2是是+删 lh2121222 2h233 n comparing the total energy at constant volume of the undistorted box L1-L2=L3) ersus the distorted box (L3*L1=L2) it can be seen that 66h2|12 LI2+L32s8mL1 as long as L32L1 c. In order to minimize the total energy expression, take the derivative of the energy with respect to Li and set it equal to zero OL 孔L8mL12+j 0 But since V=L L2L3=L12L3, then L3=V/L|2. This substitution gives (h2|66L14 aLi 8mL 0 h2(2)6(4L 24L
26 = 2h2 8mè ç æ ø ÷ ö 12 L1 2 + 12 L2 2 + 12 L3 2 + 1h2 8mè ç æ ø ÷ ö 12 L1 2 + 12 L2 2 + 22 L3 2 = 2h2 8mè ç æ ø ÷ ö 2 L1 2 + 1 L3 2 + 1h2 8mè ç æ ø ÷ ö 2 L1 2 + 4 L3 2 = 2h2 8mè ç æ ø ÷ ö 2 L1 2 + 1 L3 2 + 1 L1 2 + 2 L3 2 = 2h2 8mè ç æ ø ÷ ö 3 L1 2 + 3 L3 2 = h2 8mè ç æ ø ÷ ö 6 L1 2 + 6 L3 2 In comparing the total energy at constant volume of the undistorted box (L1 = L2 = L3) versus the distorted box (L3 ¹ L1 = L2) it can be seen that: h2 8mè ç æ ø ÷ ö 6 L1 2 + 6 L3 2 £ h2 8mè ç æ ø ÷ ö 12 L1 2 as long as L3 ³ L1. c. In order to minimize the total energy expression, take the derivative of the energy with respect to L1 and set it equal to zero. ¶Etotal ¶L1 = 0 ¶ ¶L1 è ç æ ø ÷ ö h2 8mè ç æ ø ÷ ö 6 L1 2 + 6 L3 2 = 0 But since V = L1L2L3 = L1 2L3, then L3 = V/L1 2. This substitution gives: ¶ ¶L1 è ç æ ø ÷ ö h2 8mè ç æ ø ÷ ö 6 L1 2 + 6L1 4 V2 = 0 è ç æ ø ÷ ö h2 8mè ç æ ø ÷ ö (-2)6 L1 3 + (4)6L1 3 V2 = 0 è ç æ ø ÷ ö - 12 L1 3 + 24L1 3 V2 = 0
24L 2 L16=元V2=(L12L 2 L14 L 3 d Calculate energy upon distortion distorted: V=LL3=L122L1=V2 L13 AE= Etotal L1=L2=L3)-Etotal ( L3*L1=L2) m,2 6(2)/36(2)1 2/3 Since v=8A3.V2/3=4A2=4x10- and 6 E=601x1027 erg cm44x1016cm2
27 è ç æ ø ÷ ö 24L1 3 V2 = è ç æ ø ÷ ö 12 L1 3 24L1 6 = 12V2 L1 6 = 1 2 V2 = 1 2( ) L1 2L3 2 = 1 2 L1 4L3 2 L1 2 = 1 2 L3 2 L3 = 2 L1 d. Calculate energy upon distortion: cube: V = L1 3, L1 = L2 = L3 = (V) 1 3 distorted: V = L1 2L3 = L1 2 2 L1 = 2 L1 3 L3 = 2è ç æ ø ÷ V ö 2 1 3 ¹ L1 = L2 = è ç æ ø ÷ V ö 2 1 3 DE = Etotal(L1 = L2 = L3) - Etotal(L3 ¹ L1 = L2) = h2 8mè ç æ ø ÷ ö 12 L1 2 - h2 8mè ç æ ø ÷ ö 6 L1 2 + 6 L3 2 = h2 8mè ç æ ø ÷ ö 12 V2/3 - 6(2)1/3 V2/3 + 6(2)1/3 2V2/3 = h2 8mè ç æ ø ÷ ö 12 - 9(2)1/3 V2/3 Since V = 8Å3, V2/3 = 4Å2 = 4 x 10-16 cm2 , and h2 8m = 6.01 x 10-27 erg cm2: DE = 6.01 x 10-27 erg cm2 è ç æ ø ÷ ö 12 - 9(2)1/3 4 x 10-16 cm2
△E=601x1027 erg cm cm AE=0.99x10 g AE=099×10egy2a AE=6.19eV h a. H=2m ax2+ ay2]( Cartesian coordinates) Finding ox anday from the chain rule gives a(apa a a ax ar ar + lay/ ao Evaluation of the"coefficients" gives the following Sinφ ao Cosd Sinφ,and Upon substitution of these"coefficients aSinφoSinφ Ox= Coso Or-r 89 at fixed r C ay=Sind or at fixed
28 DE = 6.01 x 10-27 erg cm2 è ç æ ø ÷ 0.66 ö 4 x 10-16 cm2 DE = 0.99 x 10-11 erg DE = 0.99 x 10-11 erg è ç æ ø ÷ 1 eV ö 1.6 x 10-12 erg DE = 6.19 eV 12. a. H = -h-2 2m î í ì þ ý ü ¶ 2 ¶x2 + ¶ 2 ¶y2 (Cartesian coordinates) Finding ¶ ¶x and ¶ ¶y from the chain rule gives: ¶ ¶x = è ç æ ø ÷ ¶rö ¶x y ¶ ¶r + è ç æ ø ÷ ¶fö ¶x y ¶ ¶f , ¶ ¶y = è ç æ ø ÷ ¶rö ¶y x ¶ ¶r + è ç æ ø ÷ ¶fö ¶y x ¶ ¶f , Evaluation of the "coefficients" gives the following: è ç æ ø ÷ ¶rö ¶x y = Cosf , è ç æ ø ÷ ¶fö ¶x y = - Sinf r , è ç æ ø ÷ ¶rö ¶y x = Sinf , and è ç æ ø ÷ ¶fö ¶y x = Cosf r , Upon substitution of these "coefficients": ¶ ¶x = Cosf ¶ ¶r - Sinf r ¶ ¶f = - Sinf r ¶ ¶f ; at fixed r. ¶ ¶y = Sinf ¶ ¶r + Cosf r ¶ ¶f = Cosf r ¶ ¶f ; at fixed r
az-( Sind sin a at fixed Cosφ s32φ a2 Cos Sind a at fixed a2Sin2φa2 Sino Cosd a Cos2φa2cosφSinφa r2+p2 (cylindrical coordinates, fixed r) 21 The Schrodinger equation for a particle on a ring then becomes H=Ey -h2a2Φ 即) The general solution to this equation is the now familiar expression d(φ)=C1e-im+C2em h
29 ¶ 2 ¶x2 = è ç æ ø ÷ ö - Sinf r ¶ ¶f è ç æ ø ÷ ö - Sinf r ¶ ¶f = Sin2f r 2 ¶ 2 ¶f2 + SinfCosf r 2 ¶ ¶f ; at fixed r. ¶ 2 ¶y2 = è ç æ ø ÷ Cosf ö r ¶ ¶f è ç æ ø ÷ Cosf ö r ¶ ¶f = Cos2f r 2 ¶ 2 ¶f2 - CosfSinf r 2 ¶ ¶f ; at fixed r. ¶ 2 ¶x2 + ¶ 2 ¶y2 = Sin2f r 2 ¶ 2 ¶f2 + SinfCosf r 2 ¶ ¶f + Cos2f r 2 ¶ 2 ¶f2 - CosfSinf r 2 ¶ ¶f = 1 r 2 ¶ 2 ¶f2 ; at fixed r. So, H = -h-2 2mr2 ¶ 2 ¶f2 (cylindrical coordinates, fixed r) = -h-2 2I ¶ 2 ¶f2 The Schrödinger equation for a particle on a ring then becomes: HY = EY -h-2 2I ¶ 2F ¶f2 = EF ¶ 2F ¶f2 = è ç æ ø ÷ -2IEö -h2 F The general solution to this equation is the now familiar expression: F(f) = C1e -imf + C2e imf , where m = è ç æ ø ÷ 2IEö -h2 1 2
Application of the cyclic boundary condition,Φ(φ)=Φ(φ+2π), results in the quantization of the energy expression: E=2I where m=0,tl, +2, #3,. It can be seen that the +m values correspond to angular momentum of the same magnitude but opposite directions. Normalization of the wavefunction(over the region 0 to 2T)corresponding to + or-m will result in a value of for the normalization constant Φ(φ) h2 6. x 10-28 erg cm2 (14x1 rg △E=(22-12)309x10-12erg=9.27x1012erg
30 Application of the cyclic boundary condition, F(f) = F(f+2p), results in the quantization of the energy expression: E = m2-h2 2I where m = 0, ±1, ±2, ±3, ... It can be seen that the ±m values correspond to angular momentum of the same magnitude but opposite directions. Normalization of the wavefunction (over the region 0 to 2p) corresponding to + or - m will result in a value of è ç æ ø ÷ 1 ö 2p 1 2 for the normalization constant. \ F(f) = è ç æ ø ÷ 1 ö 2p 1 2 eimf ¾¾ ¾¾ (±4)2-h2 2I ¾¾ ¾¾ (±3)2-h2 2I ¾¾ ¾¾ (±2)2-h2 2I ¾¾¯ ¾¾¯ (±1)2-h2 2I ¾¾¯ (0)2-h2 2I b. -h2 2m = 6.06 x 10-28 erg cm2 -h2 2mr2 = 6.06 x 10-28 erg cm2 (1.4 x 10-8 cm)2 = 3.09 x 10-12 erg DE = (22 - 12) 3.09 x 10-12 erg = 9.27 x 10-12 erg