Question

Consider the electron wave function

$\psi \left(X\right)=\left\{\begin{array}{l}0x<0nm\\ \left(1.414n{m}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)e^{-\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$\left(1.0nm\right)$}\right.}x\ge 0nm\end{array}\right.$

where x is in cm.

a. Determine the normalization constant c.

b. Draw a graph of c1x2 over the interval -2 cm $\le$x $\le$2 cm. Provide numerical scales on both axes.

c. Draw a graph of 0 c1x2 0 2 over the interval -2 cm $\le$x $\le$2 cm. Provide numerical scales.

d. If 104 electrons are detected, how many will be in the interval 0.00 cm $\le$x $\le$0.50 cm?

Short answer

(a) The value of normalization constant c is 0.866 cm^-1/2

(d) prob $\left(in0cm\le x\le 0.50cm\right)$=0.344

(e) The number of electrons detected in the interval $3440$

Step-by-step solution

Subpart (a) part 1:

$$\begin{gathered} {\int }_{-1cm}^{+1cm}{\left|\psi \left(x\right)\right|}^{2}dx=1 \\ {\int }_{-1cm}^{+1cm}{\left(c\sqrt{1-x^2}\right)}^{2}dx=1 \\ {\int }_{-1cm}^{+1cm}{c}^2\left(1-x^2\right)dx=1 \\ {c}^2{\left[x-\frac{x^3}{3}\right]}_{-1cm}^{+1cm}=1 \end{gathered}$$

Simplify it further and solve for c

$$\begin{gathered} c^2\left[\left(1cm-\left(-1cm\right)-\left(\frac{1}{3}cm-\left(\frac{-1}{3}cm\right)\right)\right)\right]=1 \\ {c}^{2\left(2cm-\frac{2}{2}cm\right)=1} \\ {c}^2\left(\frac{4}{3}cm\right)=1 \\ c=\sqrt{\frac{3}{4}}c{m}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ =0.866c{m}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \end{gathered}$$

Thus, the value of normalization constant c is 0.866 cm^-1/2

Subpart (b) part 1:

The following is the graph of $\psi \left(x\right)$over the interval $-2cm\le x\le 2cm$

From the above graph, it is clear that the values of $\psi \left(x\right)$decrease from position x=0cm to x=1cm so, the graph appears to be bowed upward in the intervalrole="math" localid="1649743294095" $0\le xx\le 1cm$and as per the condition, the value of$\psi \left(x\right)$is zero at$x\ge 1cm$

Subpart (c) part 1:

The following is the graph of $\psi \left(x\right)$over the interval $-2cm\le x\le 2cm$

From the above graph, it is clear that the values of $\psi \left(x\right)$decrease from position x=0cm to x=1cm so, the graph appears to be bowed upward in the interval$0\le xx\le 1cm$ and as per the condition, the value of$0\le x\le 1cm$

Subpart (d) step 4:

Calculate the probability of electrons in the interval $0cm\le x\le 0.50cm$
problocalid="1649757459778" role="math" $$\begin{gathered} \left(in0cm\le x\le 0.50cm\right)={\int }_{0cm}^{0.50cm}{\left|\psi \left(x\right)\right|}^{2}dx \\ ={\int }_{0cm}^{0.50cm}{c}^2{\left(1-x\right)}^{2}dx \\ ={\left(0.866c{m}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}^2\left[0.50cm-\left(\frac{{\left(0.50cm\right)}^3}{3}\right)\right] \end{gathered}$$

=0.344

Given Expression:

Substitute 10⁴ for N and 0.344 for $prob\left(in0cm\le x\le 0.50cm\right)$

$$\begin{gathered} N\left(in0cm\le x\le 0.50cm\right)=Nxprob\left(in0cm\le x\le 0.50cm\right) \\ =\left({10}^4\right)(0.344) \\ =3440 \end{gathered}$$

Thus the number of electrons detected in the interval 3440