91Ó°ÊÓ

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 $\mathrm{δ}x$at $x)=P(x)\mathrm{δ}x$

Here, $P\left(x\right)$ is the probability density and $\mathrm{δ}x$is the small width.

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

$\begin{array}{l}\mathrm{δ}x=0.010\text{mm}\\ =0.010\text{mm}\left(\frac{1\text{m}}{1×{10}^3\text{mm}}\right)\\ =1×{10}^{−5}\text{m}\end{array}$

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

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

Substitute$0.50×{10}^3{\text{m}}^{−1}$for$P\left(x\right),1×{10}^{−5}\text{m}$for$\mathrm{δ}x$in the equation (1) to solve for

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

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

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 $5.0×{10}^{−3}$.

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.

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

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

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

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 $2.5×{10}^{−3}$.

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$P\left(x\right),1×{10}^{−5}\text{m}$for$\mathrm{δ}x$in the equation (1) to solve for

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

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

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 .

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.

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

Substitute $0.25×{10}^3{\text{m}}^{−1}$for $P\left(x\right),1×{10}^{−5}\text{m}$for $\mathrm{δ}x$in the equation (1) to solve for

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

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

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 $2.5×{10}^{−3}$.