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

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))))

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)

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 .

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 

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)