91Ó°ÊÓ

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

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

The following is the graph of $\mathrm{ψ}\left(x\right)$over the interval $-2cm≤x≤2cm$

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

The following is the graph of $\mathrm{ψ}\left(x\right)$over the interval $-2cm≤x≤2cm$

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

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

=0.344

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

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

Thus the number of electrons detected in the interval 3440