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The wave function$\mathit{蠄}\mathbf{\left[}\mathit{x}\mathbf{\right]}$and its derivative $\mathit{蠄}\mathbf{\left[}\mathit{x}\mathbf{\right]}\mathbf{/}\mathit{d}\mathit{x}$must be continuous everywhere except where the potential-energy U(x) function has an infinite discontinuity. Wave functions are normalized so that the total probability of finding the particle somewhere is unity.

The normalization condition is:

${鈭�}_{{}^{-鈭�}}^{{}^{鈭�}}\left|蠄\right(x){|}^{2}dx=1$

The equation of wavefunction is:

${蠄}_n\left(x\right)=\sqrt{\frac{2}{L}}sin\left(\frac{n蟺x}{L}\right)$

${蠄}_1\left(x\right)=\sqrt{\frac{2}{L}}sin\left(\frac{蟺x}{L}\right).....\left(1\right)$

When the wave function vanishes the probability of finding the particle at any point, inside the box is zero.

So, P (x) = 0 when ${蠄}_1\left(x\right)=0$

Now, from equation

$$\begin{gathered} {蠄}_1=0 \\ \sqrt{\frac{2}{L}}sin\left(\frac{蟺x}{L}\right)=0 \\ sin\left(\frac{蟺x}{L}\right)=0.....\left(2\right) \end{gathered}$$

Since the sinusoidal function vanishes at$胃=m蟺$where

Therefore,$\frac{蟺x}{L}=m蟺$

x = mL

Hence, for the points from 0 to L the values of x at m = 0,1 is:

P(x) = 0 at x = 0 & L

$胃=\frac{m蟺}{2}$ where m = 1,3,5, . ..

So, $$\begin{gathered} \frac{蟺x}{L}=\frac{m蟺}{2} \\ x=\frac{mL}{2} \end{gathered}$$

For the value of x to at m = 1 only,

So, the P(x) has a highest value at $x=\frac{L}{2}$.