(a) Show that for the region x>L in the finite potential well of Fig. 39-7, $\mathbf{\psi }\left(x\right)\mathbf{=}{\mathbf{De}}^{\mathbf{2}\mathbf{kx}}$is a solution of Schrödinger’s equation in its one-dimensional form, where D is a constant and k is positive. (b) On what basis do we find this mathematically acceptable solution to be physically unacceptable?

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.

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

Let,$\psi ={\mathrm{De}}^{2kx}$

Then,

$\frac{d\psi }{dx}=2kD{e}^{2kx}$

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

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

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

The proposed function satisfied Schrodinger’s equation if

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

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 $x\to \infty$and $\mathrm{\psi }\to \infty$.

So, for large values of x , $\psi$becomes very large.

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