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 $x=0.999\mathrm{cm}?$

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

The area under the given graph is

Area$=2\times \frac{1}{2}\times 1\times 1=1$

which means that the wave function is indeed normalized.

.b) Every point on the graph of the wave function is the square root of its corresponding point on the curve of the probability density $\left|\psi \right(x){|}^2$. So, our plan is to choose a few points on the curve of $\left|\psi \right(x){|}^2$and try to use these points to predict the function $\left|\psi \right(x){|}^2$. After knowing $\left|\psi \right(x){|}^2,\psi (x)$can be found by taking the square root of $\left|\psi \right(x){|}^2$in each region, then we can simply draw the result.

By looking at the table above, we can deduce that $\left|\psi \right(x){|}^2=x$in the interval (0 to 1 cm) , so $\psi \left(x\right)=\pm \sqrt{x}$. Similarly, $\left|\psi \right(x){|}^2=-x$in the interval ( -1 to 0), so $\psi \left(x\right)=\pm \sqrt{-x}$.

Notice that the values of $\psi \left(x\right)$can be positive or negative and all of these different values give a right wave function. Now, we are going to choose the wave function in the region x>0 to be $\psi \left(x\right)=\sqrt{x}$, and in the region x<0 to be $\psi \left(x\right)=-\sqrt{-x}$, and of course you can choose another form if you wish. Hence, we get the graph shown in the image below.

At x = 0, we can see from FIGURE P39.32 that $\left|\psi \right(x){|}^2=0$at x = 0. Hence

Prob ( in 0.0010 at 0 ) = 0 data-custom-editor="chemistry" $\times$0.0010 cm = 0

From FIGURE P39.32, $\left|\psi \right(x){|}^2=0.5{\mathrm{cm}}^{-1}$at $x=0.5\mathrm{cm}$. Hence

{Prob ( in 0.0010 at 0.5 cm ) = (0.5$c{m}^{-1}$)$\times$0.0010 cm = 0.0005

From FIGURE P39.32,$\left|\psi \right(x){|}^2=0.999{\mathrm{cm}}^{-1}$at $x=0.999\mathrm{cm}$

Prob ( in 0.0010 at 0.999 cm ) = (0.999$c{m}^{-1}$) 0.0010 cm = 0.00999