Question

The wave function of a particle is

$\psi \left(x\right)=\sqrt{\frac{b}{\mathrm{\pi }\left({\mathrm{x}}^2+{\mathrm{b}}^2\right)}}$

where b is a positive constant. Find the probability that the particle is located in the interval -b$\le$x $\le$b

Short answer

${\int }_{-\infty }^{+\infty }\left|\psi {\left(x\right)}^2\right|dx=1$

the probability of finding the particle within the given interval$-b\le x\le b$ is 50%

Step-by-step solution

First check the normalization condition for the given wave function of the given particle within the limits -∞ to +∞

$$\begin{gathered} {\int }_{-\infty }^{+\infty }\left|\psi {\left(x\right)}^2\right|dx={\int }_{-\infty }^{+\infty }{\left(\sqrt{\frac{b}{\mathrm{\pi }\left({\mathrm{x}}^2+{\mathrm{b}}^2\right)}}\right)}^{2}dx \\ ={\int }_{-\infty }^{+\infty }\frac{b}{\mathrm{\pi }\left({\mathrm{x}}^2+{\mathrm{b}}^2\right)} \\ =\frac{b}{\mathrm{\pi }}{\left(\frac{1}{b}{\tan }^{-1}\left(\frac{x}{b}\right)\right)}_{-\infty }^{\infty } \\ =\frac{1}{\mathrm{\pi }}\left(\left(\frac{\mathrm{\pi }}{2}\right)-\left(\frac{-\mathrm{\pi }}{2}\right)\right) \\ =1 \end{gathered}$$

The probability of finding the particle within the given interval -b≤x≤b is,

$$\begin{gathered} {\int }_{-\infty }^{+\infty }\left|\psi {\left(x\right)}^2\right|dx={\int }_{-b}^{+b}{\left(\sqrt{\frac{b}{\mathrm{\pi }\left({\mathrm{x}}^2+{\mathrm{b}}^2\right)}}\right)}^{2}dx \\ ={\int }_{-b}^{+b}\frac{b}{\mathrm{\pi }\left({\mathrm{x}}^2+{\mathrm{b}}^2\right)} \\ =\frac{b}{\mathrm{\pi }}{\left(\frac{1}{b}{\tan }^{-1}\left(\frac{x}{b}\right)\right)}_{-b}^b \\ =\frac{1}{\mathrm{\pi }}\left(\left(\frac{\mathrm{\pi }}{4}\right)-\left(\frac{-\mathrm{\pi }}{4}\right)\right) \\ =\frac{1}{2} \\ =\frac{1}{2}\left(100\%\right) \\ =50\% \end{gathered}$$

Therefore, the probability of finding the particle within the given intervalrole="math" localid="1649761340195" $-b\le x\le b$is 50%