91影视

The given potential is the hybrid of the finite and infinite square wells is:

$V\left(x\right)=\{\begin{array}{l}鈭�,(x<0)\\ \frac{-32h^2}{m{a}^2},(0鈮�x鈮�a)\\ 0,(x>a)\end{array}$

In the second region,

$$\begin{gathered} -\frac{h^2}{2m}蠄"-\frac{32h^2}{m{a}^2}蠄=E蠄 \\ 蠄"=-l^2蠄 \end{gathered}$$

Where,

$$\begin{gathered} l=\sqrt{\frac{2m}{h^2}}\left(E+\frac{32h^2}{m{a}^2}\right) \\ inthethirdregion, \\ -\frac{h^2}{2m}蠄"=E蠄 \\ 蠄"=k蠄 \\ Where, \\ k=\sqrt{\frac{-2mE}{h^2}} \\ Hence,thesolutionsare, \\ 蠄\left(x\right)=\left\{\begin{array}{l}0,x<0\\ A\cos \left(lx\right)+B\sin \left(lx\right),0鈮�x鈮�a\\ C{e}^{-kx},x>a\end{array}\right. \end{gathered}$$

From the continuity of wave functions

$$\begin{gathered} Atx=0 \\ 蠄\left(0\right)=0 \\ A=0 \\ Atx=a \\ B\sin \left(la\right)=C{e}^{-ka} \end{gathered}$$

Dividing both the equations,

$$\begin{gathered} \tan \left(la\right)=-\frac{l}{k} \\ Letz=la,then: \\ {z}^0=\frac{a}{\sout{\sout{h}}}\sqrt{2m\frac{32\sout{h^2}}{m{a}^2}} \\ \left(l^2+k^2\right){\sout{h}}^2=2m\frac{32\sout{h^2}}{m{a}^2} \\ \left(l^2+k^2\right){\sout{h}}^2=\left(\frac{\sout{\sout{h}}}{a}{z}_0\right) \\ {z_0}^2=\left(l^2+k^2\right)a^2 \\ {z_0}^2=z^2+k^2{a}^2 \\ ka=\sqrt{{z_0}^2-z^2} \\ ka=\sqrt{64-z^2} \\ \sin ce, \\ \tan \left(la\right)=-\frac{l}{k} \\ \tan \left(z\right)=-\frac{la}{ka} \\ \tan \left(z\right)=-\frac{-1}{\sqrt{64/z^2-1}} \\ Thisisatranscendentalequation.Itcanbesolvedgraphicallyornumerically. \\ Hence,wecanconcludethattherearethreeboundstates. \\ {z}_1=2.98165165 \\ {z}_2=5.53899816 \\ {z}_3=7.95732149 \end{gathered}$$

Since, the energy is given by

$蠄\left(x\right)=\left\{\begin{array}{l}0,x<0\\ B\sin \left(lx\right),0鈮�x鈮�a\\ \frac{\sin \left(la\right)}{e^{-ka}}B{e}^{-kx},x>a\end{array}\right.$

Hence, the probability over its entire range from to for the second line in this equation, and from $x=a$to $x=鈭�$for the third line in this equation is:

localid="1658226702587" $\left[{鈭�}_0^a{\sin }^2\left(lx\right)dx+\frac{{\sin }^2\left(la\right)}{e^{-2ka}}{鈭�}_a^{鈭�}{e}^{-2kx}dx\right]B^2$

Therefore, the probability of the particle being outside the box,

$p_3=\frac{B^2{\displaystyle \frac{{\sin }^2\left(la\right)}{e^{-2ka}}}{鈭�}_a^{鈭�}{e}^{-2kx}dx}{\left[{鈭�}_0^a{\sin }^2\left(lx\right)dx+\frac{{\sin }^2\left(la\right)}{e^{-2ka}}{鈭�}_a^{鈭�}{e}^{-2kx}dx\right]B^2}$

Calculating the above integral, we get,

$$\begin{gathered} {鈭�}_a^{鈭�}{e}^{-2kx}dx=\frac{e^{-2ak}}{2k} \\ {鈭�}_0^a{\sin }^2\left(lx\right)dx=-\frac{\sin \left(2al\right)-2al}{4l} \end{gathered}$$

Hence, substituting these values in the probability equation:

$$\begin{gathered} p_3=\frac{{\displaystyle \frac{{\sin }^2\left(la\right)}{2k}}}{-{\displaystyle \frac{\sin \left(2al\right)-2al}{4l}}+{\displaystyle \frac{{\sin }^2\left(la\right)}{2k}}} \\ {p}_3=\frac{{\displaystyle \frac{{\sin }^2\left(z\right)}{2\sqrt{64-z^2}}}}{-{\displaystyle \frac{\sin \left(2z\right)-2z}{4z}}+{\displaystyle \frac{{\sin }^2\left(z\right)}{2\sqrt{64-z^2}}}} \\ Substitude{z}_3=7.95732149,andweget, \\ {p}_3=0.54204 \end{gathered}$$

Therefore, the probability is $0.54204$the particle being outside the box,