Answer
See in "step by step solution"
Work Step by Step
For the region $x > L$, Schrödinger’s equation is given by
$\frac{d^2\psi}{dx^2}+\frac{8\pi^2m}{h^2}(E-U_0)\psi=0$
If $\psi= De^{2kx}$,
then
$\frac{d^2\psi}{dx^2}+\frac{8\pi^2m}{h^2}(E-U_0)\psi\\
=D4k^2e^{2kx}+\frac{8\pi^2m}{h^2}(E-U_0)De^{2kx}\\
=De^{2kx}[4k^2+\frac{8\pi^2m}{h^2}(E-U_0)$]
The above equation will be zero, if
$4k^2+\frac{8\pi^2m}{h^2}(E-U_0)=0$
or, $k=\frac{\pi}{h}\sqrt{2m(U_0-E)}$
Therefore, $\psi= De^{2kx}$, is a solution of Schrödinger’s equation in its one dimensional form, where D is a constant and k is positive provided $k=\frac{\pi}{h}\sqrt{2m(U_0-E)}$.
