91影视

The state of electron in (a) is $\left(n_x,n_y,n_z\right)=\left(2,1,1\right)$

The state of electron in (b) is $\left(n_x,n_y,n_z\right)=\left(1,2,1\right)$

The state of electron in (c) is $\left(n_x,n_y,n_z\right)=\left(1,1,2\right)$

The normalization constant is the constant which is multiplied to non-negative area to become unity.

The probability of finding an electron in the region is given by:

$p\left(n_x,n_y,n_z\right)=\underset{x=0,y=L/3,z=0}{\overset{x=L,y=2L/3,z=L}{鈭�}}{\left[{\left(\frac{2}{L}\right)}^{3/2}\sin \left(\frac{n_x蟺x}{L}\right)\sin \left(\frac{n_y蟺y}{L}\right)\sin \left(\frac{n_z蟺z}{L}\right)\right]}^{2}dxdydz......\left(1\right)$

Substitute all the values in the above equation (1).

$$\begin{gathered} p\left(2,1,1\right)=\underset{x=0,y=L/3,z=0}{\overset{x=L,y=2L/3,z=L}{鈭�}}{\left[{\left(\frac{2}{L}\right)}^{3/2}\sin \left(\frac{\left(2\right)蟺x}{L}\right)\sin \left(\frac{\left(1\right)蟺y}{L}\right)\sin \left(\frac{\left(1\right)蟺z}{L}\right)\right]}^{2}dxdydz \\ p\left(2,1,1\right)={\left(\frac{2}{L}\right)}^3\left[\underset{x=0}{\overset{x=L}{鈭�}}{\sin }^2\left(\frac{2蟺x}{L}\right)dx\right]\left[\underset{y=L/3}{\overset{y=2L/3}{鈭�}}{\sin }^2\left(\frac{蟺y}{L}\right)dy\right]\left[\underset{z=0}{\overset{z=L}{鈭�}}{\sin }^2\left(\frac{蟺z}{L}\right)dz\right] \\ p\left(2,1,1\right)={\left(\frac{2}{L}\right)}^3{\left(\frac{L}{2}\right)}^2\left[\left(\frac{\left(2L/3\right)-\left(L/3\right)}{2}\right)-\frac{L}{4蟺}\left[\sin \left(\frac{2蟺\left(2L/3\right)}{L}\right)-\sin \left(\frac{2蟺\left(L/3\right)}{L}\right)\right]\right] \\ p\left(2,1,1\right)=\frac{2}{L}\left(\frac{L}{6}+\frac{L}{4蟺}\sqrt{3}\right) \end{gathered}$$

$$\begin{gathered} p\left(2,1,1\right)=\left(\frac{1}{3}-\frac{1}{4蟺}\sqrt{3}\right) \\ p\left(2,1,1\right)=0.196 \end{gathered}$$

Therefore, the probability of finding trapped electron for states (1,2,1) is 0.196.