnsition ws determined to be 65 nm Clculte the width of the box For trnsition from n1 to n, the energy difference ΔE is ΔE E E 1 A photon with n energy corresponding to ΔE would hve frequency given by E photon ΔE hν Since, for light, λν c, we cn substitute ν λ, nd obtin n expression for the energy difference, c ΔE E E 1 hc λ Substituting, ΔE E E 1 ΔE E E 1 913 1 19 J ( 66618 1 34 Js) 99793 1 8 ms 1 % ( 68 nm) 1-9 m( ' * & 1 nm ) ( ) Then, we cn use the prticle in box energies to obtin n expression for the energy difference, ΔE E E 1 h 8m 1 h 8m ΔE 3h 8m Solving for, the width of the box, yields 3h 8mΔE 1/ Substituting, 3h 8mΔE 1/ 3 ( 66618 1 34 Js) ( )( 913 1 19 J) 8 91953 1 31 kg 1/ 7866 1 1 m or 7866 Å
5 4 Determine the most probble loction for the prticle in the ground stte of the one-dimensionl prticle in box Also determine the most probble loction for the first excited stte of the prticle in box To obtin the most probble vlue, we must look t the probbility density, ψ ( x) For the ground stte of the prticle in box, the probbility density is given by ψ 1 ( x) π x sin A plot of this probbility density is shown below Note tht the mximum of this function occurs t x ; this is the most probble vlue In this cse, the mximum cn be found by inspection For more generl situtions to get the mximum vlue, we would hve to tke the derivtive, set it equl to zero, nd solve For the first excited stte of the prticle in box, the probbility density is given by ψ (x) π x sin A plot of this probbility density is given below For this function, we see from the plot tht there re two mxim, ech with identicl mplitude By inspection, these mxim occur t x 4 nd x 3 ; these two vlues re therefore the most probble vlues 4
6 5 For prticle in one-dimensionl box of width, determine the probbility of finding the prticle in the right third of the box (between /3 nd ) if the prticle is in the ground stte Since the probbility is given by ψ ( x) dx, if we wnt the totl probbility of finding the prticle between /3 nd, we must dd up the probbility for ll the points from /3 to Since x is continuous vrible, the sum is relly n integrl from /3 to, Probbility ψ x /3 ( ) dx Substituting the prticle in box wvefunction for the ground stte, ψ 1 (x) integrl becomes sin π x, the probbility Probbility sin π x dx /3 This integrl cn be evluted using tbles From the CRC Hndbook or ny other tble of integrls, we find the indefinite integrl: sin bx dx x sin bx 4b Replcing b with π yields Probbility x sin π x 4 π /3 Finlly, evluting the expression t the limits leds to Probbility 1 3 + 1 π sin 4π 3 Evluting the numericl vlue of the sine function in the expression bove, the probbility is Probbility 3333 1378 1955
7 6 The 1,3,5-hextriene molecule is conjugted molecule with 6 pi electrons Consider the pi electrons free to move bck nd forth long the molecule through the deloclized pi system Using the prticle in box pproximtion, tret the crbon chin s liner one-dimensionl "box" Allow ech energy level in the box to hold pi electrons Treting 1,3,5-hextriene s liner chin, we would hve the structure The pi electrons re free to move through the conjugted system; hence, the length of the molecule would represent the "box" We cn dd up the lengths of the single nd double bonds in the molecule to give the width of the box, For 1,3,5-hextriene, this gives where r represents the bond length C 1 C C 3 C 4 C 5 C 6 3r CC + r CC, The energy of the pi electrons in the molecule would be represented s E n n h 8m, n 1,, 3,, where is the box width defined bove nd m is the electron mss ) Clculte the energy of the highest filled level, using 154 Å s the crbon-crbon single bond length, nd 135 Å s the crbon-crbon double bond length Using r C C 135 Å nd r C C 154 Å, the width of the box cn be clculted from the eqution bove 3(135 Å) + (154 Å) 713 Å (or 713 1-1 m) Using the formul bove with 713 1 1 m for the energy leds to the energy digrm shown below n5 n4 n3 n n1
8 6 ) continued 1,3,5-hextriene hs 6 pi electrons Plcing two electrons in ech energy level fills the levels up to the n3 level, s shown in the figure bove So, n3 is the highest filled level Computing the energy of the highest filled level, E 3 ( ) ( )( 713 1 1 m) ( ) 3 h 8m 3 66618 1 34 Js 8 91953 1 31 kg 3 118511 1 19 J E 3 1666 1 18 J b) Determine the energy of the lowest unfilled level The lowest unfilled level corresponds to n4 Clculting the energy of this stte gives E 4 4 h 8m ( ) E 4 4 118511 1 19 J E 4 1896 1 18 J c) Clculte the wvelength for n electronic trnsition from the highest filled level to the lowest unfilled level, using your nswers from prts ) nd b) Compre your result to the experimentl ultrviolet bsorption mximum of 68 nm For n electronic trnsition from n3 to n4, the energy difference ΔE is ΔE E 4 E 3 A photon with n energy corresponding to ΔE would hve frequency given by E photon ΔE hν Since, for light, λν c, we cn substitute ν λ, nd solve for the wvelength to give c λ hc ΔE hc E 4 E 3 Inserting numericl vlues leds to λ ( 66618 1 34 Js) 99793 1 8 ms 1 ( ) ( 1896 1 18 J 1666 1 18 J) λ 395 1 7 m or λ 395nm This result is firly close to the experimentl wvelength, λ exp 68 nm with bout n 11% error This is surprisingly good greement given the crudeness of the model
9 7 Verify, by explicit integrtion, tht ψ nd ψ 3 for the prticle in one-dimensionl box re orthogonl For prticle in 1D box, the wvefunctions re given by ψ n (x) nπ x To show tht ψ nd ψ 3 re orthogonl, the integrl of the product of the two functions (including the complex conjugte of one of the functions if they re not rel functions) must be zero Evluting the integrl (in this cse, both functions re rel), we hve ψ * (x)ψ 3 (x) dx ψ (x) ψ 3 (x) dx sin π x sin 3π x dx L L Note tht the integrtion rnge is from x to x since the prticle in box wvefunctions vnish outside tht rnge The indefinite integrl tht we need cn be found, for exmple, in the CRC or in most ny tble of integrls: sin mx sin nx dx sin( m n)x m n ( ) sin( m + n)x m + n ( ) Setting m π nd n 3π, we hve m n π nd m + n 5π Substituting, ψ (x) ψ 3 (x) dx π sin π x L 5π x 1π L Using the identity sin( x) sin x, nd evluting the integrl t the limits yields ψ (x) ψ 3 (x) dx π sinπ 1π sin5π π sin 1π sin Since sin π sin 5π sin, the integrl simplifies to ψ (x) ψ 3 (x) dx Therefore, the functions ψ ( x) nd ψ 3 ( x) re orthogonl
1 8 Determine the verge vlue of the position x for the ground stte of the one-dimensionl prticle in box Compre your result with the most probble loction The verge vlue of the position is given by x ψ * 1 (x) ˆx ψ 1 (x) dx ψ * 1 (x) x ψ 1 (x) dx, where the definition of the position opertor, ˆ x x, hs been used In ddition, the limits of the integrl re x to x becuse the wvefunction vnishes outside this rnge The ground stte wvefunction for the prticle in box is given by ψ 1 (x) sin π x Upon substitution, the expression for the verge vlue of the position becomes x ψ 1 * (x) x ψ 1 (x) dx sin π x x x sin π x dx sin π x dx From the CRC Hndbook (or hndout of integrls), xsin x bx dx 4 xsinbx 4b cosbx 8b Replcing b by π, the verge vlue becomes x x 4 xsin π x 4 π cos π x 8 π
11 8 continued Evluting the expression t the limits yields the verge vlue of the position for the ground stte of the prticle in box: x x 4 4 sin( π ) 4π 8π cos π ( ) 8π 8π sin( ) 8π cos ( ) For the ground stte of the prticle in box, the verge vlue nd most probble vlue of x re identicl (the most probble vlue of / ws found in problem 4) This will not lwys be the cse, s we will see in next problem
1 9 Repet problem 8 for the first excited stte of the prticle in box Does your result gree this time with the most probble loction for this stte? For the first excited stte of the prticle in box, the wvefunction is ψ (x) π x The verge vlue of x for this stte is x ψ * (x) x ψ (x) dx x π x x xsin π x dx π x dx We cn use the sme integrl s we did in the previous problem; in this cse, we replce b by π : x x 4 xsin 4π x 4 π cos 4π x 8 π Evluting this expression t the limits gives the verge vlue of the position for the first excited stte, x 4 sin( 4π ) 8π cos 4π ( ) 3π sin( ) 3π cos ( ) 4 3π 3π x This is exctly the sme result tht we got for the ground stte This will not lwys hppen here we got the sme result becuse of the symmetry of the potentil energy nd s result the symmetry of the wvefunction bout x/ From problem 4, we found tht the most probble vlues for the first excited stte occur t x 4 nd x 3 4 Note tht for the first excited stte of the prticle in box, the verge vlue nd most probble vlues of x re not the sme
1 Determine the verge vlue of the momentum p x for the ground stte of the one-dimensionl prticle in box 13 The verge vlue of the momentum is given by p x ψ * 1 (x) ˆp x ψ 1 (x) dx i! ψ * 1 (x) d dx ψ 1(x) dx, where the definition of the momentum opertor, p ˆ x i! d, hs been used In ddition, the limits of the dx integrl gin re x to x becuse the wvefunction vnishes outside this rnge The ground stte wvefunction for the prticle in box is given by ψ 1 (x) sin π x Upon substitution, the expression for the verge vlue of the momentum becomes p x i! ψ 1 * (x) d dx ψ 1(x) dx i! p x i! sin π x d dx sin π x dx sin π x d sin π x dx dx Next, the derivtive must be evluted, d dx sin π x π cos π x Substituting, the verge vlue of the momentum is p x i!π sin π x cos π x dx From the CRC Hndbook (or hndout of integrls), sin bx cosbx dx sin bx b
14 1 continued Replcing b by π, the verge vlue of momentum becomes p x i!π sin π x π Evluting this expression t the limits gives the verge vlue of momentum for the ground stte, p x i!π π sin ( π ) i!π [ ] + p x + i!π i!π [ ] π sin ( ) For the ground stte of the prticle in box, the verge vlue of momentum is This is ctully true for ny stte of the prticle in box, nd reflects the ide tht motion in the positive x-direction, corresponding to positive vlues of momentum, nd motion in the negtive x-direction, corresponding to negtive vlues of momentum, re eqully probble Thus, the positive nd negtive vlues of momentum cncel, yielding n verge vlue of