半导体物理与器件第四版课后习题答案2

1、Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2By D. A. Neamen Problem Solutions_Chapter 2142.1 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-

2、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) kg-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

3、/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) kg-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 soluti

4、on toSchrodingers wave equation, then we could write which can be written as Dividing by , we find Since is a 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 m

5、eans that is, in general, not a solution to Schrodingers wave equation._2.17 so or _2.18 or _2.19 Note that Function 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(

6、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 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

7、electron in the same potential well: J or eV_2.29 Schrodingers time-independent wave equation We know that for 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 sol

8、ution: where _2.30 The 3-D time-independent wave equation in cartesian coordinates for is: Use separation of variables, 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 Appl

9、ying the boundary conditions, we find , , From Equation (1) above, we have or so that _2.31 (a) Solution is of 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 -

10、 similar to 1-D result. There can be more than one quantum state per given energy - different than 1-D result._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 re

11、presents incident wave and term with represents reflected wave. Region I, General form of the solution is where 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)

12、 Boundary conditions:(1)(2) Applying the boundary conditions to the solutions, we find Combining these two equations, 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 tra

13、nsmitted electrons. eVJ m/scm/s electrons/cm Density of incident electrons, cm_2.36 (a) For or m Then or (b) For = 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 re

14、presents a reflected wave. However, once a particle is transmitted into Region III, there will not be a reflected 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 ,

15、 we find We then find We have If we assume that , then will be large so that We can then write which becomes Substituting 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 refl

16、ected wave in Region III. The transmission coefficient is defined as: From the boundary conditions, solve for 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 I

17、II: The general solutions can be written, keeping in mind that must remain finite for , as where and (b) Boundary 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 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 obtai