《结构化学》课程PPT教学课件（讲稿）Chapter 2 The Particle in a Box

Chapter 2 The Particle in a Box s2. 1 Differential Equations Linear n(x)y)+A-1(x)y3+-…+A4y=g(x) order V(x g(x)=0: homogeneous(齐次) g(x)#0: inhomogeneous y+P(xr)y+o(x)=0 Independent y=C1V1+C2y2 n-th order c> n constants boundary conditions

Constant coefficient v+py+gy=0 assuming y =se 2 2 sx se t pse tge 0 s2+ns+q=0 auxiliary(辅助) equation V=Ge+ce 2 r Example 1+6y-7=0s2+6s-7=0 7x ce t ce

52.2 Particle in a one-dimension box to∞ to∞ Physically unreal ∈I.I (x)=0,x Figure 2. 1 Potential energy function V(x) for the particle in a one-di ional box 2a2 +v(ry(r)=ey(xr) 2m dx For region I andⅢn2 (E-∞y

For region II 21 +=E 0 2 方 s2+2mh-E=0 土(-2m)h=±(2mE)h i(2mE)2x i(2mEy2x ce 方 方 0=(2mExh-I yu=ce tce e8= cos 0+isin e-19=cos0-isin 0 Wu=C cos 6+ic, sin 8+c2 cos 6-ic, sin e =(C+c,cos 0+(ic -ic,)sin e

V =Acos h(2mEx+Bsin h(2me s Using boundary conditions v1(0)=v.(0)A=0 V()=V()=0D sin (2丌E) 7=0 (2E)1 =1 =0,±1 7=0 vm=0不合理 E n=123 8ml

E Figure 2.2 Lowest four nergy levels for the particle in a one-dimensional box n n boundary conditions o quantized energy minimum value is greater than zero ground state excited state

A particle of mass 2.00x10-26g is in a one-dimension box of length 4.00 nm Find the frequency and wavelength of the photon emitted when this particle goes from the n=3 to the n=2 level hv=e E e h 8ml- 8m/ H,-1 (32-2)(6.626×107s) 1.29×102s 8ml 8×2.00×10kg×(4.00×10m) N×m=Kg×=2×1m g A=2.32×10 久=C

nox WI= Bsin ,n=1,2,3 B=? USing normalization requirement 0 dx+ dx 0 11/m/ dr=1 Sin 2sin 6=1-2cos 20) B

(6=0 117x Sin ,n=1,2,3, 2 FIGURE 2.3 Graphs of for the three lowest-energy particle-in-a-box states

∧^∧ FIGURE 2.4 Graphs of wf for the lowest particle-in-a-box states