(************** Content-type: application/mathematica ************** CreatedBy='Mathematica 5.2' Mathematica-Compatible Notebook This notebook can be used with any Mathematica-compatible application, such as Mathematica, MathReader or Publicon. The data for the notebook starts with the line containing stars above. To get the notebook into a Mathematica-compatible application, do one of the following: * Save the data starting with the line of stars above into a file with a name ending in .nb, then open the file inside the application; * Copy the data starting with the line of stars above to the clipboard, then use the Paste menu command inside the application. Data for notebooks contains only printable 7-bit ASCII and can be sent directly in email or through ftp in text mode. Newlines can be CR, LF or CRLF (Unix, Macintosh or MS-DOS style). 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Engel also lists the five lowest eigenvalues for this box.\n\n\ 1.\tWhat are the five lowest eigenvalues (in J) for an electron in an ", FontFamily->"Arial", FontSize->10], StyleBox["infinite", FontFamily->"Arial", FontSize->10, FontVariations->{"Underline"->True}], StyleBox[" depth box of the same width?\n", FontFamily->"Arial", FontSize->10], StyleBox["Solution:", FontFamily->"Arial", FontSize->10, FontWeight->"Bold"], StyleBox[" ", FontFamily->"Arial", FontSize->10], Cell[BoxData[ \(TraditionalForm\`E\_n\ = \ \(h\^2\) n\^2/8 m\ a\^2\)]] }], "Text"], Cell[CellGroupData[{ Cell[BoxData[{ \(h = 6.626\ 10\^\(-34\)\ J\ s\), "\[IndentingNewLine]", \(me\ = \ 9.109\ 10\^\(-31\)\ kg\), "\[IndentingNewLine]", \(a\ = \ \(10\^\(-9\)\) m\), "\[IndentingNewLine]", \(n\ = \ {1, 2, 3, 4, 5}\)}], "Input"], Cell[BoxData[ \(\(6.6260000000000015`*^-34\ kg\ m\^2\)\/s\)], "Output"], Cell[BoxData[ \(9.109000000000001`*^-31\ kg\)], "Output"], Cell[BoxData[ \(m\/1000000000\)], "Output"], Cell[BoxData[ \({1, 2, 3, 4, 5}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(En\ = \ \(\(h\^2\) n\^2\)\/\(8\ me\ a\^2\)\)], "Input"], Cell[BoxData[ \({\(6.02479361071468`*^-20\ kg\ m\^2\)\/s\^2, \(2.409917444285872`*^-19\ \ kg\ m\^2\)\/s\^2, \(5.422314249643212`*^-19\ kg\ m\^2\)\/s\^2, \ \(9.639669777143487`*^-19\ kg\ m\^2\)\/s\^2, \(1.5061984026786699`*^-18\ kg\ \ m\^2\)\/s\^2}\)], "Output"] }, Open ]], Cell[TextData[{ StyleBox["2.\tWhich system has lower eigenvalues, the finite depth box or \ the infinite depth box? Why?\n", FontFamily->"Arial", FontSize->10], StyleBox["Solution:", FontFamily->"Arial", FontSize->10, FontWeight->"Bold"], StyleBox[" Engel p. 78 gives the five lowest energies of the finite depth \ box as:\n", FontFamily->"Arial", FontSize->10], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox["4.61", FontSize->10], StyleBox[" ", FontSize->10], StyleBox["\[Times]", FontSize->10], StyleBox[" ", FontSize->10], StyleBox[\(10\^\(-20\)\), FontSize->10], StyleBox[" ", FontSize->10], StyleBox["J", FontSize->10, FontSlant->"Plain"]}], StyleBox[",", FontSize->10], StyleBox["\[IndentingNewLine]", FontSize->10], RowBox[{ StyleBox["1.84", FontSize->10], StyleBox[" ", FontSize->10], StyleBox["\[Times]", FontSize->10], StyleBox[" ", FontSize->10], StyleBox[\(10\^\(-19\)\), FontSize->10], StyleBox[" ", FontSize->10], StyleBox["J", FontSize->10, FontSlant->"Plain"]}], StyleBox[",", FontSize->10], StyleBox["\[IndentingNewLine]", FontSize->10], RowBox[{ StyleBox["4.09", FontSize->10], StyleBox[" ", FontSize->10], StyleBox["\[Times]", FontSize->10], StyleBox[" ", FontSize->10], StyleBox[\(10\^\(-19\)\), FontSize->10], StyleBox[" ", FontSize->10], StyleBox["J", FontSize->10, FontSlant->"Plain"]}], StyleBox[",", FontSize->10], StyleBox["\[IndentingNewLine]", FontSize->10], RowBox[{ StyleBox["7.13", FontSize->10], StyleBox[" ", FontSize->10], StyleBox["\[Times]", FontSize->10], StyleBox[" ", FontSize->10], StyleBox[\(10\^\(-19\)\), FontSize->10], StyleBox["J", FontSize->10, FontSlant->"Plain"]}], StyleBox[",", FontSize->10], StyleBox["\[IndentingNewLine]", FontSize->10], RowBox[{ StyleBox["1.07", FontSize->10], StyleBox[" ", FontSize->10], StyleBox["\[Times]", FontSize->10], StyleBox[" ", FontSize->10], StyleBox[\(10\^\(-18\)\), FontSize->10], StyleBox[" ", FontSize->10], StyleBox["J", FontSize->10, FontSlant->"Plain"]}]}], TraditionalForm]], FontFamily->"Arial"], "\n", StyleBox["Each energy state of the finite depth box is lower than the \ corresponding state of the infinite depth box. This can be explained by the \ boundary condition for the infinite depth box that requires the wave function \ to vanish at ", FontFamily->"Arial", FontSize->10], StyleBox["x", FontFamily->"Arial", FontSize->10, FontSlant->"Italic"], StyleBox[" = ", FontFamily->"Arial", FontSize->10], StyleBox["a", FontFamily->"Arial", FontSize->10, FontSlant->"Italic"], StyleBox[". This forces the infinite depth box to curve faster over the \ interval 0 \[LessEqual] ", FontFamily->"Arial", FontSize->10], StyleBox["x", FontFamily->"Arial", FontSize->10, FontSlant->"Italic"], StyleBox[" \[LessEqual] ", FontFamily->"Arial", FontSize->10], StyleBox["a", FontFamily->"Arial", FontSize->10, FontSlant->"Italic"], StyleBox[", and larger curvature corresponds to larger energy.", FontFamily->"Arial", FontSize->10] }], "Text"], Cell[TextData[{ StyleBox["3.\tUse the formulas provided in Engel, and any ", FontFamily->"Arial", FontSize->10], StyleBox["graphing tools", FontFamily->"Arial", FontSize->10, FontVariations->{"Underline"->True}], StyleBox[" at your disposal, to determine the lowest eigenvalue (1% \ accuracy in E is adequate) for an electron confined to a finite depth box of \ the same width, but with V", FontFamily->"Arial", FontSize->10], StyleBox["o", FontFamily->"Arial", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" = 2.4 x 10", FontFamily->"Arial", FontSize->10], StyleBox["-18", FontFamily->"Arial", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" J.\n", FontFamily->"Arial", FontSize->10], StyleBox["Solution:", FontFamily->"Arial", FontSize->10, FontWeight->"Bold"], StyleBox[" Engel, p. 77 shows two equations that must be satisfied by the \ energy of a stationary state (the energy must satisfy one of the equations). \ The top equation is used to generate the lowest energy state (see figure on \ p. 78), and it is written below in two pieces (LHStop = left-hand side, RHS = \ right-hand side).", FontFamily->"Arial", FontSize->10] }], "Text"], Cell[CellGroupData[{ Cell[BoxData[{ \(Clear[Energy]\), "\[IndentingNewLine]", \(V0\ = \ 2.4\ \[Times]\ 10\^\(-18\)\), "\[IndentingNewLine]", \(me\ = \ 9.109\ \[Times]\ 10\^\(-31\)\), "\[IndentingNewLine]", \(a\ = \ 10\^\(-9\)\), "\[IndentingNewLine]", \(hbar\ = \ 1.055\ \[Times]\ 10\^\(-34\)\), "\[IndentingNewLine]", \(LHStop = \ \@\(\(me \((V0 - Energy)\)\ a\^2\)\/\(2\ hbar\^2\)\)\), "\ \[IndentingNewLine]", \(RHStop\ = \ \(\@\(\(me\ Energy\ a\^2\)\/\(2\ hbar\^2\)\)\) Tan[\@\(\(me\ Energy\ a\^2\)\/\(2\ hbar\^2\)\)]\)}], "Input"], Cell[BoxData[ \(2.4`*^-18\)], "Output"], Cell[BoxData[ \(9.109000000000001`*^-31\)], "Output"], Cell[BoxData[ \(1\/1000000000\)], "Output"], Cell[BoxData[ \(1.0550000000000002`*^-34\)], "Output"], Cell[BoxData[ \(6.396875430527865`*^9\ \@\(2.4`*^-18 - Energy\)\)], "Output"], Cell[BoxData[ \(6.396875430527865`*^9\ \@Energy\ Tan[ 6.396875430527865`*^9\ \@Energy]\)], "Output"] }, Open ]], Cell[TextData[{ StyleBox["Any E that satisfies this equation, i.e., makes LHStop = RHStop, \ is acceptable, but we want the lowest energy possible. Unfortunately, the \ equations cannot be solved algebraically, so we must 1) plot the functions on \ the left-hand side for a range of possible energies, and 2) plot the \ functions on the right-hand side for the same energy range. The state's \ energy is given by the crossing point of these two graphs.\n\nNote #1: We \ already know from the previous problems that the ground state energy of the \ finite depth box must be less than the ground state energy of the infinite \ depth box (6.02 \[Times] ", FontFamily->"Arial", FontSize->10], Cell[BoxData[ FormBox[ RowBox[{ SuperscriptBox[ StyleBox["10", FontSize->10], \(-20\)], " ", StyleBox["J", FontSize->10, FontSlant->"Plain"]}], TraditionalForm]], FontFamily->"Arial"], StyleBox["). Since ", FontFamily->"Arial", FontSize->10], StyleBox["V", FontFamily->"Arial", FontSize->10], StyleBox["o", FontFamily->"Arial", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" for this problem is larger than V", FontFamily->"Arial", FontSize->10], StyleBox["o", FontFamily->"Arial", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" for the previous problem, we also know that the answer must be ", FontFamily->"Arial", FontSize->10], StyleBox["greater", FontFamily->"Arial", FontSize->10, FontSlant->"Italic"], StyleBox[" than ground state energy given in the previous problem (4.61 \ \[Times] ", FontFamily->"Arial", FontSize->10], Cell[BoxData[ FormBox[ SuperscriptBox[ StyleBox["10", FontFamily->"Arial", FontSize->10], StyleBox[\(-20\), FontFamily->"Arial"]], TraditionalForm]]], StyleBox["J). So we need to graph the energy range 4.61-6.02 \[Times] ", FontFamily->"Arial", FontSize->10], Cell[BoxData[ \(TraditionalForm\`10\^\(-20\)\)], FontFamily->"Arial", FontSize->10], StyleBox[" J. The following graph was arrived at by using trial-and-error \ to adjust the horizontal (Energy) and vertical ranges. 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ImageRangeCache->{{{151, 438}, {290.375, 113.438}} -> {4.93172*^-20, \ 9.85741, 1.5815*^-24, 0.00243708}}], Cell[BoxData[ TagBox[\(\[SkeletonIndicator] Graphics \[SkeletonIndicator]\), False, Editable->False]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(FindRoot[LHStop == RHStop, {Energy, 5\ *\ 10\^\(-20\)}]\)], "Input"], Cell[BoxData[ \({Energy \[Rule] 4.9719225864514765`*^-20}\)], "Output"] }, Open ]], Cell[TextData[{ StyleBox["Referring to Engel\[CloseCurlyQuote]s original finite depth box \ (V", FontFamily->"Arial", FontSize->10], StyleBox["o", FontFamily->"Arial", FontSize->10, FontVariations->{"CompatibilityType"->"Subscript"}], StyleBox[" = 1.2 x 10", FontFamily->"Arial", FontSize->10], StyleBox["-18", FontFamily->"Arial", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" J, a = 1 x 10", FontFamily->"Arial", FontSize->10], StyleBox["-9", FontFamily->"Arial", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" m) and the general formulas that he provides for the \ eigenfunctions,\n\n4a.\tWhat are the ", FontFamily->"Arial", FontSize->10], StyleBox["normalized", FontFamily->"Arial", FontSize->10, FontVariations->{"Underline"->True}], StyleBox[" eigenfunctions for E = 4.61 x 10", FontFamily->"Arial", FontSize->10], StyleBox["-20", FontFamily->"Arial", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" J and 4.09 x 10", FontFamily->"Arial", FontSize->10], StyleBox["-19", FontFamily->"Arial", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" J?\n", FontFamily->"Arial", FontSize->10], StyleBox["Solution:", FontFamily->"Arial", FontSize->10, FontWeight->"Bold"], StyleBox[" The general formulas for the wave functions in regions I-III are \ given on Engel p. 77. Engel states that A' = B' = 0, which greatly simplifies \ the formulas for region I and region III. To take the next step, it is \ helpful to note the symmetry of the potential function. The box has a point \ of symmetry at ", FontFamily->"Arial", FontSize->10], StyleBox["x", FontFamily->"Arial", FontSize->10, FontSlant->"Italic"], StyleBox[" = 0, and all wave functions must have either even or odd \ symmetry with respect to this point. The ground state should be described by \ an even function (no nodes). The first and second excited states should be \ described by odd (one node) and even (two nodes) functions, respectively.\n\n\ The symmetry requirements introduce additional restrictions on A, B, C, and \ D. An even wave function has the same value at ", FontFamily->"Arial", FontSize->10], StyleBox["x", FontFamily->"Arial", FontSize->10, FontSlant->"Italic"], StyleBox[" = ", FontFamily->"Arial", FontSize->10], StyleBox["a", FontFamily->"Arial", FontSize->10, FontSlant->"Italic"], StyleBox["/2 and -", FontFamily->"Arial", FontSize->10], StyleBox["a", FontFamily->"Arial", FontSize->10, FontSlant->"Italic"], StyleBox["/2. This means C = 0 and A = B. As it happens, ", FontFamily->"Arial", FontSize->10], StyleBox["E = 4.61 x 10", FontFamily->"Arial", FontSize->10], StyleBox["-20", FontFamily->"Arial", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" J", FontFamily->"Arial", FontSize->10], StyleBox[" is the ground state, and ", FontFamily->"Arial", FontSize->10], StyleBox["E = 4.09 x 10", FontFamily->"Arial", FontSize->10], StyleBox["-19", FontFamily->"Arial", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" J is the second excited state, so both energies correspond to \ states described by even wave functions.\n\nUsing the formulas in Engel, p. \ 77, and letting A = B, we can obtain the relative values of A, B, and D:\n\n", FontFamily->"Arial", FontSize->10], StyleBox["2A ", FontFamily->"Arial", FontSize->10], Cell[BoxData[ \(TraditionalForm\`e\^\(-\(\[Kappa](a/2)\)\)\)], FontFamily->"Arial", FontSize->10], StyleBox[" = 2D ", FontFamily->"Arial", FontSize->10], Cell[BoxData[ \(TraditionalForm\`cos(k a\/2)\)], FontFamily->"Arial", FontSize->10], "\n\n", Cell[BoxData[ FormBox[ StyleBox[\(A\/D\), FontFamily->"Arial", FontSize->10], TraditionalForm]]], "= ", Cell[BoxData[ \(TraditionalForm\`\(e\^\(\[Kappa](a/2)\)\) \(cos(k a\/2)\)\)], FontFamily->"Arial", FontSize->10], "\n\n", StyleBox["and also", FontFamily->"Arial", FontSize->10], "\n\n", StyleBox["2A \[Kappa] ", FontFamily->"Arial", FontSize->10], Cell[BoxData[ \(TraditionalForm\`e\^\(-\(\[Kappa](a/2)\)\)\)], FontFamily->"Arial", FontSize->10], StyleBox[" = 2D ", FontFamily->"Arial", FontSize->10], StyleBox["k", FontFamily->"Arial", FontSize->10, FontSlant->"Italic"], StyleBox[" sin", FontFamily->"Arial", FontSize->10], Cell[BoxData[ \(TraditionalForm\`\((k a\/2)\)\)], FontFamily->"Arial", FontSize->10], "\n\n", Cell[BoxData[ FormBox[ StyleBox[\(A\/D\ = \ \(k\/\[Kappa]\) \(e\^\(\[Kappa](a/2)\)\) \(sin( k a\/2)\)\), FontFamily->"Arial", FontSize->10], TraditionalForm]]] }], "Text"], Cell[CellGroupData[{ Cell[BoxData[{ \(V0\ = \ 1.2\ \[Times]\ 10\^\(-18\)\), "\[IndentingNewLine]", \(me\ = \ 9.109\ \[Times]\ 10\^\(-31\)\), "\[IndentingNewLine]", \(a\ = \ 10\^\(-9\)\), "\[IndentingNewLine]", \(hbar\ = \ 1.055\ \[Times]\ 10\^\(-34\)\), "\[IndentingNewLine]", \(Energy\ = \ {4.61\ \[Times]\ 10\^\(-20\), \ 4.09\ \[Times]\ 10\^\(-19\)}\), "\[IndentingNewLine]", \(k\ = \ \@\(\(2\ me\ Energy\)\/hbar\^2\)\), "\[IndentingNewLine]", \(\[Kappa]\ = \ \@\(\(2\ me\ \((V0 - \ Energy)\)\)\/hbar\^2\)\)}], \ "Input"], Cell[BoxData[ \(1.2`*^-18\)], "Output"], Cell[BoxData[ \(9.109000000000001`*^-31\)], "Output"], Cell[BoxData[ \(1\/1000000000\)], "Output"], Cell[BoxData[ \(1.0550000000000002`*^-34\)], "Output"], Cell[BoxData[ \({4.61`*^-20, 4.0899999999999995`*^-19}\)], "Output"], Cell[BoxData[ \({2.7469348038256445`*^9, 8.182001282556644`*^9}\)], "Output"], Cell[BoxData[ \({1.374301358862926`*^10, 1.1378529269020603`*^10}\)], "Output"] }, Open ]], Cell[TextData[{ StyleBox["I obtained ", FontFamily->"Arial", FontSize->10], StyleBox["two", FontFamily->"Arial", FontSize->10, FontVariations->{"Underline"->True}], StyleBox[" formulas for A/D above. I evaluate both formulas below, but I \ will just use the first formula to give me \"AoverD\" for future work.", FontFamily->"Arial", FontSize->10] }], "Text"], Cell[CellGroupData[{ Cell[BoxData[{ \(AoverD\ = \ \(\[ExponentialE]\^\(\((\[Kappa]\ a)\)/2\)\) Cos[\(k\ a\)\/2]\), "\[IndentingNewLine]", \(\((k\/\[Kappa])\) \(\[ExponentialE]\^\(\((\[Kappa]\ a)\)/2\)\) Sin[\(k\ a\)\/2]\)}], "Input"], Cell[BoxData[ \({189.07150800363155`, \(-172.13214680659945`\)}\)], "Output"], Cell[BoxData[ \({189.02227085393932`, \(-172.86939802636314`\)}\)], "Output"] }, Open ]], Cell[TextData[{ StyleBox["The small discrepancies between the two estimates of A/D \ (\"AoverD\") can be blamed on round-off errors in our energies.\n\nThe wave \ function will be normalized if the following is true:\n\n", FontFamily->"Arial", FontSize->10], Cell[BoxData[ \(TraditionalForm\`B\^2\ \ \(\[Integral]\_\(-\[Infinity]\)\%\(\(-a\)/2\)\(\[ExponentialE]\^\(2 \[Kappa]\ \ x\)\) \[DifferentialD]x\)\ + \ \(D\^2\) \(\[Integral]\_\(\(-a\)/2\)\%\(a/2\ \)\(\(cos\^2\)( k\ x)\) \[DifferentialD]x\)\ + \ \(A\^2\) \ \(\[Integral]\_\(a/2\)\%\[Infinity]\( \[ExponentialE]\^\(\(-2\) \[Kappa]\ x\)\ \) \[DifferentialD]x\)\ = \ 1\)], FontFamily->"Arial", FontSize->10], StyleBox["\n\nwhich can be rearranged to give:\n\n", FontFamily->"Arial", FontSize->10], Cell[BoxData[ \(TraditionalForm\`\(D\^2\)\ [ 2 \(\((A\/D)\)\^2\) \(\[Integral]\_\(a/2\)\%\[Infinity]\( \ \[ExponentialE]\^\(\(-2\)\ \[Kappa]\ x\)\) \[DifferentialD]x\)\ + \ \ \[Integral]\_\(\(-a\)/2\)\%\(a/2\)\(\(cos\^2\)( k\ x)\) \[DifferentialD]x]\ = \ 1\)], FontFamily->"Arial", FontSize->10], StyleBox["\n\nSo once we know the values of the integrals, we can solve for \ D (\"Dcoef\") and then use this and A/D (\"AoverD\" from above) to obtain A (\ \"Acoef\"), which is also B.", FontFamily->"Arial", FontSize->10] }], "Text"], Cell[CellGroupData[{ Cell[BoxData[{ \(int1\ = \ \[Integral]\_\(a/2\)\%\[Infinity]\( \ \[ExponentialE]\^\(\(-2\)\ \[Kappa]\ x\)\) \[DifferentialD]x\), "\ \[IndentingNewLine]", \(int2 = \[Integral]\_\(\(-a\)/2\)\%\(a/2\)\(Cos[ k\ x]\^2\) \[DifferentialD]x\)}], "Input"], Cell[BoxData[ RowBox[{\(Integrate::"gener"\), \(\(:\)\(\ \)\), "\<\"Unable to check \ convergence. \\!\\(\\*ButtonBox[\\\"More\[Ellipsis]\\\", \ ButtonStyle->\\\"RefGuideLinkText\\\", ButtonFrame->None, \ ButtonData:>\\\"Integrate::gener\\\"]\\)\"\>"}]], "Message"], Cell[BoxData[ RowBox[{\(Integrate::"gener"\), \(\(:\)\(\ \)\), "\<\"Unable to check \ convergence. \\!\\(\\*ButtonBox[\\\"More\[Ellipsis]\\\", \ ButtonStyle->\\\"RefGuideLinkText\\\", ButtonFrame->None, \ ButtonData:>\\\"Integrate::gener\\\"]\\)\"\>"}]], "Message"], Cell[BoxData[ \({3.9117677521727356`*^-17, 5.026334642889423`*^-16}\)], "Output"], Cell[BoxData[ \({5.699857095739242`*^-10, 5.578515068867071`*^-10}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(\(\(Dcoef\)\(\ \)\(=\)\(\@\(1\/\(2\ \(AoverD\^2\) int1\ + \ int2\)\)\)\(\ \)\)\)], "Input"], Cell[BoxData[ \({41783.53128906588`, 41252.03894298143`}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[{ \(Acoef\ = \ AoverD*Dcoef\), "\[IndentingNewLine]", \(k\), "\[IndentingNewLine]", \(\[Kappa]\)}], "Input"], Cell[BoxData[ \({7.900075270540609`*^6, \(-7.100802023404838`*^6\)}\)], "Output"], Cell[BoxData[ \({2.7469348038256445`*^9, 8.182001282556644`*^9}\)], "Output"], Cell[BoxData[ \({1.374301358862926`*^10, 1.1378529269020603`*^10}\)], "Output"] }, Open ]], Cell[TextData[{ StyleBox["Now that ", FontFamily->"Arial", FontSize->10], StyleBox["A", FontFamily->"Arial", FontSize->10, FontSlant->"Italic"], StyleBox[" and ", FontFamily->"Arial", FontSize->10], StyleBox["D", FontFamily->"Arial", FontSize->10, FontSlant->"Italic"], StyleBox[" have been calculated, it is possible to write the wave functions \ for all three regions (recall ", FontFamily->"Arial", FontSize->10], StyleBox["B", FontFamily->"Arial", FontSize->10, FontSlant->"Italic"], StyleBox[" = ", FontFamily->"Arial", FontSize->10], StyleBox["A", FontFamily->"Arial", FontSize->10, FontSlant->"Italic"], StyleBox[").\n\t\tE = 4.61 \[Times] ", FontFamily->"Arial", FontSize->10], Cell[BoxData[ \(TraditionalForm\`10\^\(-20\)\)], FontFamily->"Arial", FontSize->10], StyleBox[" J\n\t\tRegion I:\t\[Psi] = 7.90008 \[Times] ", FontFamily->"Arial", FontSize->10], Cell[BoxData[ \(TraditionalForm\`10\^6\ \[ExponentialE]\^\(1.3743\ \[Times]\ 10\^10\ \ x\)\)], FontFamily->"Arial", FontSize->10], StyleBox["\n\t\tRegion II:\t\[Psi] = 41783.5 cos", FontFamily->"Arial", FontSize->10], Cell[BoxData[ \(TraditionalForm\`\((2.74693\ \[Times]10\^9\ x)\)\)], FontFamily->"Arial", FontSize->10], StyleBox["\n\t\tRegion III:\t\[Psi] = 7.90008 \[Times] ", FontFamily->"Arial", FontSize->10], Cell[BoxData[ \(TraditionalForm\`10\^6\ \[ExponentialE]\^\(\(-1.3743\)\ \[Times]\ \ 10\^10\ x\)\)], FontFamily->"Arial", FontSize->10], StyleBox["\n\t\t\n\t\tE = 4.09 \[Times] ", FontFamily->"Arial", FontSize->10], Cell[BoxData[ \(TraditionalForm\`10\^\(-19\)\)], FontFamily->"Arial", FontSize->10], StyleBox[" J\n\t\tRegion I:\t\[Psi] = -7.1008 \[Times] ", FontFamily->"Arial", FontSize->10], Cell[BoxData[ \(TraditionalForm\`10\^6\ \[ExponentialE]\^\(1.13785\ \[Times]\ 10\^10\ \ x\)\)], FontFamily->"Arial", FontSize->10], StyleBox["\n\t\tRegion II:\t\[Psi] = 41252 cos", FontFamily->"Arial", FontSize->10], Cell[BoxData[ \(TraditionalForm\`\((8.182\ \[Times]10\^9\ x)\)\)], FontFamily->"Arial", FontSize->10], StyleBox["\n\t\tRegion III:\t\[Psi] = -7.1008 \[Times] ", FontFamily->"Arial", FontSize->10], Cell[BoxData[ \(TraditionalForm\`10\^6\ \[ExponentialE]\^\(\(-1.13785\)\ \[Times]\ 10\ \^10\ x\)\)], FontFamily->"Arial", FontSize->10] }], "Text"], Cell[TextData[{ StyleBox["The following calculations double-check the coefficient values by \ testing whether the boundary condition is satisfied at ", FontFamily->"Arial", FontSize->10], StyleBox["x", FontFamily->"Arial", FontSize->10, FontSlant->"Italic"], StyleBox[" = ", FontFamily->"Arial", FontSize->10], StyleBox["a", FontFamily->"Arial", FontSize->10, FontSlant->"Italic"], StyleBox["/2.", FontFamily->"Arial", FontSize->10] }], "Text"], Cell[CellGroupData[{ Cell[BoxData[{ \(Acoef*\[ExponentialE]\^\(\(-\ \[Kappa]\)\ \((a/2)\)\)\), "\ \[IndentingNewLine]", \(Dcoef*Cos[k \((a/2)\)]\)}], "Input"], Cell[BoxData[ \({8191.694474526482`, \(-24015.47425447899`\)}\)], "Output"], Cell[BoxData[ \({8191.694474526483`, \(-24015.474254478988`\)}\)], "Output"] }, Open ]], Cell[TextData[{ StyleBox["4b.\tWhat are the probabilities of finding an electron in Regions \ I, II, and III, respectively, when the electron\[CloseCurlyQuote]s energy E = \ 4.61 x 10", FontFamily->"Arial", FontSize->10], StyleBox["-20", FontFamily->"Arial", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], StyleBox[" J and 4.09 x 10", FontFamily->"Arial", FontSize->10], StyleBox["-19", FontFamily->"Arial", FontSize->10, FontVariations->{"CompatibilityType"->"Superscript"}], 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