FIGURE P39.32 shows $\left|\psi \right(x){|}^2$for the electrons in an experiment.

a. Is the electron wave function normalized? Explain.

b. Draw a graph of $\psi \left(x\right)$over this same interval. Provide a numerical scale on both axes. (There may be more than one acceptable answer.)

c. What is the probability that an electron will be detected in a $0.0010-\mathrm{cm}$-wide region at $x=0.00\mathrm{cm}$? At $x=0.50\mathrm{cm}$? At $x=0.999cm$?

d. If ${10}^4$electrons are detected, how many are expected to land in the interval $-0.30\mathrm{cm}\le x\le 0.30\mathrm{cm}$?

Formula Used:

One possible graph can be made by plotting

$$\begin{gathered} \psi \left(x\right)=\sqrt{-x} \\ \psi \left(x\right)=\sqrt{x}\text{for}-1<x<1 \end{gathered}$$

From the graph:

$$\begin{gathered} \left|\psi \right(x){|}^2=-x\text{for}-1<x<0 \\ |\psi \left(x\right){|}^2=x\text{for}0<x<1 \end{gathered}$$

We need to show that ${\int }_{-\infty }^{*}\psi \left(x\right){\psi }^{*}\left(x\right)dx=1$

$$\begin{gathered} {\int }_{-1}^1\psi \left(x\right){\psi }^{*}\left(x\right)dx \\ {\int }_{-1}^0-xdx+{\int }_0^{1}xdx \\ =-{\left.\frac{x^2}{2}\right|}_{-1}^0+{\left.\frac{x^2}{2}\right|}_0^1=0+\frac{(-1{)}^2}{2}+\frac{(1{)}^2}{2} \\ =\frac{1}{2}+\frac{1}{2}=1 \end{gathered}$$

Conclusion: Therefore, the wavefunction is normalized.

Formula Used:

$$\begin{gathered} \left|\psi \right(x){|}^2=-x\text{for}-1<x<0 \\ |\psi \left(x\right){|}^2=x\text{for}0<x<1 \end{gathered}$$

One possible graph can be made by plotting

$$\begin{gathered} \psi \left(x\right)=\sqrt{-x} \\ \psi \left(x\right)=\sqrt{x}\text{for}-1<x<1 \end{gathered}$$

One possible graph can be made by plotting

localid="1650893249997" $$\begin{gathered} \psi \left(x\right)=\sqrt{-x} \\ \psi \left(x\right)=\sqrt{x}\text{for}-1<x<1 \end{gathered}$$

Conclusion: A graph for has been made.

Formula Used:

The probability function is given by:

$P\left(x\right)=\left|\psi \right(x){|}^2\delta x$

The probability that an electron is detected in a $0.0010\mathrm{cm}$wide region i.e. $\delta x=0.0010\mathrm{cm}$

For $x=0.00\mathrm{cm}$

$$\begin{gathered} P\left(x\right)=\left|\psi \right(0.00){|}^{2}0.0010 \\ =0.00 \end{gathered}$$

For $x=0.5\mathrm{cm}$

$$\begin{gathered} P\left(x\right)=\left|\psi \right(0.50){|}^{2}0.0010 \\ =0.0005 \\ P(x)=|\psi (0.999){|}^{2}0.0010 \\ =0.000999\approx 0.0010 \end{gathered}$$

Conclusion: The probability that an electron is detected in a $0.0010\mathrm{cm}$wide region at $x=0.00\mathrm{cm}$and $x=$$0.50\mathrm{cm}$and $x=0.999\mathrm{cm}$is $0.00,0.0005$and $0.00010$respectively.

Total number of electrons$={10}^4$

Formula Used:

The probability function is given by:

$P\left(x\right)=\left|\psi \right(x){|}^2$

Calculation:

Probability to find an electron in the interval $-0.30\mathrm{cm}<x<0.30\mathrm{cm}$is:

$$\begin{gathered} {\int }_{-0.3}^0-xdx+{\int }_0^{0.3}xdx \\ ={\int }_0^{-0.3}xdx+{\int }_0^{0.3}xdx={\left.\frac{x^2}{2}\right|}_0^{-0.3}+{\left.\frac{x^2}{2}\right|}_0^{0.3} \\ =\frac{(-0.3{)}^2}{2}+\frac{(0.3{)}^2}{2}=0.09 \end{gathered}$$

Number of electrons that would land in the interval $-0.30\mathrm{cm}<x<0.30\mathrm{cm}$are:

role="math" localid="1650894290307" $$\begin{gathered} N=N_{o}P\left(x\right) \\ N={10}^4\times 0.09=900 \end{gathered}$$electrons

Conclusion:900 electrons are expected to land in the interval $-0.30\mathrm{cm}<x<0.30\mathrm{cm}$.