Question

Question: An electron is trapped in a cubic 3D well. In the states $\left(n_x,n_y,n_z\right)$= (a) (2,1,1) (b) (1,2,1)(c) (1,1,2), what is the probability of finding the electron in the region $\left(\mathbf{0}⩽\mathbf{x}⩽\mathbf{L},\mathbf{L}\mathbf{/}\mathbf{3}⩽\mathbf{y}⩽\mathbf{2}\mathbf{L}\mathbf{/}\mathbf{3},\mathbf{0}⩽\mathbf{z}⩽\mathbf{L}\right)$. Discus any difference in these results.

Short answer

Answer

(a) The probability of finding trapped electron for states (2,1,1) is 0.609.

(b) The probability of finding trapped electron for states (1,2,1) is 0.196.

(c) The probability of finding trapped electron for states (1,1,2) is 0.609.

Step-by-step solution

Identification of given data

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.

Step 2(a): Determination of probability of finding electron for (2,1,1) 

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

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

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

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

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

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