Question

Câu 12

Cho 2 hàm sóng: \(\psi_{1s}=\frac{1}{\sqrt{\pi}}.a^{\frac{3}{2}}_0.e^{-\frac{r}{a_0}}\)và \(\psi_{2s}=\frac{1}{4\sqrt{2}}.a^{\frac{3}{2}}_0.\left(2-\frac{r}{a_0}\right).e^{-\frac{r}{2a_0}}\)

a) Hãy chứng minh hai hàm sóng trên trực giao nhau

b) Tìm hàm mật độ xác suất trong mỗi trường hợp và chỉ ra những vị trí mà mật độ xác suất đạt giá trị cực đại.

Answer 1

a, Ta có:

Hai hàm sóng trực giao nhau khi  \(I=\int\psi_{1s}.\psi_{2s}d\psi=0\) \(\Leftrightarrow I=\iiint\psi_{1s}.\psi_{2s}dxdydz=0\)

Chuyển sang tọa độ cầu ta có:  \(\begin{cases}x=r.\cos\varphi.sin\theta\\y=r.\sin\varphi.sin\theta\\z=r.\cos\theta\end{cases}\)

\(\Rightarrow\)\(I=\frac{a^3_o}{4.\sqrt{2.\pi}}\int\limits^{\infty}_0\left(2-\frac{r}{a_o}\right).e^{-\frac{3.r}{2.a_o}}.r^2.\sin\theta dr\int\limits^{2\pi}_0d\varphi\int\limits^{\pi}_0d\theta\)

       \(=a^3_o.\sqrt{\frac{\pi}{2}}\)(.\(2.\int\limits^{\infty}_0r^2.e^{-\frac{3.r}{2.a_o}}dr-\frac{1}{a_o}.\int\limits^{\infty}_0r^3.e^{-\frac{3.r}{2.a_o}}dr\))

         \(=a_o.\sqrt{\frac{\pi}{2}}.\left(2.I_1-\frac{1}{a_o}.I_2\right)\)  

Tính \(I_1\):

Đặt \(r^2=u\); \(e^{-\frac{3r}{2a_o}}dr=dV\)

\(\Rightarrow\begin{cases}2.r.dr=du\\-\frac{2a_o}{3}.e^{-\frac{3r}{2a_o}}=V\end{cases}\)    \(\Rightarrow I_1=-r^2.\frac{2a_o}{3}.e^{-\frac{3r}{2a_o}}+\frac{4.a_o}{3}.\int\limits^{\infty}_0r.e^{-\frac{3r}{2a_o}}dr\)\(=0+\frac{4a_o}{3}.I_{11}\)

Tính \(I_{11}\):

Đặt r=u; \(e^{-\frac{3r}{2a_o}}dr=dV\)\(\Rightarrow\begin{cases}dr=du\\-\frac{2a_o}{3}.e^{-\frac{3r}{2a_o}}=V\end{cases}\)\(\Rightarrow I_{11}=0+\frac{2a_0}{3}.\int\limits^{\infty}_0e^{-\frac{3r}{2a_o}}dr=\frac{4a^2_o}{9}\)

\(\Rightarrow2.I_1=2.\frac{4a_o}{3}.\frac{4a_o^2}{9}=\frac{32a^3_o}{27}\)

Tính \(I_2\):

Đặt \(r^2=u;e^{-\frac{3r}{2a_o}}dr=dV\) \(\Rightarrow\)\(3r^2dr=du;-\frac{2a_o}{3}.e^{-\frac{3r}{2a_o}}=V\)

\(\Rightarrow I_2=0+2.a_o.\int\limits^{\infty}_0r^2.e^{-\frac{3r}{2a_o}}dr\)\(\Rightarrow\frac{1}{a_o}.I_2=2a_o.\frac{16a^3_o}{27}.\frac{1}{a_o}=\frac{32a^3_o}{27}\)

\(\Rightarrow I=a^3_o.\sqrt{\frac{\pi}{2}}.\left(\frac{32a^3_o}{27}-\frac{32a^3_o}{27}\right)=0\)

Vậy hai hàm sóng này trực giao với nhau.

b,

Xét hàm \(\Psi_{1s}\):

Hàm mật độ sác xuất là: \(D\left(r\right)=\Psi^2_{1s}=\frac{1}{\pi}.a^3_o.e^{-\frac{2r}{a_o}}\)

\(\Rightarrow D'\left(r\right)=-\frac{2.a_o^2}{\pi}.e^{-\frac{2r}{a_o}}=0\)

\(\Rightarrow\)Hàm đạt cực đại khi \(r\rightarrow o\) nên hàm sóng có dạng hình cầu.

Xét hàm \(\Psi_{2s}\):

Hàm mật độ sác xuất: \(D\left(r\right)=\Psi_{2s}^2=\frac{a^3_o}{32}.\left(2-\frac{r}{a_o}\right)^2.e^{-\frac{r}{a_0}}\)\(\Rightarrow D'\left(r\right)=\left(2-\frac{r}{a_o}\right).e^{-\frac{r}{a_o}}.\left(-4+\frac{r}{a_o}\right)=0\)

\(\Rightarrow r=2a_o\Rightarrow D\left(r\right)=0\); \(r=4a_o\Rightarrow D\left(r\right)=\frac{a^3_o}{8}.e^{-4}\)

Vậy hàm đạt cực đại khi \(r=4a_o\), tại \(D\left(r\right)=\frac{a^3_o}{8}.e^{-4}\)

 

                

 

Answer 2

hai hàm trực giao: I=\(\int\)\(\Psi\)*\(\Psi\)d\(\tau\)=0

Ta có: I=\(\int\limits^{ }_x\)\(\int\limits^{ }_y\)\(\int\limits^{ }_z\)\(\Psi\)*\(\Psi\)dxdydz=0

           =\(\int\limits^{ }_r\)\(\int\limits^{ }_{\theta}\)\(\int\limits^{ }_{\varphi}\)\(\Psi\)_1s\(\Psi\)_2sr²sin\(\theta\)drd\(\theta\)d\(\varphi\)

          =\(\int\limits^{\infty}_0\)\(\int\limits^{\pi}_0\)\(\int\limits^{2\pi}_0\)(2-\(\frac{r}{a_0}\)).e^-3r/a₀r²sin\(\theta\)drd\(\theta\)d\(\varphi\)

          =C.\(\int\limits^{\infty}_0\)(2-\(\frac{r}{a_0}\)).e^-3r/a₀r2dr.\(\int\limits^{\pi}_0\)sin\(\theta\)\(\int\limits^{2\pi}_0\)d\(\varphi\)

 với C=\(\frac{1}{4\sqrt{2\pi}}\)a₀^-3

 Xét tích phân: J=\(\int\limits^{\infty}_0\)(2-\(\frac{r}{a_0}\)).e^-3r/a₀r²dr

 =\(\int\limits^{\infty}_0\)(2r²- \(\frac{r^3}{a_0}\)).e^-3r/a₀dr

 =\(\int\limits^{\infty}_0\)(2r²- \(\frac{r^3}{a_0}\)).\(\frac{-2a_0}{3}\)de^-3r/a₀

  =\(\frac{-2a_0}{3}\).((2r²-\(\frac{r^3}{a_0}\))e^-3r/a₀\(-\)\(\int\)(4r-\(\frac{3r^2}{a_0}\))e^-3r/adr)

 =\(\frac{-2a_0}{3}\)((2r²-\(\frac{r^3}{a_0}\))e^-3r/a₀ - \(\int\)(4r-\(\frac{3r^2}{a_0}\)).\(\frac{-2a_0}{3}\)de^-3r/a)

 =\(\frac{-2a_0}{3}\)((2r²-\(\frac{r^3}{a_0}\))e^-3r/a₀ +\(\frac{2a_0}{3}\).((4r-\(\frac{3r^2}{a_0}\))e^-3r/a₀ - \(\int\)(4 - \(\frac{6r}{a_0}\))e^-3r/a₀dr))

 =\(\frac{-2a_0}{3}\)((2r²-\(\frac{r^3}{a_0}\))e^-3r/a₀ +\(\frac{2a_0}{3}\).((4r-\(\frac{3r^2}{a_0}\))e^-3r/a₀- \(\int\)(4 - \(\frac{6r}{a_0}\))\(\frac{-2a_0}{3}\).de^-3r/a₀))

 =\(\frac{-2a_0}{3}\)(((2r²-\(\frac{r^3}{a_0}\))e^-3r/a₀ +\(\frac{2a_0}{3}\).((4r-\(\frac{3r^2}{a_0}\))e^-3r/a₀+\(\frac{2a_0}{3}\)((4-\(\frac{6r}{a_0}\)).e^-3r/a₀ + \(\int\)(\(\frac{6}{a_0}\)e^-3r/a₀dr)))

=\(\frac{-2a_0}{3}\)(((2r²-\(\frac{r^3}{a_0}\))e^-3r/a₀ +\(\frac{2a_0}{3}\).((4r-\(\frac{3r^2}{a_0}\))e^-3r/a₀+\(\frac{2a_0}{3}\)((4-\(\frac{6r}{a_0}\)).e^-3r/a₀ + \(\int\)(\(\frac{6}{a_0}\).\(\frac{-2a_0}{3}\)de^-3r/a₀)))

=\(\frac{-2a_0}{3}\)((((2r²-\(\frac{r^3}{a_0}\))e^-3r/a₀ +\(\frac{2a_0}{3}\).((4r-\(\frac{3r^2}{a_0}\))e^-3r/a₀+\(\frac{2a_0}{3}\)((4-\(\frac{6r}{a_0}\)).e^-3r/a₀ - 4.e^-3r/a₀))))

=\(\frac{-2a_0}{3}\)e^-3r/a₀.\(\frac{-r^3}{a_0}\)=2/3.e^-3r/a₀.r³Thế cận tích phân 0 và \(\infty\)            J= 0 suy ra I=0. Vậy 2 hàm số trực giao

Answer 3

b1.\(\Psi\)_1s=\(\frac{1}{\sqrt{\pi}}\).a₀^3/2.e^-r/a₀.

Hàm mật độ xác suất :

      D_r=|\(\Psi\)²|r²

              =\(\frac{1}{\pi}\)a₀³.e^-2r/a₀.r²

xét \(\frac{dD_r}{r}\)= \(\frac{1}{\pi}\)a₀³.(r².\(\frac{-2}{a_0}\)e^-2r/a₀+2r.e^-2r/a₀)

           = \(\frac{1}{\pi}\)a₀^3.e^-2r/a₀.2r.(1\(-\)\(\frac{r}{a_0}\))

        \(\frac{dD_r}{r}\)=0 \(\Leftrightarrow\)r=a₀.

tại r=a₀ D_r đạt cực đại. D_max=\(\frac{1}{\pi}\)a₀³.e^-2.a₀²=\(\frac{1}{\pi}\)a₀⁵.e^-2

                                    

b2. \(\Psi\)_2s=\(\frac{1}{4\sqrt{2}}\)a₀^3/2.(2-\(\frac{r}{a_0}\)).e^-r/2a_o.

Hám mật độ xác suất:

    D_r=|\(\Psi\)²|r².

       =\(\frac{1}{32}\).a₀³.e^-r/a_o.(2-\(\frac{r}{a_0}\))²

       =\(\frac{1}{32}\).a₀³.e^-r/a_o.(4-4.\(\frac{r}{a_0}\)+\(\frac{r^2}{a^2_0}\))

  Xét \(\frac{dDr}{dr}\)= \(\frac{1}{32}\).a₀³.((e^-r/a_o.\(\frac{-1}{a_0}\).(2-\(\frac{r}{a_0}\))²+e^-r/a_o.(-.\(\frac{4}{a_0}\)+\(\frac{2r}{a^2_0}\))

              =- \(\frac{1}{32}\).a₀³.e^-r/a_o.\(\frac{1}{a_0}\).(2-\(\frac{r}{a_0}\))(\(\frac{r}{a_0}\)-4)

\(\frac{dDr}{dr}\)=0 \(\Rightarrow\)r=2a₀ hoặc r=4a₀.tại r=2a₀ D_r đạt cực đại. D_rmax=0.

     

             

Answer 4

a) Điều kiện để hai hàm trực giao là:

\(I=\int\left(\psi_{1s}.\psi_{2s}\right)d_C=0\)\(\)

\(\Leftrightarrow I=\int\limits^{ }_r\int\limits^{ }_{\theta}\int\limits^{ }_{\varphi}\left(\psi_{1s}.\psi_{2s}\right).r^2\sin^2\theta d_rd_{\theta}d_{\varphi}\)

\(\Leftrightarrow I=C\int\limits^{\infty}_0\int\limits^{\pi}_0\int\limits^{2\pi}_0\left(2-\frac{r}{a_0}\right)e^{\frac{-r}{a_0}-\frac{r}{2a_0}}.r^2\sin^2\theta d_rd_{\theta}d_{\varphi}\)

\(\Leftrightarrow I=C\int\limits^{\infty}_0\left(2-\frac{r}{a_0}\right)e^{\frac{-3r}{2a_0}}.r^2d_r\int\limits^{\pi}_0\sin^2\theta d_{\theta}\int\limits^{2\pi}_0d_{\varphi}\)

Xét tích phân :

A=\(\int\limits^{\infty}_0\left(2-\frac{r}{a_0}\right)e^{\frac{-3r}{2a_0}}.r^2d_r\)

áp dụng công thức tích phân từng phần ta có :

A=\(\frac{-2a_0}{3}e^{\frac{-3r}{2a_0}}.\left(2r^2-\frac{r^3}{a_0}\right)+\)\(\int\limits^{\infty}_0\frac{2a_0}{3}e^{\frac{-3r}{2a_0}}\left(4r-\frac{3r^2}{a_0}\right)d_r\)

=\(\frac{-2a_0}{3}e^{\frac{-3r}{2a_0}}.\left(2r^2-\frac{r^3}{a_0}\right)+\)\(\left(4r-\frac{3r^2}{a_0}\right)\frac{-4a^2_0}{9}e^{\frac{-3r}{2a_0}}+\frac{4a^2_0}{9}\int\limits^{\infty}_0\left(4-\frac{6r}{a_0}\right)e^{\frac{-3r}{2a_0}}d_r\)

=\(\frac{-2a_0}{3}e^{\frac{-3r}{2a_0}}.\left(2r^2-\frac{r^3}{a_0}\right)+\)\(\left(4r-\frac{3r^2}{a_0}\right)\frac{-4a^2_0}{9}e^{\frac{-3r}{2a_0}}\)+\(\frac{4a^2_0}{9}.\left(\left(4-\frac{6r}{a_0}\right).\frac{-2a_0}{3}.e^{\frac{-3r}{2a_0}}-4.\int\limits^{\infty}_0e^{\frac{-3r}{2a_0}}\right)\)

=\(\frac{-2a_0}{3}e^{\frac{-3r}{2a_0}}.\left(2r^2-\frac{r^3}{a_0}\right)+\)\(\left(4r-\frac{3r^2}{a_0}\right)\frac{-4a^2_0}{9}e^{\frac{-3r}{2a_0}}\)+\(\frac{4a^2_0}{9}\left(\left(4-\frac{6r}{a_0}\right)\frac{-2a_0}{3}e^{\frac{-3r}{2a_0}}-4.\frac{2a_0}{3}.e^{\frac{-3r}{2a_0}}\right)\)

=\(\frac{2}{3}e^{\frac{-3r}{2a_0}}.r^3\) Thê cận từ \(0\)đến \(\infty\) bằng 0 

\(\Rightarrow I=0.\int\limits^{\pi}_0\sin^2\theta d_{\theta}\int\limits^{2\pi}_0d_{\varphi}=0\) Vậy hai hàm trên trực giao .

Answer 5

Để hai hàm trực giao:

\(\int\limits^{ }_{ }\psi_{1s}\psi_{2s}d_k=0\)\(\Leftrightarrow I=\int\limits^{ }_r\int\limits^{ }_{\theta}\int\limits^{ }_{\varphi}\left(\psi_{1s}\psi_{2s}\right)r^2\sin^2\theta d_rd_{\theta}d_{\varphi}=0\)

\(\Leftrightarrow I=C.\int\limits^{\infty}_0\int\limits^{\pi}_0\int\limits^{2\pi}_0\left(2-\frac{r}{a_0}\right)e^{\frac{-r}{a_0}-\frac{r}{2a_0}}r^2\sin^2\theta d_rd_{\theta}d_{\varphi}=0\)

\(\Leftrightarrow I=\int\limits^{\infty}_0\left(2-\frac{r}{a_0}\right).r^2.e^{\frac{-3r}{2a_0}}d_r\int\limits^{\pi}_0\sin^2\theta d_{\theta}\int\limits^{2\pi}_0d_{\varphi}=0\)

Xét tính phân 

\(J=\int\limits^{\infty}_0\left(2-\frac{r}{a_0}\right)r^2e^{\frac{-3r}{2a_0}}d_r\)=\(2\int\limits^{\infty}_0r^2e^{\frac{-3r}{2a_0}}d_r-\frac{1}{a_0}\int\limits^{\infty}_0r^3e^{\frac{-3r}{2a_0}}d_r\)

áp dụng tích phân \(\int\limits^{\infty}_0x^ne^{-ax}d_x=\frac{n!}{a^{n+1}}\)

\(\Rightarrow J=2.\frac{2!}{\left(\frac{3}{2a_0}\right)^3}-\frac{1}{a_0}\frac{3!}{\left(\frac{3}{2a_0}\right)^4}\)\(=\frac{4.\frac{3}{2a_0}-\frac{6}{a_0}}{\left(\frac{3}{2a_0}\right)^4}=\frac{\frac{6}{a_0}-\frac{6}{a_0}}{\left(\frac{3}{2a_0}\right)^4}=0\)

\(\Rightarrow I=J.\int\limits^{\pi}_0\sin^2\theta d_{\theta}\int\limits^{2\pi}_0d_{\varphi}=0.\int\limits^{\pi}_0\sin^2\theta d_{\theta}\int\limits^{2\pi}_0d_{\varphi}=0\)

Vậy hai hàm sóng trên trực giao 

Answer 6

câu a> 2 hàm sóng \(\Psi_{1s}=\frac{1}{\Pi}.a^{\frac{3}{2}}_0e^{\frac{-r}{a_0}}\)   và    \(\Psi_{2s}=\frac{1}{4\sqrt{2}}.\left(2-\frac{r}{a_0}\right).a^{\frac{3}{2}}_0e^{\frac{-r}{2a_0}}\)

Đk 2 hàm trực giao là : I = \(\int\)\(\Psi^*\Psi d\tau\) = 0

I = \(\int_x\int_y\int_z\)\(\Psi^*\Psi dxdydz\)    =   \(\int_r\int_{\theta}\int_{\varphi}\Psi_{1s}\Psi_{2s}r^2\sin\theta drd\theta d\varphi\)      =        C\(\int\limits^{\infty}_0\)\(\int\limits^{\pi}_0\)\(\int\limits^{2\pi}_0\)\(\left(2-\frac{r}{a_0}\right)e^{-\frac{r}{a_0}-\frac{r}{2a_0}}r^2sin\theta drd\theta d\varphi\)

= \(\int\limits^{\infty}_0\)\(\left(2-\frac{r}{a_0}\right)e^{-\frac{r}{a_0}-\frac{r}{2a_0}}r^2dr\)\(\int\limits^{\pi}_0\)\(sin\theta d\theta\)\(\int\limits^{2\pi}_0\)\(d\varphi\)    =    \(\int\limits^{\infty}_0\)\(\left(2r^2-\frac{r^3}{a_0}\right)e^{-\frac{3r}{2a_0}}dr\)\(\int\limits^{\pi}_0\)\(sin\theta d\theta\)\(\int\limits^{2\pi}_0\)\(d\varphi\)

Với C = \(\frac{1}{4\sqrt{2\Pi}}a^3_0\)  ;ta thấy \(\int\limits^{\pi}_0\)\(sin\theta d\theta\)\(\int\limits^{2\pi}_0\)\(d\varphi\)   \(\ne0\)     nên ta chỉ xét A= \(\int\limits^{\infty}_0\)\(\left(2r^2-\frac{r^3}{a_0}\right)e^{-\frac{3r}{2a_0}}dr\)

nếu A = 0 thì 2 hàm sóng trên trực gIao

Đặt k = -\(\frac{3}{2}\)

A = \(\frac{1}{k}\)\(\int\limits^{\infty}_0\)\(\left(2r^2-\frac{r^3}{a_0}\right)de^{kr}\)     =     \(\frac{1}{k}\)\(\left\{\left(2r^2-\frac{r^3}{a_0}\right)e^{kr}-\frac{1}{k}\left[\left(4r-\frac{3r^2}{a_0}\right)e^{kr}-\int\limits^{\infty}_0e^{kr}\left(4-\frac{6r}{a_0}\right)dr\right]\right\}\)  \(|_0^\infty\)

=    \(\frac{2}{3}r^3e^{-\frac{3r}{2a_0}}\)

thay cận vào  A ta được     A = 0 

suy ra I = 0 vậy 2 hàm đã cho ở trên trực giao với nhau (dpcm)