91Ó°ÊÓ

The given data is listed below as-

The probability density for the region is, 0 < x < L

Thesinusoidal functionis given by-

$\mathbf{ψ}\left(x\right)\mathbf{=}\mathbf{B}\mathbf{}{\mathbf{sin}}^{\mathbf{2}\mathbf{}}\mathbf{kx}$

Here, B is the constant.

The region is 0 < x <L .

The Schrodinger’s equation for this region is $\frac{d^2\mathrm{ψ}}{d{x}^2}+\frac{8{\mathrm{π}}^2\mathrm{m}}{h^2}E\mathrm{ψ}=0$

Here, E > 0 .

Now, if ${\mathrm{ψ}}^2\left(x\right)=B{\sin }^{2}kx$

Then, $\mathrm{ψ}\left(x\right)=B\text{'}\sin kx$

Here, B’ is another constant.

Satisfying $B{\text{'}}^2=B$

Therefore,

$$\begin{gathered} \frac{d^2\mathrm{ψ}}{d{x}^2}=-k^{2}B\text{'}\sin kx \\ =-k^2\mathrm{ψ}\left(x\right) \end{gathered}$$

And, $\frac{d^2\mathrm{ψ}}{d{x}^2}+\frac{8{\mathrm{π}}^2\mathrm{m}}{h^2}E\mathrm{ψ}=-k^2\mathrm{ψ}\left(x\right)+\frac{8{\mathrm{π}}^2\mathrm{m}}{h^2}E\mathrm{ψ}$

The above equation tends to zero provided that $k^2=\frac{8{\mathrm{Ï€}}^2\mathrm{mE}}{h^2}$

Thus, the proposed function satisfies the Schrodinger’s equation as the right side of the above equation is positive and so k is real.

The value of k can be positive or negative.

$k^2=\frac{8{\mathrm{Ï€}}^2\mathrm{mE}}{h^2}$

Solving the above will give final value of k.

$k=Â\pm \frac{2\mathrm{Ï€}}{h}\sqrt{2mE}$

Thus, value of $k=Â\pm \frac{2\mathrm{Ï€}}{h}\sqrt{2mE}$.