Consider the electron wave function

$\psi \left(x\right)=\left\{\begin{array}{l}cx\left|x\right|\le 1nm\\ \frac{c}{x}\left|x\right|\ge 1nm\end{array}\right.$

where x is in nm.

a. Determine the normalization constant c.

b. Draw a graph of $\psi \left(x\right)$over the interval role="math" localid="1650907186096" $-5nm\le x\le 5nm$.

Provide numerical scales on both axes.

c. Draw a graph of ${\left|\psi \left(x\right)\right|}^2$over the interval role="math" localid="1650907657944" $-5nm\le x\le 5nm$.

Provide numerical scales.

d. If ${10}^6$ electrons are detected, how many will be in the interval

role="math" localid="1650908765290" $-1.0nm\le x\le 1.0nm$?

An electron wave function is given for the interval of $\left|x\right|\le 1nm$,

$\psi \left(x\right)=cx$and for the interval of $\left|x\right|\ge 1nm$is, $\psi \left(x\right)=\frac{c}{x}$.

Any wave function should satisfy the equation,known as the normalization condition which is the probability of finding a particle at position $x$.

Substituting the values of $\psi \left(x\right)$in the given interval, we get,

$$\begin{gathered} {\int }_{-\infty }^{-1}\frac{c^2}{x^2}dx+{\int }_{-1}^0{c}^2{x}^{2}dx+{\int }_0^1{c}^2{x}^{2}dx+{\int }_1^{\infty }\frac{c^2}{x^2}dx=1 \\ 2{\int }_1^{\infty }\frac{c^2}{x^2}dx+2{\int }_0^1{c}^2{x}^{2}dx=1 \\ 2c^2{\left[-\frac{1}{x}\right]}_1^{\infty }+2c^2{\left[\frac{x^3}{3}\right]}_0^1=1 \\ \frac{2}{3}{c}^2+2c^2=1 \\ \frac{8}{3}{c}^2=1 \\ c=\sqrt{\frac{3}{8}} \end{gathered}$$

Therefore, the value of normalization constant is$c=\sqrt{\frac{3}{8}}$

Plotting the values of wave function $\psi \left(x\right)$along y-axis and the position $x$along x-axis we get the graph as:

Within the given limit, we can write the probability density, substituting the value of normalization constant as,

${\left|\psi \left(x\right)\right|}^2=\frac{3}{8}{x}^2$when $\left|x\right|\le 1nm$and

${\left|\psi \left(x\right)\right|}^2=\frac{3}{8x^2}$when $\left|x\right|\ge 1$

Therefore, the ${\left|\psi \left(x\right)\right|}^2$versus $x$graph should be as following:

The probability of electron density within the given limit can be written as,

$$\begin{gathered} P\left(-1nm\le x\le 1nm\right)={\int }_{-1}^1{\left|\psi \left(x\right)\right|}^{2}dx \\ P\left(-1nm\le x\le 1nm\right)={\int }_{-1}^1{c}^2{x}^{2}dx \\ P\left(-1nm\le x\le 1nm\right)=c^2{\left[\frac{x^3}{3}\right]}_{-1}^1 \\ P\left(-1nm\le x\le 1nm\right)=\frac{2}{3}{c}^2 \\ P\left(-1nm\le x\le 1nm\right)=\frac{2}{3}\times \frac{3}{8} \\ P\left(-1nm\le x\le 1nm\right)=\frac{1}{4} \end{gathered}$$

The number of electrons detected is$\left({10}^6\times \frac{1}{4}\right)=2.5\times {10}^5.$