半导体物理与器件第四版课后习题答案2

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1、Chapter 22.1 Sketch_2.2 Sketch_2.3 Sketch_2.4 From Problem 2.2, phase = constant Then From Problem 2.3, phase = constant Then _2.5 Gold: eV J So, cm or m Cesium: eV J So, cm or m_2.6 (a) kg-m/s m/s or cm/s (b) kg-m/s m/s or cm/s (c) Yes_2.7(a) (i) kg-m/s m or (ii) kg-m/s m or (iii) kg-m/s m or (b) k

2、g-m/s m or _2.8 eV Now or kg-m/s Now m or _2.9 Now and Set and Then which yields J keV_2.10(a) kg-m/s m/sor cm/s Jor eV(b) Jor eV kg-m/s mor _2.11(a) J Now VkV(b) kg-m/s Then m or _2.12 kg-m/s_2.13(a) (i) kg-m/s(ii) Now kg-m/sso Jor eV(b) (i) kg-m/s (ii) kg-m/s J or eV_2.14 kg-m/s m/s_2.15(a) s(b) k

3、g-m/s_2.16(a) If and are solutions to Schrodingers wave equation, then and Adding the two equations, we obtain which is Schrodingers wave equation. So is also a solution.(b) If were a solution toSchrodingers wave equation, then we could write which can be written as Dividing by , we find Since is a

4、solution, then Subtracting these last two equations, we have Since is also a solution, we have Subtracting these last two equations, we obtain This equation is not necessarily valid, which means that is, in general, not a solution to Schrodingers wave equation._2.17 so or _2.18 or _2.19 Note that Fu

5、nction has been normalized.(a) Now or which yields (b) or which yields (c) which yields _2.20 (a) or (b) or (c) or _2.21(a) or (b) or (c) or _2.22(a) (i) m/s or cm/s m or (ii) kg-m/s Jor eV(b) (i) m/sor cm/s m or (ii) kg-m/s eV_2.23(a)(b) so m/scm/sFor electron traveling in direction, cm/s kg-m/s m

6、m or rad/s_2.24(a) kg-m/s m m rad/s(b) kg-m/s m m rad/s_2.25 J or or eV Then eV eV eV_2.26(a) Jor eVThen eV eV eV(b) J mor nm_2.27(a) or (b) mJ(c) No_2.28 For a neutron and : J or eV For an electron in the same potential well: J or eV_2.29 Schrodingers time-independent wave equation We know that for

7、 and We have for so in this region The solution is of the form where Boundary conditions: at First mode solution: where Second mode solution: where Third mode solution: where Fourth mode solution: where _2.30 The 3-D time-independent wave equation in cartesian coordinates for is: Use separation of v

8、ariables, so let Substituting into the wave equation, we obtain Dividing by and letting , we find(1) We may set Solution is of the form Boundary conditions: and where Similarly, let and Applying the boundary conditions, we find , , From Equation (1) above, we have or so that _2.31 (a) Solution is of

9、 the form: We find Substituting into the original equation, we find: (1) From the boundary conditions, , where So , Also , where So , Substituting into Eq. (1) above (b)Energy is quantized - similar to 1-D result. There can be more than one quantum state per given energy - different than 1-D result.

10、_2.32(a) Derivation of energy levels exactly the same as in the text(b) For Then (i) For J or eV (ii) For cm J or eV_2.33(a) For region II, General form of the solution is where Term with represents incident wave and term with represents reflected wave. Region I, General form of the solution is wher

11、e Term involving represents the transmitted wave and the term involving represents reflected wave: but if a particle is transmitted into region I, it will not be reflected so that . Then (b) Boundary conditions:(1)(2) Applying the boundary conditions to the solutions, we find Combining these two equ

12、ations, we find The reflection coefficient is The transmission coefficient is _2.34 where m(a) For m (b) For m (c) For m _2.35 where or m(a) For m (b) For m (c) , where is the density of transmitted electrons. eVJ m/scm/s electrons/cm Density of incident electrons, cm_2.36 (a) For or m Then or (b) F

13、or = or m Then or _2.37 where m (a) (b) or m_2.38 Region I , ; Region II , Region III , (a) Region I: (incident) (reflected) where Region II: where Region III: (b) In Region III, the term represents a reflected wave. However, once a particle is transmitted into Region III, there will not be a reflec

14、ted wave so that . (c) Boundary conditions: At : At : The transmission coefficient is defined as so from the boundary conditions, we want to solve for in terms of . Solving for in terms of , we find We then find We have If we assume that , then will be large so that We can then write which becomes S

15、ubstituting the expressions for and , we find and Then Finally, _2.39 Region I: incident reflected where Region II: transmitted reflected where Region III: transmitted where There is no reflected wave in Region III. The transmission coefficient is defined as: From the boundary conditions, solve for

16、in terms of . The boundary conditions are: At : At : But Then, eliminating , , from the boundary condition equations, we find _2.40(a) Region I: Since , we can write Region II: , so Region III: The general solutions can be written, keeping in mind that must remain finite for , as where and (b) Bound

17、ary conditions At : At : or (c) and since , then From , we can write or This equation can be written as or This last equation is valid only for specific values of the total energy . The energy levels are quantized._2.41 (J) (eV) or (eV) eV eV eV eV_2.42 We have and or To find the maximum probability

18、 which gives or is the radius that gives the greatest probability._2.43 is independent of and , so the wave equation in spherical coordinates reduces to where For Then so We then obtain Substituting into the wave equation, we have where Then the above equation becomes or which gives 0 = 0 and shows that is indeed a solution to the wave equation._2.44 All elements are from the Group I column of the periodic table. All have one valence electron in the outer shell._