Question

FIGURE EX39.13 shows the probability density for an electron that has passed through an experimental apparatus. What is the probability that the electron will land in a $0.010-\text{mm}$-wide strip at (a) $x=0.000\text{mm}$, (b)$x=0.500\text{mm}$, (c) $x=1.000\text{mm}$, and (d) $x=2.000\text{mm}$?

Short answer

a. The probability of electron will land in a $0.010\text{mm}$wide string at a position of $x=0.000\text{mm}$is $5.0\times {10}^{-3}$.

b. The probability of electron will land in a $0.010\text{mm}$wide string at a position of $x=0.500\text{mm}$is $2.5\times {10}^{-3}$.

c. The probability of electron will land in a $0.010\text{mm}$wide string at a position of $x=1.000\text{mm}$is 0.

d. The probability of electron will land in a $0.010\text{mm}$wide string at a position of $x=2.000\text{mm}$is $2.5\times {10}^{-3}$.

Step-by-step solution

Part.a.

Determine the probability of electron will land in a $0.010\text{mm}$wide string at a position of $x=0.000\text{mm}$using the formula.

The probability of any electron ends up in the strip at position is,

$\mathrm{Prob}($in $\delta x$at $x)=P(x)\delta x$

Here, $P\left(x\right)$ is the probability density and $\delta x$is the small width.

Step.2

Convert the units for the small width of the strip from $\text{mm}$to $\text{m}$.

$\begin{array}{l}\delta x=0.010\text{mm}\\ =0.010\text{mm}\left(\frac{1\text{m}}{1\times {10}^3\text{mm}}\right)\\ =1\times {10}^{-5}\text{m}\end{array}$

From the figure EX39.13, the probability density of the electron at the position $x=0.000\text{mm}$is as follows.

$\begin{array}{l}P\left(x\right)=0.50{\text{mm}}^{-1}\\ =0.50{\text{mm}}^{-1}\left(\frac{1{\text{m}}^{-1}}{{10}^{-3}{\text{mm}}^{-1}}\right)\\ =0.50\times {10}^3{\text{m}}^{-1}\end{array}$

Substitute$0.50\times {10}^3{\text{m}}^{-1}$for$P\left(x\right),1\times {10}^{-5}\text{m}$for$\delta x$in the equation (1) to solve for

Prob (in data-custom-editor="chemistry" $0.010\text{mm}$at $0.000\text{mm}$).

$\begin{array}{l}\mathrm{Prob}(\text{in}0.010\text{mmat}0.000\text{mm})=\left(0.50\times {10}^3{\text{m}}^{-1}\right)\left(1\times {10}^{-5}\text{m}\right)\\ =5.0\times {10}^{-3}\end{array}$

Therefore, the probability of electron will land in a $0.010\text{mm}$wide string at a position of $x=0.000\text{mm}$ is $5.0\times {10}^{-3}$.

Part.b.

Determine the probability of electron will land in a $0.010\text{mm}$wide string at a position of $x=0.500\text{mm}$using the equation (1) .

From the figure EX39.13, the probability density of the electron at the position $x=0.500\text{mm}$is as follows.

$\begin{array}{l}P\left(x\right)=0.25{\text{mm}}^{-1}\\ =0.25{\text{mm}}^{-1}\left(\frac{1{\text{m}}^{-1}}{{10}^{-3}{\text{mm}}^{-1}}\right)\\ =0.25\times {10}^3{\text{m}}^{-1}\end{array}$

Substitute $0.25\times {10}^3{\text{m}}^{-1}$for $P\left(x\right),1\times {10}^{-5}\text{m}$for in the equation (1) to solve for Prob $($ in $0.010\text{mm}$at $0.000\text{mm}$) .

$\begin{array}{l}\mathrm{Prob}(\text{in}0.010\text{mmat}0.500\text{mm})=\left(0.25\times {10}^3{\text{m}}^{-1}\right)\left(1\times {10}^{-5}\text{m}\right)\\ =2.5\times {10}^{-3}\end{array}$

Therefore, the probability of electron will land in a $0.010\text{mm}$wide string at a position of $x=0.500\text{mm}$is $2.5\times {10}^{-3}$.

Part.c.

Determine the probability of electron will land in a$0.010\text{mm}$wide string at a position of$x=1.000\text{mm}$using the equation (1) .

From the figure EX $39.13$, the probability density of the electron at the position $x=1.000\text{mm}$is as follows.

$\begin{array}{l}P\left(x\right)=0.0{\text{mm}}^{-1}\\ =0.0{\text{mm}}^{-1}\left(\frac{1{\text{m}}^{-1}}{{10}^{-3}{\text{mm}}^{-1}}\right)\\ =0.0{\text{m}}^{-1}\end{array}$

Substitute$0.0{\text{m}}^{-1}$for$P\left(x\right),1\times {10}^{-5}\text{m}$for$\delta x$in the equation (1) to solve for

Prob (in $0.010\text{mm}$at $1.000\text{mm}$).

$\begin{array}{l}\mathrm{Prob}(\text{in}0.010\text{mmat}1.000\text{mm})=\left(0.0{\text{m}}^{-1}\right)\left(1\times {10}^{-5}\text{m}\right)\\ =0\end{array}$

Therefore, the probability of electron will land in a $0.010\text{mm}$wide string at a position of $x=1.000\text{mm}$is 0 .

Part.d.

Determine the probability of electron will land in a $0.010\text{mm}$wide string at a position of $x=2.000\text{mm}$using the equation (1) .

From the figure EX $39.13$, the probability density of the electron at the position$x=2.000\text{mm}$is as follows.

$\begin{array}{l}P\left(x\right)=0.25{\text{mm}}^{-1}\\ =0.25{\text{mm}}^{-1}\left(\frac{1{\text{m}}^{-1}}{{10}^{-3}{\text{mm}}^{-1}}\right)\\ =0.25\times {10}^3{\text{m}}^{-1}\end{array}$

Substitute $0.25\times {10}^3{\text{m}}^{-1}$for $P\left(x\right),1\times {10}^{-5}\text{m}$for $\delta x$in the equation (1) to solve for

Prob (in $0.010\text{mm}$at $2.000\text{mm}$).

$\begin{array}{l}\mathrm{Prob}(\text{in}0.010\text{mmat}0.500\text{mm})\\ =\left(0.25\times {10}^3{\text{m}}^{-1}\right)\left(1\times {10}^{-5}\text{m}\right)\\ =2.5\times {10}^{-3}\end{array}$

Therefore, the probability of electron will land in a $0.010\text{mm}$wide string at a position of $x=2.000\text{mm}$is $2.5\times {10}^{-3}$.