Question

Consider the electron wave function

$\psi \left(X\right)=\left\{\begin{array}{l}c\sin \left(\frac{2\mathrm{\pi x}}{L}\right)0\le x\le L\\ 0x<0orx>L\end{array}\right.$

a. Determine the normalization constant c. Your answer will be in terms of L.

b. Draw a graph of $\psi \left(x\right)$over the interval -L$\le$x $\le$2L.

c. Draw a graph of ${\left|\psi \left(x\right)\right|}^2$over the interval -L $\le$x $\le$2L. d. What is the probability that an electron is in the interval 0 $\le$x $\le$L/3?

Short answer

The Value of Probability is$40.2\%$.

Step-by-step solution

Use the equations of normalization to determine the value of constant and the formula of probability to determine its value

Sub part (a) step1:

For the probability interpretation of $\psi \left(x\right)$to make sense the wave function must satisfy the following equation

${\int }_{-\infty }^{+\infty }{\left|\psi \left(x\right)\right|}^{2}dx=1$

The above integrals is expanded as follows${\int }_{-\infty }^0{\left|\psi \left(x\right)\right|}^{2}dx+{\int }_0^1{\left|\psi \left(x\right)\right|}^{2}dx+{\int }_1^{+\infty }{\left|\psi \left(x\right)\right|}^{2}dx=1$

The wave function is defined only in the region $0\le x\le L$therefore substitute csin$\left(\frac{2\mathrm{\pi r}}{L}\right)$for thye second integral and zero for the rest of regions

$$\begin{gathered} 0+{\int }_0^L{\left[c\sin \left(\frac{2\mathrm{\pi x}}{L}\right)\right]}^{2}dx+0=1 \\ {c}^2{\int }_0^L{\left[c\sin \left(\frac{2\mathrm{\pi x}}{L}\right)\right]}^{2}dx=1 \end{gathered}$$

consider the following trigonometric relation

${\sin }^2\theta =\frac{1-\cos \theta }{2}$

Hence substitute $\frac{1-\cos \left[2\left({\displaystyle \frac{2\mathrm{\pi x}}{L}}\right)\right]}{2}$ for sin²$\left(\frac{2\mathrm{\pi x}}{L}\right)$AND SOLVE FOR C

$$\begin{gathered} C^2{\int }_0^L\frac{1-\cos \left[2\left({\displaystyle \frac{2\mathrm{\pi x}}{L}}\right)\right]}{2}dx=1 \\ {C}^2{\int }_0^{L}1-\cos \left(\frac{4\mathrm{\pi x}}{L}\right)dx=1 \end{gathered}$$

Thus the step was

$$\begin{gathered} {\left[x\right]}_0^L-{\left[\frac{\sin \left({\displaystyle \frac{4\mathrm{\pi x}}{L}}\right)}{{\displaystyle \frac{4\mathrm{\pi }}{L}}}\right]}_0^L=\frac{2}{C^2} \\ L-\frac{L}{4\mathrm{\pi }}\left[\sin \left(\frac{4\mathrm{\pi x}}{L}\right)-\sin 0\right]=\frac{2}{C^2} \\ L-0=\frac{2}{C^2} \\ C=\sqrt{\frac{2}{L}} \end{gathered}$$

Sub part (b) step2:The graph of ψx over the interval-L≤X≤2L is represented as follows

Subpart (c) step 3:The graph of ψx2 over the interval -L≤x≤2L is represented as follows

$$\begin{gathered} P=\int {\left|\psi \left(x\right)\right|}^{2}dx \\ P={\int }_0^{\frac{L}{3}}{\left[\sqrt{\frac{2}{L}\sin \left(\frac{2\mathrm{\pi x}}{l}\right)}\right]}^{2}dx \\ =\frac{1}{L}\left[\frac{L}{3}-\frac{L}{4\mathrm{\pi }}\left(-0.866\right)\right] \\ =0.402 \\ =40.2\% \end{gathered}$$

THEREFORE, THE VALUE OF PROBABILITY IS 40.2%