Negative Potential Step - What happens when it isn't sharp (there is a width $a$)? I'm wondering what happens when you take a normal negative potential step, but then give it a width $a$ instead of a straight drop in potential. As the width, $a$, gets bigger, what would happen to incident electrons? I know if $a=0$, then a small percentage of incident electrons would be reflected back at the boundary. As $a$ gets bigger would this happen at an exponentially increasing rate?

 A: Given you are looking at the case where most of the incoming beam
is transmitted, the kinetic energy must be positive on both sides of
the well, so I won't worry about which is the incoming direction of
the particle.
Your problem has an analytic result in terms of Airy functions.
However, in general, the reflections go to zero when the width gets
large for well behaved smooth potentials.
The step barrier is obviously a mathematical approximation of the
true potential. Looking at the Schrödinger equation, the approximation
for using a step instead of the physical potential
requires that the de Broglie wavelength of the particle be much
larger than the width of the step. You can see this in the usual
step result where the reflection probability doesn't depend on $\hbar$.
If you take a smooth potential (rather than the discontinuous slope
potential you have), the reflections go to zero exponentially in
$a$ over the de Broglie wavelength. This is also the limit where
the semiclassical WKB solution can be applied. Obviously the classical
limit has no reflection.
I wrote a quick Maple script to plot your case. I think it is correct.
with(ScientificConstants);
eV := evalf(Constant(e)):
hbar := evalf(Constant(hbar)):
m := evalf(Constant(m[e])):
E := 4.0*eV:
V0 := 3.0*eV:
alpha := (hbar^2*a/(2*m*V0))^(1/3):
k := sqrt(2*m*E/hbar^2):
kp := sqrt(2*m*(E-V0)/hbar^2):
psi1 := exp(I*kp*x)+r*exp(-I*kp*x);
psi2 := C*AiryAi((a-x)/alpha-k^2*alpha^2)+D*AiryBi((a-x)/alpha-k^2*alpha^2);
psi3 := t*exp(I*k*x);
res1 := (-hbar^2/(2*m)*diff(psi2,x$2)+V0*(1-x/a)*psi2)/psi2;
eq1 := subs(x=0,psi1-psi2);
eq2 := subs(x=a,psi2-psi3);
eq3 := subs(x=0,diff(psi1-psi2,x));
eq4 := subs(x=a,diff(psi2-psi3,x));
sol := solve([eq1,eq2,eq3,eq4],{r,C,D,t}):
s1 := subs(sol,r);
R := abs(subs(sol,r))^2;
evalf(subs(a=1.0e-15,R)); #check it's 1/9
plotsetup(postscript,plotoutput=`rampscat.eps`);
plot(R,a=1.0e-15..5.0e-9);

It produces the plot of the reflection probability versus $a$ in meters

