91Ó°ÊÓ

Here,

${\left|\mathrm{ψ}\left(x\right)\right|}^2$is the probability density

According to the problem, the electrons are uniformly distributed over an interval of $0cm≤x≤2cm$. so the probability ${\left|\mathrm{ψ}\left(x\right)\right|}^2$is constant over the interval $0cm≤x≤2cm$and thus ${\left|\mathrm{ψ}\left(x\right)\right|}^2$is a square wave function

The condition for normalization is

${\left|\mathrm{ψ}\left(x\right)\right|}^2=1$

Determine the probability of an electron to be found over the interval $0cm≤x≤2cm$

${\left|\mathrm{ψ}\left(x\right)\right|}^2\mathrm{δ}x=1$

Substitute 2cm for $\mathrm{δ}x$

${\left|\mathrm{ψ}\left(x\right)\right|}^2\mathrm{δ}x=1$

${\left|\mathrm{ψ}\left(x\right)\right|}^2=0.5c{m}^{-1}$

Thus the probability of an electron to be found over the interval 0.79 cm to 0.81cm is 1%

$N\left(in\mathrm{δ}xatx\right)=Nxprob\left(in\mathrm{δ}xatx\right)$

Calculate the number of electrons detected in the interval 0.79cm to 0.81 cm

$N\left(in\mathrm{δ}xatx=0.80cm\right)=NxProb\left(in\mathrm{δ}xatx=0.80cm\right)$

Thus the number of electrons detected in the interval 0.79cm to 0.81cm is 10⁴

prob$$\begin{gathered} \left(x=0.80cm\right)={\left|\mathrm{ψ}=0.80cm\right|}^2 \\ Prob\left(x=0.80cm\right)=0.5c{m}^{-1} \end{gathered}$$

Thus the probability density at x-0.80 cm is 0.5 cm^-1