91Ó°ÊÓ

(a) The wave function is given in the question which is

$\mathrm{Ψ}\left(x\right)=\frac{\sqrt{2/\mathrm{Π}}}{x^2−x+1.25}$

Here we can see that the probability to find the particle somewhere is 1.

So, we can write it as:

$\begin{array}{l}\underset{−\mathrm{∞}}{\overset{\mathrm{∞}}{∫}}\mathrm{Ψ}\left(x\right){\mathrm{Ψ}}^0\left(x\right)dx=\underset{−\mathrm{∞}}{\overset{\mathrm{∞}}{∫}}\frac{\sqrt{2/\mathrm{Π}}}{x^2−x+1.25}\frac{\sqrt{2/\mathrm{Π}}}{x^2−x+1.25}dx\\ =\frac{2}{\mathrm{Π}}\underset{−\mathrm{∞}}{\overset{\mathrm{∞}}{∫}}\frac{1}{{(x^2−x+1.25)}^2}dx\\ =\frac{2}{\mathrm{Π}}\underset{−\mathrm{∞}}{\overset{\mathrm{∞}}{∫}}\frac{1}{{({(x−0.5)}^2+1)}^2}dx\end{array}$

Now,

By using trigonometry transformation,$x=0.5+\tan \mathrm{θ}$

The equation can be written as:

$\underset{−\mathrm{∞}}{\overset{\mathrm{∞}}{∫}}\frac{1}{{({(x−0.5)}^2+1)}^2}dx=\underset{−\mathrm{Π}/2}{\overset{\mathrm{Π}/2}{∫}}\frac{1}{{({\tan }^2\mathrm{θ}+1)}^2}{\sec }^2\mathrm{θ}d\mathrm{θ}$

$\begin{array}{c}=\underset{−\mathrm{Π}/2}{\overset{\mathrm{Π}/2}{∫}}\frac{1}{{\sec }^2\mathrm{θ}}d\mathrm{θ}\\ =\underset{−\mathrm{Π}/2}{\overset{\mathrm{Π}/2}{∫}}{\cos }^2\mathrm{θ}d\mathrm{θ}\\ =\frac{\mathrm{Π}}{2}\end{array}$

Here,

$\begin{array}{c}\underset{−\mathrm{Π}/2}{\overset{\mathrm{Π}/2}{∫}}{\cos }^2\mathrm{θ}d\mathrm{θ}=\underset{−\mathrm{Π}/2}{\overset{\mathrm{Π}/2}{∫}}\frac{1+\cos 2\mathrm{θ}}{2}d\mathrm{θ}\\ =\frac{1}{2}\mathrm{Π}+\underset{−\mathrm{Π}/2}{\overset{\mathrm{Π}/2}{∫}}\frac{\cos 2\mathrm{θ}}{2}d\mathrm{θ}\\ =\frac{\mathrm{Π}}{2}+0\\ =\frac{\mathrm{Π}}{2}\end{array}$

The probability is.$\frac{2}{\mathrm{Î }}\frac{\mathrm{Î }}{2}=1$ The normalization is right,

Therefore the normalization constant will be $\sqrt{\frac{2}{\mathrm{Î }}}$.

Now by finding the maximum for this we will use this equation,

$\mathrm{ÒÏ}=\frac{2}{\mathrm{Ï€}}\frac{1}{{(x^2−x+1.25)}^2}$

Now by Differentiating above equation with respect to $x$,

$\begin{array}{c}\frac{d\mathrm{ÒÏ}}{dx}=\frac{2}{\mathrm{Ï€}}\frac{−2(x^2−x+1.25)(2x−1)}{{(x^2−x+1.25)}^4}\\ =\frac{−4(2x−1)}{\mathrm{Ï€}{(x^2−x+1.25)}^3}\end{array}$

The derivative will be$0$when it’s maximum, so

$\begin{array}{c}\frac{−4(2x−1)}{\mathrm{π}{(x^2−x+1.25)}^3}=0\\ ∴2x−1=0\\ 2x=1\\ x=\frac{1}{2}\end{array}$

Therefore, the most probable location of particle is .$x=\frac{1}{2}$

Now for the probability per unit length:

$\begin{array}{l}\mathrm{ÒÏ}\left(0.5\right)=\frac{2}{\mathrm{Ï€}}\frac{1}{{({0.5}^2−0.5+1.25)}^2}\\ =\frac{2}{\mathrm{Ï€}}\end{array}$

Therefore, the probability per unit length for the particle at the location is$=\frac{2}{\mathrm{Ï€}}$.$=\frac{2}{\mathrm{Ï€}}$