An electron that is confined to x Ú 0 nm has the normalized wave function c1x2 = b 0 x 6 0 nm 11.414 nm-1/2 2e-x/11.0 nm2 x Ú 0 nm where x is in nm. a. What is the probability of finding the electron in a 0.010-nmwide region at x = 1.0 nm? b. What is the probability of finding the electron in the interval 0.50 nm … x … 1.50 nm?

The wave $\psi \left(x\right)$is the probability of finding particle at a given point x the article probability of finding the particle in a region by the product of the wave function with is complex

The expression of the probability of the normalized wave function is

$p={\int }_{x_1}^{x_2}\psi *\left(x\right)\psi \left(x\right)$

$x_1=x-\frac{∆x}{2}$

and

$x_1=x+\frac{∆x}{2}$

substitute 0.01nm for $∆x$and 1nm, for X in the above expression to find the limits

$$\begin{gathered} x_1=\left(1nm\right)-\frac{0.01}{2}nm \\ =0.995nm \end{gathered}$$

and,

$$\begin{gathered} x_2=\left(1nm\right)+\frac{0.01}{2}nm \\ =1.005nm \end{gathered}$$

The expression of the probability of a normalized wavefunction is given as follows

role="math" localid="1649741312917" $P={\int }_{x_1}^{x_2}\psi *\left(x\right)\psi \left(x\right)dx$

Substitute ${\sqrt{1.414nme}}^{\frac{-x}{1.0nm}}$for $\psi \left(x\right)$and $\psi \left(x\right),1.005nm$ for x₁ and 0.995 nm for x₂ in the above expression

$$\begin{gathered} P={\int }_{0.995nm}^{1.005nm}\left({\left(1.414n{m}^{\frac{1}{2}}\right)}^{\frac{-x}{e^{1.0nm}}}\right)\left({\left(1.414n{m}^{\frac{1}{2}}\right)}^{\frac{-x}{e^{1.0nm}}}\right)dx \\ =\left(2.00n{m}^{-1}\right){\int }_{0.995nm}^{1.005nm}{e}^{\frac{2x}{1.0nm}}dx \\ =\left(2.00n{m}^{-1}\right)\frac{e^{{\displaystyle \frac{-2x}{1.0nm}}}}{\left({\displaystyle \frac{-2}{1.0nm}}\right)} \\ =-e^{\frac{-2x}{1.0nm}} \end{gathered}$$

Substitute the limits and find the value of probability

$$\begin{gathered} P=-\left(e^{\frac{2\left(1.005nm\right)}{1.0nm}}-e^{\frac{2\left(0.995nm\right)}{1.0nm}}\right) \\ =-\left(0.1339886-\left(0.1366954\right)\right) \\ =0.002706 \end{gathered}$$

In percentage:

%P=Px100%

substitute 0.002706 in P in the above expression

%P=(0.002706)x100%

=0.27%

Thus the probability of finding a particle in the region 1nm is 0.27%

$P={\int }_{x_1}^{x_2}\psi *\left(x\right)\psi \left(x\right)dx$

Substitute ${\sqrt{1.414nme}}^{\frac{-x}{1.0nm}}$for $\psi \left(x\right)$and $\psi \left(x\right),1.005nm$ for x₁ and 0.995 nm for x₂ in the above expression

$$\begin{gathered} P={\int }_{0.5nm}^{1.5nm}\left({\left(1.414n{m}^{\frac{1}{2}}\right)}^{\frac{-x}{e^{1.0nm}}}\right)\left({\left(1.414n{m}^{\frac{1}{2}}\right)}^{\frac{-x}{e^{1.0nm}}}\right)dx \\ =\left(2.00n{m}^{-1}\right){\int }_{0.995nm}^{1.005nm}{e}^{\frac{2x}{1.0nm}}dx \\ =\left(2.00n{m}^{-1}\right)\frac{e^{{\displaystyle \frac{-2x}{1.0nm}}}}{\left({\displaystyle \frac{-2}{1.0nm}}\right)} \\ =0.25-e^{\frac{-2x}{1.0nm}} \end{gathered}$$

substitute the limits and find the value of probability

$$\begin{gathered} P=-\left(e^{\frac{2\left(1.5nm\right)}{1.0nm}}-e^{\frac{2\left(0.5nm\right)}{1.0nm}}\right) \\ =-\left(0.049787-\left(0.367879\right)\right) \\ =0.318 \end{gathered}$$

In percentage:

%P=Px100%

substitute 0.002706 in P in the above expression

%P=(0.318)x100%

=32%

Thus the probability of finding a particle in the region 1nm is 32%