Physics 43 Homework Set 9 Chapter 40 Key


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1 Physics 43 Homework Set 9 Chpter 4 Key. The wve function for n electron tht is confined to x nm is. Find the normliztion constnt. b. Wht is the probbility of finding the electron in. nmwide region t x =. nm? ) b) + Normlize: dx = gives b =.559nm.5nm.5nm 7% P = dx = (.559) exp( x / 3.) dx =.86 =.3%.995nm.995nm πx x = sin for x nd is zero otherwise.. Sketch the probbility density function for this stte. b. Find the most probble position(s) of the electron in the well. c. Find the expecttion vlue of x. d. Why or why not should the results of b nd c be the sme or different? Tht is, should the most probble position(s) be the expecttion vlue? (Hint: Use your sketch). An electron in n infinite squre well hs wve function tht is given by ( ) Solution in glss cse. 3. A prticle in n infinite squre well hs wve function tht is given by πx ( x) = sin for x nd is zero otherwise. () Determine the probbility of finding the prticle between x = nd x = /3. (b) Use the result of this clcultion nd symmetry rguments to find the probbility of finding the prticle between x = /3 nd x = /3. Do not reevlute the integrl. (c) Wht If? Compre the result of prt () with the clssicl probbility. () The probbility is πx πx P = dx = sin dx cos dx = 3 x π x π 3 P = sin = sin = =.96 π 3 π 3 3 4π.
2 (b) The probbility density is symmetric bout x =. Thus, the probbility of finding the prticle between x = nd x = is the sme.96. Therefore, the 3 probbility of finding it in the rnge x is P =. (.96) = FIG. P4.(b) (c) Clssiclly, the electron moves bck nd forth with constnt speed between the wlls, nd the probbility of finding the electron is the sme for ll points between the wlls. Thus, the clssicl probbility of finding the electron in ny rnge equl to onethird of the vilble spce is P clssicl =. 3 πx sin 4. A prticle in n infinitely deep squre well hs wve function given by ( ) for x nd zero otherwise. () Determine the expecttion vlue of x. (b) Determine the probbility of finding the prticle ner /, by clculting the probbility tht the prticle lies in the rnge.49 x.5. (c) Wht If? Determine the probbility of finding the prticle ner /4, by clculting the probbility tht the prticle lies in the rnge.4 x.6. (d) Argue tht the result of prt () does not contrdict the results of prts (b) nd (c). x = () π x 4π x x = x sin dx = x cos dx x x 4π x 4π x 4π x = + = π sin cos 6 (b).5.5 π x 4π x Probbility = sin dx = x sin π 4.49 Probbility ( π π) 4π.49 =. sin.4 sin.96 = (c).6 x 4π x Probbility sin = π (d) In the n =, it is more probble to find the prticle either ner x = or = 3 x thn t the center, where the 4 4 probbility density is zero. Nevertheless, the symmetry of the distribution mens tht the verge position is.
3 5. An electron is trpped somewhere in molecule which is 7.3 nm long. Clculte the minimum kinetic energy of the electron. 34 h h 6.33x Js 7 x p p = = 7.6x kgm / s 9 4π 4π x 4 π(7.3x m) 7 p (7.6x kgm / s) 3 K = = =.74x J =.ev 3 m (9.x kg) 6. The wve function for prticle is ( x) = π ( x + ) for > nd < x < +. Determine the probbility tht the prticle is locted somewhere between x = nd x = +. P4. Probbility ( ) tn x P = x = dx = π π ( x + ) π π P = tn tn ( ) π = π = The wve function of prticle is given by ( x) = Acos ( kx) + Bsin( kx) where A, B, nd k re constnts. Show tht is solution of the Schrödinger eqution, ssuming the prticle is free (U = ), nd find the corresponding energy E of the prticle.
4 8. Consider prticle moving in onedimensionl box for which the wlls re t x = / nd x = /. () Write the wve functions nd probbility densities for n =, n =, nd n = 3. (b) Sketch the wve functions nd probbility densities. A prticle is confined in rigid onedimensionl box of length. Visulize: Solve: () (x) is zero becuse it is physiclly impossible for the prticle to be there becuse the box is rigid. (b) The potentil energy within the region / x / is U(x) = J. The Schrödinger eqution in this region is where β = me. ( ) d x m = E x = x dx ( ) β( ) (c) Two functions (x) tht stisfy the bove eqution re sinβx nd cosβx. A generl solution to the Schrödinger eqution in this region is (x) = Asinβ x + Bcosβ x where A nd B re constnts to be determined by the boundry conditions nd normliztion. (d) The wve function must be continuous t ll points. = just outside the edges of the box. Continuity requires tht lso be zero t the edges. The boundry conditions re (x = /) = nd (x = /) =. (e) The two boundry conditions re β β β β ( ) = Asin + Bcos = Asin + Bcos = β β ( ) = Asin + Bcos = These re two simultneous equtions. Unlike the boundry conditions in the prticle in box problem of Section 4.4, there re two distinct wys to stisfy these equtions. The first wy is to dd the equtions. This gives To finish stisfying the boundry conditions, β Bcos = B = (x) = Asinβx β sin = β = π, 4π, 6π, = nπ with n =,, 3, With this restriction on the vlues of β, the wve function becomes ( x) Asin( nπx) from prt (b), the energy is ( nπ ) h ( n) En = = n =,, 3, m 8m The second wy is to subtrct the second eqution from the first. This gives β Acos = A = (x) = Bcosβ x =. Using the definition of β
5 To finish stisfying the boundry conditions, β cos = β = π, 3π, 5π, = (n )π n =,, 3, With this restriction on the vlues of β, the wve function becomes ( x) Bcos( ( n ) πx ) β, the energy is ( n ) π h ( n ) En = = n =,, 3, m 8m Summrizing this informtion, the llowed energies nd the corresponding wve functions re ( x) ( ) n π x h Bcos En = ( n ) = E, 9 E, 5 E, 8m = ( n) π x h Asin En = ( n) = 4 E, 6 E,36 E, 8m =. Using the definition of where E = h /8m. (f) The results re ctully the sme s the results for prticle locted t x. Tht is, the energy levels re the sme nd the shpes of the wve functions re the sme. This hs to be, becuse neither the prticle nor the potentil well hve chnged. All tht is different is our choice of coordinte system, nd physiclly meningful results cn t depend on the choice of coordinte system. The new coordinte system forces us to use both sines nd cosines, wheres before we could use just sines, but the shpes of the wve functions in the box hven t chnged. 9. An electron is contined in onedimensionl box of length. nm. () Drw n energylevel digrm for the electron for levels up to n = 4. (b) Find the wvelengths of ll photons tht cn be emitted by the electron in mking downwrd trnsitions tht could eventully crry it from the n = 4 stte to the n = stte. () We cn drw digrm tht prllels our tretment of stnding mechnicl wves. In ech stte, we mesure the distnce d from one node to nother (N to N), nd bse our solution upon tht: Since λ h dn to N = nd λ = p h h p λ d Next, p h K = = = m 8 3 e md e d 8 9. kg Evluting, J m K = d ev m K =. d In stte, d =. m K = 37.7 ev. In stte, d = 5. m K = 5 ev. In stte 3, d = 3.33 m K 3 = 339 ev. In stte 4, d =.5 m K 4 = 63 ev. 34 ( 6.66 J s) ( ). FIG. P4.5
6 . A prticle is described by the wve function () Determine the normliztion constnt A. (b) Wht is the probbility tht the prticle will be found between x = nd x = /8 if its position is mesured? () dx = becomes 4 4 π x π x 4π x π A cos dx = A + = = π sin A π or A = nd A =. 4 (b) The probbility of finding the prticle between nd 8 is 8 8 π x dx = A cos dx = + =.49 4 π
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