FIGURE P39.31 shows the wave function of a particle confined

between x = 0 nm and x = 1.0 nm. The wave function is zero

outside this region.

a. Determine the value of the constant c, as defined in the figure.

b. Draw a graph of the probability density$P\left(x\right)={\left|\psi \left(x\right)\right|}^2$

c. Draw a dot picture showing where the first 40 or 50 particles

might be found.

d. Calculate the probability of finding the particle in the interval

$0nm\le x\le 0.25nm$.

Any wave function $\psi \left(x\right)$should satisfy the equation${\int }_{-\infty }^{\infty }{\left|\psi \left(x\right)\right|}^2=1...\left(1\right)$

This equation states that the total area under the probability density curve must be 1.

Now, according to the question, wave function of a particle is confined between x = 0 nm and x = 1.0 nm

Consider the interval, $0\le x\le 0.75nm$, the wave function can be written as,

$\psi \left(x\right)=\frac{4}{3}cx$, $x$is in $nm.$

In the interval of $0.75nm\le x\le 1nm$, he wave function can be written as, $\psi \left(x\right)=4c\left(1nm-x\right)$

Therefore, from equation $\left(1\right)$, we can write,role="math" localid="1650896503981" $$\begin{gathered} {\int }_0^{0.75}{\left(\frac{4}{3}cx\right)}^{2}dx+{\int }_{0.75}^1{\left[4c\left(1-x\right)\right]}^{2}dx=1 \\ {\int }_0^{0.75}{\left(\frac{4}{3}cx\right)}^{2}dx+{\int }_{0.75}^{1}16c^2\left[\left(1-2x+x^2\right)\right]dx=1 \\ \frac{16}{9}{c}^2{\left[\frac{x^3}{3}\right]}_0^{0.75}+16c^2{\left[x-\frac{2x^2}{2}+\frac{x^3}{3}\right]}_{0.75}^1=1 \\ \frac{6.75}{27}{c}^2+\left[\frac{16c^2}{3}-5.25c^2\right]=1 \\ 0.25c^2+0.0833c^2=1 \\ 0.333c^2=1 \\ {c}^2=\frac{1}{0.333} \\ c=\sqrt{3}n{m}^{-\frac{1}{2}} \end{gathered}$$

Therefore, the constant$c$has a value of$\sqrt{3}n{m}^{-\frac{1}{2}}$

Putting the value of $c$, we can write the probability density of the given wave function as,

localid="1650899720458" $\psi \left(x\right)=\frac{4}{\sqrt{3}}x...\left(2\right)$for localid="1650899725784" $0\le x\le 0.75nm$and,

localid="1650899733838" $\psi \left(x\right)=4\sqrt{3}\left(1nm-x\right)...\left(3\right)$for localid="1650899738891" $0.75nm\le x\le 1nm$

Now, equation 2 and 3 gives,

localid="1650898727768" ${\left|\psi \left(x\right)\right|}^2=\frac{16}{3}{x}^2$and localid="1650899746880" ${{\left|\psi \left(x\right)\right|}^2=48\left(1-x\right)}^2$, where localid="1650899754028" ${\left|\psi \left(x\right)\right|}^2$has a unit of localid="1650899762684" $n{m}^{-1}.$

Therefore, the graph of the probability density should be as follows:

The particle is most likely to be found at the points where ${\left|\psi \left(x\right)\right|}^2$is a maximum. We can draw the dot picture according to the graph of probability density as shown earlier:

The probability of the particle in the interval of $0\le \le x\le 0.25nm$, can be determined by taking the equation ${\left|\psi \left(x\right)\right|}^2=\frac{16}{3}{x}^2$as,

$$\begin{gathered} P\left(0\le x\le 0.25\right)={\int }_0^{0.25}{\left|\psi \left(x\right)\right|}^{2}dx \\ P\left(0\le x\le 0.25\right)={\int }_0^{0.25}\left(\frac{16}{3}{x}^2\right)dx \\ P\left(0\le x\le 0.25\right)=\frac{16}{3}{\left[\frac{x^3}{3}\right]}_0^{0.25} \\ P\left(0\le x\le 0.25\right)=\frac{1}{36} \\ P\left(0\le x\le 0.25\right)=0.028 \end{gathered}$$