91Ó°ÊÓ

An electron in a finite potential well is one in which the potential energy of an electron outside the well is not infinitely great but has a finite positive value called well depth.

Schrodinger wave equation for the region x>L is given by the relation as below.

$\frac{d^2\mathrm{ψ}}{d{x}^2}+\frac{8{\mathrm{π}}^2\mathrm{m}}{h^2}\left(E-v_0\right)\mathrm{ψ}=0$ ….. (1)

Let,$\mathrm{ψ}={\mathrm{De}}^{2kx}$

Then,

$\frac{d\mathrm{ψ}}{dx}=2kD{e}^{2kx}$

role="math" localid="1661771248398" $$\begin{gathered} \frac{d^2\mathrm{ψ}}{d{x}^2}=4k^{2}D{e}^{2kx} \\ 4k^2\mathrm{ψ}+\frac{8{\mathrm{π}}^2\mathrm{m}}{h}\left(E-V_0\right)\mathrm{ψ}=0 \end{gathered}$$

If $K=\frac{\mathrm{Ï€}}{h}\sqrt{2m\left(V_0-E\right)}$, then the above equation will be is satisfied.

Hence, $\mathrm{ψ}\left(x\right)=D{e}^{2kx}$is the solution of the equation (1).

The proposed function satisfied Schrodinger’s equation if

$K=\frac{\mathrm{Ï€}}{h}\sqrt{2m\left(V_0-E\right)}$

As in the region of x>L, the value of $V_0>E$.

Hence, it is real.

But if is position then the function is unrealistic. This is because if k is positive, then $x→∞$and $\mathrm{ψ}→∞$.

So, for large values of x , $\mathrm{ψ}$becomes very large.

Hence, the probability density becomes greater than unity. This is impossible.